Haryana Board Class 10 Biology Solutions For Chapter 1 Life Processes

Chapter 1 Life Processes Multiple Choice Questions And Answers

Question 1. Which statement can be concluded from the image?

HBSE Solutions Class 10 Biology Chapter 1 Process Of Photosynthesis In Plants

  1. Plants absorb CO2 from air and H2O from the soil as raw materials and convert them into glucose
  2. Plants absorb CO2 from the soil and H2O from air as raw materials and convert them into glucose
  3. plants absorb O2 from air and glucose from the soil as raw materials and convert them into light energy
  4. Plants absorb O2 from air and minerals from the soil as raw materials and convert them into heat energy.

Answer: 1. plants absorb CO2 from air and H2O from the soil as raw materials and convert them into glucose

Question 2. A student sets up an experiment to study the importance of nutrition in plants. The student takes 2 pots, pot 1 and pot 2 each with the same healthy plant. Both the pots were placed in the garden and watered properly. Pot1 was kept as such, while pot 2 was kept in an air tight glass box with caustic soda. Caustic soda absorbs carbon dioxide present in the surrounding. After 2 days, the student observes that the plant kept in the garden is healthy while the plant placed in container shed leaves and droops. What is the likely reason for this observation?

  1. Lack of nutrients in the soil
  2. Absence of oxygen for survival
  3. Inability to perform photosynthesis
  4. Absorption of light by caustic soda restricting growth

Answer: 3. Inability to perform photosynthesis

Question 3. Which of the equation show correct conversion of CO2 and H2O into carbohydrates in plants?

HBSE Solutions Class 10 Biology Chapter 1 CO₂ And H₂O Into Carbohydrates In Plants

Answer: 3. Carbohydrates

Question 4. A student sets up an experiment to study the photosynthesis in plants. The student destarched a potted plant by keeping it in a dark room for 3 days. Half of the portion of destarched leaf was placed in a bottle containing caustic potash (absorbs CO2) as shown:

HBSE Solutions Class 10 Biology Chapter 1 Effect of CO₂ on Photosynthesis

The student then places the plant in light and tests the leaf after 5 hours for the presence of starch. The portions inside the bottle shows negative starch test by reflecting no change in colour when react with iodine, however, other upper portions of the leaf gave positive starch test showing blue-black colour with iodine. What can be evaluated from this experiment?

  1. Carbon dioxide is directly linked with the colour of leaf
  2. Carbon dioxide is necessary for preparing carbohydrate
  3. Lack of carbon dioxide increases amount of starch in plant
  4. Lack of carbon dioxide slows the process of photosynthesis

Answer: 2. Carbon dioxide is necessary for preparing carbohydrate

Question 5. The image shows the bread moulds on a bread.

HBSE Solutions Class 10 Biology Chapter 1 Bread Moulds On A Bread

 

How these fungi obtain nutrition?

  1. By eating the bread on which it is growing
  2. By using nutrients from the bread to prepare their own food
  3. By breaking down the nutrients of bread and then absorbing them
  4. By allowing other organisms to grow on the bread and then consuming them

Answer: 3. By breaking down the nutrients of bread and then absorbing them

Question 6. The image shows how Amoeba obtains nutrition.

HBSE Solutions Class 10 Biology Chapter 1 Amoeba Obtains Nutrition

How this process is advantageous for Amoeba?

  1. Capturing of food takes less time
  2. Complex food can be digested easily
  3. More amount of food can be consumed
  4. Fast distribution of nutrition within the body

Answer: 4. Fast distribution of nutrition within the body

Question 7. The image shows the human digestive system.

HBSE Solutions Class 10 Biology Chapter 1 Human Digestive System

Digestion of food starts from which organ of the digestive system?

  1. Mouth due to the presence of saliva
  2. Oesophagus that moves the food in gut
  3. That releases juices for fat breakdown
  4. Which helps in mixing food with digestive juices

Answer: 1. Mouth due to the presence of saliva

Question 8. The image shows a cross section of small intestine.

HBSE Solutions Class 10 Biology Chapter 1 Cross Section Of Small Intestine

What will be the likely happen if the number of villi increases in the intestine?

  1. Increase in the absorption of food
  2. Fast elimination of waste from the body
  3. Increase in flow of blood in the small intestine
  4. Fast breakdown of larger food particles into smaller ones

Answer: 1. Increase in the absorption of food

Question 9. An incomplete equation for the digestion of starch using saliva is shown as:

Saliva + Starch (in test tube)→ What will be the likely outcome of this?

  1. Saliva will convert starch into complex fat molecules.
  2. Saliva will convert starch into complex sugar molecules.
  3. Saliva will breakdown starch into simple sugar molecules.
  4. Saliva will breakdown starch into simple protein molecules.

Answer: 3. Saliva will breakdown starch into simple sugar molecules.

Question 10. A student sets up an experiment to study the role of enzymes in digestion of food. In which test tube, the digestion of protein will occur?

HBSE Solutions Class 10 Biology Chapter 1 Effect Of Pepsin And HCL On Protein Digestion

  1. Test tube A as pepsin will breakdown into simple molecules.
  2. Test tube B as HCI will breakdown protein into simple molecules.
  3. Test tubes A as pepsin will breakdown protein into simple molecules.
  4. Test tube B as HCI will activate pepsin for breakdown of protein into simple molecules.

Answer: 2. Test tube B as HCI will breakdown protein into simple molecules.

Question 11. The image shows the flow diagram for the breakdown of glucose in yeast.

Breakdown Of Glucose In yeast

Under which condition these types of products are obtained?

  1. In the presence of oxygen
  2. In the absence of oxygen
  3. The presence of carbon dioxide
  4. In the absence of carbon dioxide

Answer: 2. In the absence of oxygen

Question 12. Which pathway will occur in the cell of an athlete who is performing 100m sprint?

Cell Of An Athlete

Answer: 3. Athlete

Question 13. A student setup an experiment to study the human respiratory system. In the experiment, the student places candle and a living cockroach in the flask A, while a candle and a dead cockroach in flask B. The burning of candle needs oxygen.

HBSE Solutions Class 10 Biology Chapter 1 Student Places Candle And A Living Cockroach In The Flask

After 10 minutes, the student observes that the candle in flask A extinguish faster while candle in flask B keeps burning for a longer time. What can be evaluated from this experiment?

  1. Candle produces high amount of carbon dioxide
  2. Living beings consumes oxygen during respiration
  3. Burning of candle decreases the life span of cockroach
  4. Water vapours produced by living beings prevents burning of candle

Answer: 2. Living beings consumes oxygen during respiration

Question 14. A student sets up an experiment to study human respiration using lime water, test tube and a straw. Lime water is colourless in the absence of CO2 and is milky in its presence. The student fills a freshly prepared limewater in a test tube and blows air through straw into the limewater. It was observed that the solution turns cloudy as shown.

HBSE Solutions Class 10 Biology Chapter 1 Human Respiration Using Lime Water, Test Tube And A Straw

What can be evaluated from this observation?

  1. Oxygen is exhaled during respiration
  2. Glucose is produced during respiration
  3. Dioxide is exhaled during respiration
  4. Water vapours are produced during respiration

Answer: 3. Dioxide is exhaled during respiration

Question 15. The image shows the transport of gases in body through heart and lungs.

HBSE Solutions Class 10 Biology Chapter 1 Transport Of Gases In Body Through Heart And Lungs

Which option correctly shows the transport of oxygen to the cell?

  1. Lungs →pulmonary vein →left atrium →left ventricle →aorta → body cells
  2. Lungs vein →right atrium →right ventricle → aorta body cells
  3. Lungs →pulmonary artery →left atrium → left ventricle→ venacava → body cells
  4. Lungs →pulmonary artery →right atrium → right ventricle→ venacava → body cells

Answer: 1. Lungs →pulmonary vein→left atrium →left ventricle →aorta → body cells

Question 16. The image shows oxygenated and de-oxygenated blood in the human heart.

HBSE Solutions Class 10 Biology Chapter 1 Oxygenated And De-Oxygenated Blood In The Human Heart

What is the direction of deoxygenated blood from right atrium of the heart?

  1. Towards the lungs
  2. Towards the lower body
  3. Towards the upper body
  4. Towards the left atrium of heart

Answer: 1. Towards the lungs

Question 17. The image shows the circulation of blood in fishes.

HBSE Solutions Class 10 Biology Chapter 1 Circulation Of Blood In Fishes

Which option correctly traces the pathway of blood flow in fish body?

  1. Gill capillaries→ oxygenated blood→heart → body cells→ deoxygenated blood-*gills
  2. Gill capillaries→oxygenated blood→ body cells→ deoxygenated blood→ heart→gills
  3. capillaries→ heart blood→ body cells→ deoxygenated blood→heart→gills
  4. Gill capillaries→ oxygenated blood→heart → body cells→ deoxygenated blood→heart→gills

Answer: 2. Gill capillaries→ oxygenated blood→ body cells→ deoxygenated blood→ heart→gills

Question 18. The image shows the circulation of blood in fishes and humans.

HBSE Solutions Class 10 Biology Chapter 1 Circulation Of Blood In Fishes And Humans

How is the circulations of blood in fish different from that in humans?

  1. The heart in fish is bigger in size.
  2. The flow of blood in fish is unidirectional.
  3. The blood goes through heart only once in fishes.
  4. The heart of fish has more chambers compared to that of a human.

Answer: 3. The blood goes through heart only once in fishes.

Question 19. The image shows the structure of an artery.

HBSE Solutions Class 10 Biology Chapter 1 Structure Of An Artery

Which statement supports the likely reason for thick walls in arteries?

  1. To Carry Large Amount Of Blood
  2. To Allow Easy Exchange Of Gases With Cells
  3. To Ensure Blood Flows In Only One Direction
  4. To Sustain The High-Pressure Blood From The Heart

Answer: 4. To Sustain The High-Pressure Blood From The Heart

Question 20. The image shows the healing of a wound.

HBSE Solutions Class 10 Biology Chapter 1 Healing Of A Wound

Based on the image, what explains the process?

  1. Platelets form clot by plugging the site of injury
  2. Platelets uses component of broken vessel to form clot
  3. Red blood cells divide and replace the broken vessel at the site of injury
  4. Red Blood Cells And Platelets Migrate To Site Of Injury And Secrete Substance That Forms New Vessel

Answer: 1. Platelets form clot by plugging the site of injury

Question 21. How water is taken up from soil to the xylem tissue of the plant roots?

  1. Xylem attracts water molecules
  2. Roots act as a suction pump for taking water
  3. Soil expels the water with pressure to the xylem
  4. Difference in the ion concentration creates a gradient for water movement

Answer: 4. Difference in the ion concentration creates a gradient for water movement

Question 22. A student performs an experiment using a balsam plant with intact stem, leaves, roots and flowers. The plant was kept in a test tube containing eosin solution (a pink colour dye). The test tube mouth was covered using cotton plug as shown

HBSE Solutions Class 10 Biology Chapter 1 Demonstration Of Water Transport In Plants Using Eosin Solution

The student kept the plant undisturbed in the lab. After 2-3 hours, a transverse section of stem was obtained using sharp scissors and studied under microscope. The studies reveal the presence of pink colour in the vessels of xylem. What does this observation explain?

  1. Eosin solution gets stored in the xylem
  2. Water moves through xylem in the plant
  3. Xylem reacts with eosin and gives colour
  4. Most portion of the plant stem is occupied by xylem

Answer: 2. Water moves through xylem in the plant

Question 23. The loss of water from the leaves of the plant is transpiration. How this process is advantageous for the plant?

  1. It helps in the downward movement of the water.
  2. It helps the plant to maintain temperature in hot sunny days.
  3. It acts as a driving force for distribution of food in plant’s body.
  4. Helps maintain a constant level of water in the soil around the plant.

Answer: 2. It helps the plant to maintain temperature in hot sunny days.

Question 24. A student setup an experiment using a well-watered plant. The plant’s roots and soil were covered with a rubber sheet. The plant was then kept in a glass bell jar and sealed with Vaseline at the bottom part to prevent the flow of air. The student keeps the apparatus in the light and observes water drops inside the jar after 2 hours as shown in the image.

HBSE Solutions Class 10 Biology Chapter 1 Demonstration Of Transpiration In Plants Using Bell Jar

What can be evaluated about transpiration from this experiment?

  1. Plant leaves give off water in form of vapours.
  2. Heat from the outside warms the jar which melts the vaseline into vapours.
  3. Plant absorbs water from environment thus extra water appears on the inside of jar.
  4. Covered roots and stem of the plant decreases the temperature of jar resulting in condensation of moisture into vapours.

Answer: 1. Plant leaves give off water in form of vapours.

Question 25. The image shows the transport of food material inside plant body with the help of phloem.

HBSE Solutions Class 10 Biology Chapter 1 Transport Of Food Material Inside Plant Body With The Help Of Phloem

How is food transported from phloem to the tissues according to plants need?

  1. Food is transported along with the water in plant’s body.
  2. Food is transported in only direction like water in the plant body through xylem.
  3. Food is transported from a region with low concentration to higher concentration.
  4. Food is transported from a region where it is produced to other parts of the plants.

Answer: 4. Food is transported from a region where it is produced to other parts of the plants.

Question 26. The image shows the movement of sucrose into phloem against the concentration gradient which also leads to the movement of water due to osmotic difference. This osmotic pressure allows movement of material in plant body.

HBSE Solutions Class 10 Biology Chapter 1 Movement Of Sucrose Into Phloem

How the movement of sucrose into phloem takes place initially?

  1. With the help of transpiration
  2. With the help of water gradient
  3. With the help of atp molecules
  4. With the help of adp molecules

Answer: 3. With the help of atp molecules

Question 27. The image shows the excretory system in humans.

HBSE Solutions Class 10 Biology Chapter 1 Excretory System in Humans

What is the importance of the labelled part in the excretory system?

  1. It produces urine.
  2. It filters waste from the blood.
  3. It stores the urine till urination.
  4. It carries urine from kidney to outside.

Answer: 3. It stores the urine till urination.

Question 28. The image shows the structure of a nephron.

HBSE Solutions Class 10 Biology Chapter 1 Structure Of A Nephron

Nephron is a unit of filtration in kidneys that filters waste material. It selectively reabsorbs or excretes water with the help of capillaries that surround it. What is the likely benefit of this?

  1. It makes the process of filtration at Bowman’s capsule easier.
  2. It helps keep the output of urine constant throughout the day.
  3. It helps to uptake and store excess amount of water in the body for later use.
  4. It maintains the concentration of urine based on the amount of water present in the body.

Answer: 4. It maintains the concentration of urine based on the amount of water present in the body.

Question 29. The image shows the process of photosynthesis in plants

 

HBSE Solutions Class 10 Biology Chapter 1 Photosynthesis

Based on the image, which component is excreted by plants during photosynthesis?

  1. Carbon dioxide
  2. Glucose
  3. Light energy
  4. Oxygen

Answer: 4. Oxygen

Question 30. A plant gets rid of excess water through transpiration. Which is a method used by plants to get rid of solid waste products?

  1. Shortening of stem
  2. Dropping down of fruits
  3. Shedding of yellow leaves
  4. Expansion of roots into the soil

Answer: 3. Shedding of yellow leaves

Question 31. The given graph indicates the effect of exercise intensity on carbohydrate consumption.

HBSE Solutions Class 10 Biology Chapter 1 The Effect Of Exercise Intensity On Carbohydrate Consumption.

At high intensity of exercise

  1. The anaerobic consumption of sugars increases
  2. The aerobic consumption of sugars increases
  3. The anaerobic consumption of sugars decreases
  4. No consumption of sugars takes place

Answer: 1. The anaerobic consumption of sugars increases

Chapter 1 Life Processes Very Short Questions And Answers

Question 1. In which of the following group/groups of animals, heart does not pump oxygenated blood to different parts of the body?

  1. Pisces
  2. Amphibians
  3. Amphibians and reptiles
  4. Pisces and amphibians

Answer: 1. Pisces

Question 2. What processes would you consider essential for maintaining life?

Answer: Nutrition, respiration, transportation and excretion.

Question 3. Which organ of male reproductive system acts as thermoregulator?

Answer: Scrotum acts as thermoregulator.

Question 4. Why is DNA copying an essential part of the process of reproduction?

Answer: DNA copying is essential for inheritance of features from parents to the next generation.

Question 5. What is dialysis?

Answer: The process by which blood is cleared of metabolic wastes in case of a kidney failure is called dialysis.

Question 6. Where do plants get each of the raw materials required for photosynthesis?

Answer: Plants get carbon dioxide and sunlight from atmosphere. It gets water and minerals from the soil.

Question 7. What is the mode of nutrition in fungi and plasmodium?

Answer: Modes of nutrition: Fungi- Saprophytic; Plasmodium- Parasitic

Question 8. Which of the following carries only deoxygenated blood?

  1. Pulmonary artery
  2. Pulmonary vein
  3. Capillary
  4. Aorta

Answer: 1. Pulmonary artery

Question 9. Which of the following makes a 3-chambered heart less efficient as compared to a 4- chambered heart?

Answer: In a 3-chambered heart, due to mixing of blood in a single ventricle, the parts of the body do not get blood saturated with oxygen. Therefore, a 3-chambered heart is less efficient as compared to a 4- chambered heart.

Question 10. What regulates the opening and closing of stomata?

Answer: The opening and closing of stomata is regulated by kidney-shaped cells called guard cells

Question 11. What will happen in the absence of anther?

Answer: If anther is absent, then pollen grain will not be produced.

Question 12. How do woody plants carry out gaseous exchange?

Answer: Woody plants carry out gaseous exchange through the lenticels.

Question 13. Name the excretory organ in Amoeba.

Answer: Cell membrane.

Question 14.Where does digestion of starch begin in human body?

Answer: The digestion of starch begins inside the mouth.

Question 15. Name the structure through which pollen tube enters the ovule.

Answer: The pollen tube enters the ovule through the stigma.

Question 16. During contraction of heart, what prevents the backflow of blood?

Answer: During contraction of heart, the valves in heart prevent the backflow of blood.

Question 17. Why is potassium hydroxide used in photosynthesis experiments?

  1. It reacts with carbon dioxide.
  2. It reacts with oxygen produced.
  3. It helps in destarching the leaf.
  4. It eliminates the interference of sunlight

Answer: 1. It reacts with carbon dioxide.

Chapter 1 Life Processes Short Questions And Answers

Question 1. Why do fish die when taken out of water?

Answer: Fish utilise the oxygen dissolved in water for their respiration. When fish are taken out of water, the supply of oxygen to the fish is cut as they cannot absorb and breathe using the oxygen present in the atmosphere. Hence, the fish die after some time.

Question 2. What is the role of HCI in our stomach?

Answer: HCI makes the medium inside the stomach acidic, which is necessary for the activation of enzyme pepsin. HCI also kills the bacteria, which may enter the stomach along with food.

Question 3. How is oxygen and carbon dioxide transported in human beings?

Answer: Oxygen and carbon dioxide are transported in human beings with the help of transportation system. Transport of oxygen: The air present in the alveolar sacs have high concentration of oxygen, while the blood capillaries surrounding the alveolar sacs are deficient in oxygen.

Oxygen diffuses from the alveoli to the blood capillaries where it combines with the haemoglobin to form oxyhaemoglobin. Then the blood reaches the tissues where oxyhaemoglobin breaks into haemoglobin and oxygen. This oxygen enters the cells.

Transport of carbon dioxide: The tissues have high concentration of carbon dioxide than the blood entering them. Therefore, carbon dioxide diffuses from the tissues into the blood. Then it reaches the lungs where it diffuses into the alveoli, and expelled out into the atmosphere through the respiratory tract.

Question 4. Differentiate between single and double circulation found in vertebrates.

Answer:

HBSE Solutions Class 10 Biology Chapter 1 Differentiate Between Single And Bouble Circulation

Question 5. How is food transported in plants?

Answer: The food is transported in plants both in upward and downward direction through phloem tissues, which consist of sieve tubes and companion cells.

Question 6. The diagram given below represents a system in the human body. Study the diagram and answer the following questions:

HBSE Solutions Class 10 Biology Chapter 1 Female Reproductive System – Labeled Diagram

  1. Label the parts marked 5 and 6.
  2. Name the two hormones secreted by 1.

Answer:

  1. Part 5: Oviduct; Part 6: Cervix
  2. Hormones secreted by 1 (ovary): Oestrogen and Progesterone

Question 7. What is the significance of the testes being located in the scrotal sac outside the abdomen?

Answer: The production and survival of sperm require a temperature which is lower than the normal body temperature. So, the testes are located in the scrotal sac which is outside the abdomen to maintain the temperature at 3*ÿ below the normal body temperature.

Question 8. Why does blood in the arteries flow with jerks and is under pressure?

Answer: Blood in the arteries moves because of the pressure of blood from the heart. Each time the heart pumps, it pushes the blood a little further. Due to this reason it flows with jerks.

Question 9. Do veins rely on heart for movement of blood? Justify your answer.

Answer: Veins do not rely on the heart to move blood. Veins have a system of valves to keep the blood from not moving backward. Movement of blood in veins occurs due to contraction of muscles.

Question 10. What is translocation? Why is it essential for plants?

Answer: Transportation of organic solutes in plants is called translocation. It occurs in the upward as well as downward directions and in the storage organs of roots, fruits, seeds and growing organs.Translocation is necessary because all the plant cells need food to carry out their vital functions.

Question 11. What are villi? Mention their functions.

Answer: Villi are tiny finger-like projections present in the walls of small intestine. The function of villi is to increase the surface area of intestines for efficient food absorption.

Question 12. What is saliva? State its role in the digestion of food.

Answer: Saliva is a watery fluid secreted by the salivary glands in the mouth. Functions of Saliva:

  1. Moistening food, and helping to create a food bolus, so it can be swallowed easily.
  2. Saliva contains the enzyme amylase that breaks down starch into simple sugars.

Question 13. How are the alveoli designed to maximise the exchange of gases?

Answer: Alveoli have thin walls and a close network of blood arteries to allow gas exchange between blood and air-filled alveoli. They have a balloon-like structure to maximize surface area in exchange for gas.

Question 14. The diagrams given below are cross-sections of blood vessels:

HBSE Solutions Class 10 Biology Chapter 1 Cross-Sections Of Blood Vessels

  1. Identify the blood vessels A, B and C.
  2. Mention one structural difference between A and B.

Answer:

  1. A- Artery, B- Vein, C- Blood capillary
  2. Structural difference between A (artery) and B (vein): In artery valves are not present, whereas in veins, valves are present.

Question 15. State the role of the placenta in the development of the embryo.

Answer: Placenta is a disc-like tissue which develops between the uterus wall and the embryo. Role Of Placenta: Exchange of water between the mother and the foetus; Exchange of nutrients; Exchange of respiratory gases; Removal of nitrogenous wastes from the foetus.; Antibodies cross the placenta and provide immunity to the baby.

Question 16. Explain the cause of cramps after excessive physical exercise.

Answer: During heavy exercise, the demand for energy is high but the supply of oxygen to produce energy is limited. Therefore, anaerobic respiration takes places in the muscles cells to fulfil the demand for energy. This anaerobic breakdown of glucose leads to the formation of lactic acid in muscles. The accumulation of lactic acid in muscles leads to muscle cramps.

Question 17. Pre-natal sex determination has been prohibited by law. State two reasons.

Answer:

  1. To prevent female foeticide in desire for a male child.
  2. Declining female-male sex ratio.

Question 18. Rahul complained of acidity on reaching home after a marriage. Explain the reason for acidity.

Answer: When we eat more food or spicy food, our digestive system has to work more by releasing more enzymes for digestion. The stomach releases more HCI to digest more food because of which a lot of acid is formed. This may cause acidity.

Question 19. Most of the CO2 produced in a tissue enters the red blood cells by diffusion. What happens to this CO2?

Answer: The pigment haemoglobin present in RBCs binds with CO2 and gets transported to the lungs through blood from where it is released out through the nostrils.

Question 20. Study the diagram given below and answer the following questions

HBSE Solutions Class 10 Biology Chapter 1 The Opening And Closing Of Stomata

  1. Name the parts labelled 1, 2, 3, and 4.
  2. State the function of (2).

Answer:

  1. 1- Epidermal cell, 2- guard cells, 3- stoma, 4 – chlorophyll.
  2. Guard cells help in the opening and closing of stomata

Question 21. What is excretion? How do unicellular organisms remove their wastes?

Answer: The biological process of removing harmful metabolic wastes from the body is called excretion. Unicellular organisms remove their wastes through simple diffusion.

Question 22. If you consume butter during lunch, how will it get digested in your body?

Answer: Butter consists of fat, which is digested by bile released from the liver.

  1. Fats are present in the intestines in the form of large globules, making it difficult for enzymes to act on them.
  2. Bile salts present in the bile break fats into smaller globules to increase the action of enzymes. This process is known as emulsification.
  3. Later, lipase acts on the emulsified fats and breaks them down into fatty acids and glycerol.

Question 23. What is the role of the following in the human digestive system?

  1. Mucus
  2. Bicarbonate
  3. Trypsin

Answer:

  1. Mucus: It protects the inner lining of the stomach from HCI.
  2. Bicarbonate: It makes the acidic food alkaline so that pancreatic enzymes can act on it.
  3. Trypsin: It converts proteins into amino acids.

Question 24. You have visited a zoo where you have observed that a fish is respiring at a faster rate as compared to a dog. Explain the reason for this.

Answer: Fish is an aquatic animal, while dog is a terrestrial animal. Aquatic organisms obtain oxygen dissolved in water, while terrestrial organisms use oxygen present in the air for respiration. As compared to air, the availability of oxygen in water is low. Hence, aquatic organisms like fish have to breathe faster as compared to terrestrial organisms like dog. A faster rate of breathing provides more oxygen to aquatic animals.

Question 25. What are the consequences of deficiency of haemoglobin in our body?

Answer: Deficiency of haemoglobin in our body causes anaemia. In anaemia, the blood is unable to transport sufficient amount of oxygen required by the body. As a result, respiration will be less, and hence, less energy will be available for the body. A haemoglobin-deficient person will feel weak, pale, lethargic, and will not be able to do heavy physical work.

Question 5.

  1. Write the three main steps which take place in the chloroplast during photosynthesis.
  2. How do stomata open and close?

Answer:

  1. Steps which take place inside the chloroplast during photosynthesis:
    1. Absorption of sunlight energy by chlorophyll.
    2. Conversion of light energy to chemical energy, and splitting of water into hydrogen and oxygen by light energy.
    3. Reduction of carbon dioxide by hydrogen to form carbohydrates like glucose by utilising chemical energy.
  2. The opening and closing of stomata is controlled by guard cells. When water flows into the guard cells, they swell, become curved and cause the stomata to open. When the guard cells lose water, they shrink, become straight and the stomata close.

HBSE Solutions Class 10 Biology Chapter 1 Opening And Closing Of Stomata

Question 26. What are the methods used by plants to get rid of excretory products?

Answer:

Methods used by plants to get rid of excretory products:

  1. Oxygen produced during photosynthesis is removed by diffusion through stomata and lenticels.
  2. Excess water is removed by transpiration.
  3. Many waste products are stored in cell vacuoles.
  4. Some waste products are removed in the falling leaves.
  5. Some wastes are stored as resins and gums, especially in old xylem.
  6. Plants also get rid of some waste products by excreting into the surrounding soil.

Question 27. Tooth enamel is one of the hardest substances in our body. How does it undergo damage due to eating chocolates and sweets?

Answer:

Dental caries or tooth decay occurs when bacteria acting on sugars produce acids due to which food particles stick to the teeth to form dental plaque. When plaque covers the teeth, saliva (which is alkaline in nature) cannot reach the tooth surface to neutralise the acid. This may cause gradual softening of enamel.

Question 28. Give reasons for the following:

  1. Diffusion is insufficient to meet the oxygen requirements of multicellular organisms.
  2. People living in the mountains have more red corpuscles in their blood than people living in the plains.
  3. Energy requirement is less for amphibians than for birds.

Answer:

  1. Diffusion is insufficient to meet the oxygen requirement of multicellular organisms because the volume of the human body is so large that oxygen cannot diffuse into all the cells of the body quickly.
  2. People living in the mountains have more RBCs in their blood than people living in the plains because the low air pressure requires more RBCs to supply the body cells with oxygen.
  3. Amphibians are cold-blooded animals whose body temperature depends on the temperature in the environment. They do not need energy to maintain their body temperature, and hence, their requirement of energy is less.

Question 29.

  1. What is regeneration of an organism? With a neat diagram, describe regeneration in Planaria.
  2. How does the embryo get nourishment inside the mother’s body?
  3. List the changes seen in the ovule and ovary after fertilisation.

Answer:

1. Planaria can be cut into any number of pieces, and each piece grows into a complete organism. This process is called regeneration. It is carried out by specialised cells in the organism.

Regeneration in Planaria

HBSE Solutions Class 10 Biology Chapter 1 Regeneration In Planaria

2. The embryo gets nutrition from the mother’s blood with the help of a special tissue called the placenta. It provides a large space and area for glucose and oxygen to pass from the mother to the embryo. The developing embryo also produces waste substances which can be moved by transferring them into the mother’s blood through the placenta.

3. The ovule develops a tough coat and is gradually converted to a seed. The ovary grows rapidly and ripens to form a fruit.

Question 30. Name the end products formed during:

  1. Oxidation of glucose in the muscles
  2. Oxidation of glucose in body cells
  3. Breakdown of glucose anaerobically

Answer:

  1. Lactic acid and ATP.
  2. Water, carbon dioxide and ATP.
  3. Ethanol, carbon dioxide and ATP.

Chapter 1 Life Processes Long Questions And Answers

Question 1. What happens to the pollen which falls on a suitable stigma?

Answer:

Reproduction in flowering plants begins with pollination, the transfer of pollen from anther to stigma of the flower. If the transfer takes place to the stigma of flower of the same plant, it is called self-pollination. If the transfer takes place to the stigma of flower of another plant, it is called cross-pollination. Once the pollen grain reaches the stigma, a pollen tube grows from the pollen grain to an ovule. Two sperm nuclei develop. One of them unites with the egg nucleus and produces a zygote. The other unites with two polar nuclei to produce an endosperm that stores food for the development of a plant. Zygote develops into seed and seedling.

Question 2. What is the advantage of having a four-chambered heart? Support your answer with a diagram of the section of the human heart.

Answer:

In a four-chambered heart, the left half is completely separated from the right half by septa. This prevents oxygenated and deoxygenated blood from mixing. It allows a highly efficient supply of oxygenated blood to all parts of the body. This is useful in animals that have energy needs, such as birds and mammals.

Question 3. What could be the possible reason for the declining female-male sex ratio in our country? Suggest two measures to achieve the 1:1 ratio.

Answer:

The reason for declining females in India is sex-selective abortions of the female foetus through surgeries (female foeticides). This can be avoided by banning prenatal sex determination. Everyone in society needs to be educated about the equality of gender and the health of women.

Question 4. Where does fertilisation occur in the female reproductive tract?

Answer:

Fertilisation occurs in the fallopian tube or oviduct of the female reproductive tract.

Question 5.

  1. What happens if an egg is not fertilised?
  2. Why do we need to adopt contraceptive measures?
  3. Name one bacterial and one viral sexually transmitted disease.
  4. How does the embryo get nourishment inside the mother’s body?
  5. The embryo gets its nourishment inside the mother’s body with the help of the placenta.

Answer:

  1. If an egg is not fertilised by a sperm, then blood along with cellular debris comes out through the vagina; this process is called menstruation.
  2. Need to adopt contraceptive measures:
    • To prevent unwanted pregnancies
    • To prevent sexually transmitted diseases
    • Spacing between children
    • Sound health
  3. Sexually transmitted diseases: Bacterial- Gonorrhea; Viral- AIDS
  4. The embryo gets its nourishment inside the mother’s body with the help of the placenta

Question 6. Study this diagrammatic representation of the process of fertilisation and answer the questions which follow.

HBSE Solutions Class 10 Biology Chapter 1 The Process Of Fertilization

  1. Name the parts labelled 1, 2, and 4.
  2. What happens to the ovary and the ovule after fertilisation?
  3. What part does the stigma play in the process of fertilisation?

Answer:

  1. 1 : Style; 2 : Pollen tube; 4 : Embryo sac;
  2. After fertilisation, the ovary enlarges to form the fruit and the ovarian wall forms the fruit wall. The ovule becomes the seed.
  3. Pollen grain is transferred to the stigma during pollination. Germination of pollen grain takes place only if it falls on the stigma. After germination, the pollen tube grows through the stigma and reaches the ovary for the fertilisation of the egg cell.

Haryana Board Class 10 Biology Solutions For Chapter 5 Our Environment

Chapter 5 Our Environment Multiple Choice Questions And Answers

Question 1. The table lists some components of an ecosystem.

Rain, Water, Butterfly, Air, Grass, Bacteria, Fungi, Sunlight. A student wants to classify these into abiotic components from biotic components separately. Which option correctly shows the classification done by the student?

HBSE Solutions Class 10 Biology Chapter 5 Components Of An Ecosystem

Answer: Components Of An Ecosystems

Question 2. Which statement shows interaction of an abiotic component with a biotic component in an ecosystem?

  1. A grasshopper feeding on a leaf,
  2. Rainwater running down into the lake.
  3. An earthworm making a burrow in the soil.
  4. A mouse fighting with another mouse for food.

Answer: 3. An earthworm making a burrow in the soil.

Question 3. The table shows some organisms including plants, animals and how they get energy.

HBSE Solutions Class 10 Biology Chapter 5 Organisms Including Plants, Animals And How They Get Energy

Which option shows the correct model made based on the table?

HBSE Solutions Class 10 Biology Chapter 5 Organisms Including Plants, Animals And How They Get Energys

Answer: 1.

Question 4. The image shows a food web.

HBSE Solutions Class 10 Biology Chapter 5 A Food Web

  1. Fox feeds on hawk obtain energy,
  2. Hawk feeds on oak acorn to obtain energy.
  3. Squirrel feeds on pine borer to obtain energy,
  4. Salamander feeds on pine borer to obtain energy.

Answer: 4. Salamander feeds on pine borer to obtain energy.

Question 5. The manufacturing of Chlorofluorocarbons free refrigerators is mandatory throughout the world. How this help prevent ozone depletion?

  1. This will help convert oxygen molecules into ozone.
  2. This will help convert the CFCs into ozone molecules.
  3. This will reduce the production of CFC from oxygen molecules.
  4. This will reduce the release of CFCs that reacts with ozone molecule

Answer: 4. This will reduce the release of CFCs that reacts with ozone molecule

Question 6. Ozone forms by combination of free oxygen atoms and oxygen molecules. How do free oxygen atoms form at higher levels of atmosphere?

  1. By splitting of molecular oxygen into free oxygen atoms in the presence of low energy uv radiations
  2. By splitting of a molecular oxygen into free oxygen atoms in presence of high energy uv radiations
  3. By the combination of two molecular oxygen in the presence of high energy uv radiations
  4. By the combination of two free oxygen atoms in the presence of lower energy UV radiations

Answer: 2. By splitting of a molecular oxygen into free oxygen atoms in presence of high energy uv radiations

Question 7. Some waste products are listed below:

Grass cutting; Polythene bags; Plastic bags; Used tea bags; Paper straw; Old clothes Which group of waste materials can be classified as non-biodegradable?

  1. Plant waste, used tea bags
  2. Polyethene bags, plastic toys
  3. Used tea bags, paper straw
  4. Old clothes, broken footwear

Answer: 2. Polyethene bags, plastic toys

Question 8. The table shows some waste materials that changed and remain unchanged when buried in the soil. Plastic box; Bubble wrap; Vegetable peels; Rubber tyre; Empty carton.

What materials correctly classified biodegradable and non-biodegradable materials?

HBSE Solutions Class 10 Biology Chapter 5 Biodegradable And Non-Biodegradable Materials

Answer: Biodegradable And Non-Biodegradable Material

Question 9. Every food chain in the ecosystem begins with ___ which are the original source of food.

  1. Saprophytes
  2. Parasites
  3. Producers
  4. Herbivores

Answer: 4. Herbivores

Question 10. We should reduce the use of the plastic bags, bottles etc. because:

  1. They are not durable
  2. They are non-biodegradable
  3. They are made of toxic materials
  4. They react with the atmospheric gases

Answer: 2. They are non-biodegradable

Question 11. Which among the following statements is incorrect in view of the plants?

  1. They convert the solar energy into mechanical energy
  2. They prepare their food from organic compounds
  3. They are also called producers
  4. They are the initial source of energy in a food chain

Answer: 2. They prepare their food from organic compounds

Question 12. In a food chain the second trophic level is occupied by:

  1. Carnivores
  2. Autotrophs
  3. Herbivores
  4. Producers

Answer: 3. Herbivores

Question 13. The process of accumulation of harmful chemical substances like pesticides, in the body of living organisms at each trophic level of a food chain is known as:

  1. Biological magnification
  2. Biological accumulation
  3. Chemical magnification
  4. Chemical accumulation

Answer: 1. Biological magnification

Question 14. Which of the following may be a conclusion of the excessive exposure of humans to sun’s ultraviolet rays?

  • Peptic ulcers
  • Eye disease like cataract
  • Damage to lungs
  • Skin cancer
  1. (1) and (4);
  2. (2), (3) and (4);
  3. (2) and (4);
  4. Only (4)

Answer: 3. (2) and (4);

Question 15. Which among the following is a correct full form for DDT?

  1. Dichlorodiphenyltrichloroethane
  2. Dichlorodiphenyltetrachloroethane
  3. Dichlorodecaphenyltrichloroethane
  4. Dichlorodiethyltrichloroethane

Answer: 1. Dichlorodiphenyltrichloroethane

Question 16. Which of the following radiations is responsible for the conversion of atmospheric oxygen to ozone?

  1. Gamma radiations
  2. Cosmic radiations
  3. Infrared radiations
  4. Ultraviolet radiations

Answer: 4. Ultraviolet radiations

Chapter 5 Our Environment Very Short Answer Questions

Question 1. How much percentage of solar radiation is absorbed by the green plants during photosynthesis?

Answer: 1% of solar radiation is absorbed by the green plants during photosynthesis.

Question 2. Why plastic is not degraded by bacteria?

Answer: Plastic is not degraded by bacteria because they do not have enzymes to degrade plastic.

Question 3. List any one eco-friendly practice.

Answer: Carrying cloth bag while shopping is an eco-friendly practice.

Question 4. Name the process which is a direct outcome of excessive burning of fossil fuels.

Answer: Global warming is a direct outcome of excessive burning of fossil fuels.

Question 5. What will happen if deer is missing in this food chain?

Grass →Deer → Tiger

Answer: If deer is missing in the given food chain, the population of tiger will decrease and the population of grass will increase.

Question 6. Construct a food chain comprising of a snake, hawk, rat and plant.

Answer: Plant → Rat → Snake→ Hawk.

Question 7. Write one aquatic food chain.

Answer: Phytoplankton → Zooplankton → Small fishes → Large fishes

Question 8. Which of the following belong to the same trophic level: grasshopper, spider, grass, hawk and lizard?

Answer: Grasshopper and spider.

Question 9. Why some materials are biodegradable whereas others are non-biodegradable?

Answer: It is due to the property of decomposer organisms of being specific in their action that some waste materials are biodegradable whereas others are nonbiodegradable.

Question 10. In a food chain consisting of snake, insect, grass and frog, assign an appropriate trophic level to frog.

Answer: Grass -> insect -> frog -> snake

Question 11. Ozone is deadly poisonous; still it performs an essential function. How?

Answer: Ozone performs an essential function by forming an ozone layer which shields the surface of the earth from the entry of UV rays from the sun.

Question 12. How is depletion of ozone layer in the atmosphere responsible for causing skin cancer?

Answer: Ultraviolet rays can pass through ozone hole and causes skin cancer.

Question 13. We do not clean ponds or lakes, but an aquarium needs to be cleaned. Why?

Answer: An aquarium is an artificial and incomplete ecosystem compared to pond or lakes which are natural, self-sustaining and complete ecosystem where there is a perfect recycling of materials. Therefore, it needs to be cleaned.

Question 14. Which methodology is preferred for reducing the problem of waste disposal?

Answer: The 5R approach

Question 15. Give reason why a food chain cannot have more than four trophic levels.

Answer: Only 10% of the energy gets transferred from one trophic level to the next. So after 3 or 4 trophic levels, the energy available for passing on is too less to support another trophic level. Very little usable energy remains after 4 trophic levels. Hence the number of trophic levels in a food chain is limited.

Question 16. How is the increase in demand for energy affecting our environment adversely?

Answer: The increase in demand for energy affects our environment adversely. Due to this increase, pollutants like CO, CO2 and SO2 are released into the atmosphere which leads to greenhouse effect.

Question 17. Explain how ozone being a deadly poison can still perform an essential function for our environment.

Answer: The ozone layer is a thin part of the Earth’s atmosphere that absorbs almost all of the sun’s harmful ultraviolet light. It protects earth from the ultraviolet radiations which come from the sun.

Question 18. Which of the following belong to the same trophic level? Tree, Crow, Lion, Grass, Deer

Answer: Tree and Grass (both are producers)

Chapter 5 Short Answer Type Questions And Answers

Question 1. Why is there a need to ban the use of polythene bags?

Answer: Polythene bags need to be banned because they are non-biodegradable; microorganisms are unable to decompose them. So, they keep on accumulating in the environment and cause land pollution.

Question 2. Why a vegetarian food habit help us in getting more energy?

Answer: A person having a vegetarian food habit is close to the producer level and gets maximum amount of energy as compared to the organisms at a higher trophic level as only 10% of energy is passed on from one trophic level to another.

Question 3.

We do not clean ponds or lakes on a regular basis, but an aquarium needs to be cleaned regularly. Explain.

Answer:

A pond is an example of a natural ecosystem, whereas an aquarium is an example of an artificial ecosystem. Ponds do not need to be cleaned on a regular basis because they have natural flora and fauna present in them which helps in cleaning the pond ecosystem.

However, an aquarium does not contain soil and decomposing bacteria which help in degrading complex organic substances into simpler inorganic substances. Therefore, an aquarium needs to be cleaned regularly.

Question 4. “Industrialization is one of the main causes of deterioration of environment”. List any four reasons in favour of this statement.

Answer:

Industrialisation causes following effects on the environment- Noise and air pollution caused due to industrialization disturbs the environment; Waste released from industries leads to water pollution; SO2, NO2 etc. emitted by the industries, are toxic; Radioactive radiations emitted by nuclear power stations are toxic to living organisms.

Question 5. Pesticides are useful to farmers yet considered as pollutants. Given reasons.

Answer:

Pesticides kill insects and pests thereby protect the crops but these pesticides remain on the crops which enter the food chain and get accumulated in the organisms at the top most tropic level that causes diseases. When these are washed away by rain river water, it also causes pollution.

Question 6. Damage to the ozone layer is a cause for concern.” Justify this statement. Suggest any two steps to limit this damage.

Answer:

Ozone layer prevents the harmful ultraviolet radiation to enter the atmosphere and reach the earth’s surface. Depletion of ozone layer has become a cause for concern because it can cause serious effects on human body and other organisms of the environment like fatal diseases such as skin cancer, changes in genetic material DNA and eye damage.

Two steps to limit this damage are:

  1. Judicious use of aerosol spray propellants such as fluorocarbon and chlorofluorocarbons which cause depletion or hole in the ozone layer,
  2. Control over large scale nuclear explosions and limited use of supersonic planes.

Question 7. What will happen if we kill all the organisms in one trophic level?

Answer:

Killing all the organisms in one trophic level would decrease the availability of food at the next trophic level. It will also result in excess of organisms at the previous trophic level. This would cause an imbalance in the food chain.

Question 8. In the following food chain, plants provide 500 J of energy to rats. How much energy will be available to hawks from snakes?

Plants → Rats → Snakes → Hawks

Answer:

On applying the 10% law to the given food chain- Plants -> Rats → Snakes → Hawks

Energy available to snakes from rats = 10% of 500 = 10/100 x 500 = 50 J

So, energy available to hawks from snakes = 10% of 50 J

= 10/100 x 50 = 5J

Question 9. Which one of the following food chain is better and why?

Plant → Man

Plant → Goat → Man

Answer:

Food chain

  1. Is better than food chain
  2. In a shorter food chain maximum transfer of energy takes place according to ten percent law of nature.

Question 10. Observe the given food chain. Plant (1000 KJ) → Goat → Lion

If autotrophs occupying the first tropic level are called producers what are herbivores called?

How much energy does the lion get in the above food chain?

Answer:

  1. Primary consumers;
  2. When 10% law is applied, the goat gets 100 kJ and lion gets 10 kJ.

Question 11. Plastic cups were used to serve tea in trains in early days- these could be returned to the vendors, cleaned and reused. Later, Kulhads were used instead of plastic cups. Now, paper cups are used for serving tea. What are the reasons for the shift from Plastic to Kulhads and then finally to paper cups?

Answer:

Use of plastic cups raised the concern towards hygiene, and thus they were replaced by disposable plastic cups. Plastic cups are non-biodegradable and harm the environment.

They were thus replaced by Kulhads. Making Kulhad made of clay on a large scale resulted in the loss of top fertile soil.

Now, disposable paper cups are used because paper can be recycled, it is biodegradable and is ecofriendly material which does not cause environmental pollution.

Question 12.

  1. Give two examples of decomposers present in an ecosystem.
  2. How is the presence of decomposers crucial in the ecosystem?

Answer:

Bacteria and fungi are examples of decomposers present in an ecosystem,

Decomposers breakdown or decompose the dead remains of plants and animals and their waste organic products into simpler, inorganic substances. The latter are released into the environment for their reuse as raw materials by producers. Thus, decomposers provide space for new life to settle in the biosphere. Hence, their presence is crucial to the functioning of the ecosystem.

Question 13. Refer to the given food web.

HBSE Solutions Class 10 Biology Chapter 5 Food Web

  1. What will be the effect on food web if population of water fleas gets eliminated?
  2. In the given food web, which organism operates at both primary and tertiary consumer level?

Answer:

  1. In the given food web, water fleas feed on insect larvae and are in turn fed by water beetles. So, if water fleas get eliminated, then population of insect larvae will increase and that of water beetles will decrease. As small fish are dependent on water beetles for food, a decrease in population of water fleas will cause a decrease in their population as well. Population of frogs remain unaffected as frog also depend on insect larvae for food.
  2. Small fish operates at both primary and tertiary consumer level in the given food web. Aquatic plant – Small fish – Large fish – Aquatic plant – Insect larvae – Water fleas – Small fish – Large fish

Question 14. A modern insecticide has been introduced with certain new properties like accumulation in the bodies of predators, broken down by soil bacteria, easily washed into lakes and rivers and taken up by plant roots. Among all these properties which one will help in reducing or keeping the level of environment pollution to lowest.

Answer:

The property of newly developed insecticide includes that it can easily get decomposed into simpler components by soil bacteria, i.e. biodegradable, which will help in the reduction of environmental pollution.

Question 15. In the following food chain, 5 J of energy is available to man. How much energy was available at producer level?

Plant-Sheep-Man

Answer:

Energy available to man in the above food chain is J5. According to 10% law, energy available to sheep is 50 J. Again according to 10% law, energy available to plants, i.e., at produce level is 500 J.

Question 16. Rearrange the following according to their trophic levels in a food chain.

Fish, zooplankton, seal, phytoplankton

HBSE Solutions Class 10 Biology Chapter 5 Trophic Levels In A Food Chain

Answer:

Since a food chain always starts with producers followed by primary, secondary and tertiary consumer, thus, the food chain will be Phytoplankton, zooplankton, fish, seal.

Chapter 5 Long Questions And Answers

Question 1.

  1. Our food grains such as wheat and rice, vegetables and fruits, and even meat are found to contain varying amounts of pesticide residues. State the reason to explain how and why it happens?
  2. Make food chains in
    1. Forest
    2. Grassland. How does food chain differ from a food web?

Answer:

  1. Harmful pesticides like DDT enter the plant body on being absorbed from the soil. When these are eaten by animals they get accumulated in the animal’s body as they are neither metabolized nor excreted out of their body. The concentration of these harmful chemicals gets increased at successive levels and are thus found in food grains as well as meat,
  2. The food chains are:
    1. Forest: Plants → Deer Lion
    2. Grassland: Plants → Grasshopper → Frog → Snake → Eagle

Food chain is a series of organisms feeding on one another while the food web is a network of food chains consisting of a number of interlinked food chains.

Question 2. Draw a sequence of suitable methods of disposal of waste produced at your home to minimise environmental pollution.

Answer:

The household waste produced from various activities is called garbage and its proper disposal is done in such a way that it does not cause environmental pollution. Methods of waste disposal include:

Recycling: The processing of certain wastes to form new products is called recycling, For Example. paper, glass
etc., are recyclable,

Composting: It is the process of collecting biodegradable wastes like leftovers of food items, peels, etc. and burying them in a pit and using it as manure.

Incineration: It is burning of a substance at high temperature to reduce it to ashes.

Landfills: Dumping of non-biodegradable waste in low-lying areas is called landfill,

Sewage treatment: In sewage treatment plants, the sewage is
processed and decomposed into simpler inorganic chemicals.

Question 3. The activities of man had adverse effects on all forms of living organisms in the biosphere. Unlimited exploitation of nature by man disturbed the delicate ecological balanced between the living and non-living components of the biosphere. The unfavourable conditions created by man himself threatened the survival not only of himself but also of the entire living organisms on the mother earth. One of your classmates is an active member of ‘Eco dub’ of your school, which is creating enviromental awareness amongst the school students, spreading the same in the sociey and also working hard for preventing environmental degradation of the surroundings.

  1. Why is it necessary to conserve our environment?
  2. State the importance of green and blue dustbins in the safe disposal of the household waste.
  3. List two values exhibited by your classmate who is an active member of Eco-club of your school?

Answer:

  1. It is necessary to conserve our environment for the sustenance of the present life and also for the future generations. We exist because of support and supply from nature our sole existence is not possible.
  2. Green dustbin (for biodegradable waste) and blue dustbin (for non-biodegradable waste) should be used in household for the proper segregation of waste. If both type of waste are segregated, they can be treated accordingly,
  3. My classmate reflects eco-friendly nature, concern towards the
    environment and knowledge of environmental issues.

Question 4.

  1. Why energy transfer is said to be unidirectional whereas biochemical transfer is said to be cyclic?
  2. Suggest any three activities in daily life which are eco-friendly.

Answer:

  1. Energy transfer is said to be unidirectional because when the energy is absorbed by autotrophs from the sun, it is never reabsorbed by it. Similarly when consumers eat up the producers directly or indirectly the energy transferred in this process can never be reversed in the food chain. In biogeochemical cycles chemical elements move from environment to organism and back to the environment.
  2. Activities in daily life which are eco-friendly: Separation of biodegradable and nonbiodegradable substances; Gardening; Use of gunny bags or paper bags in place of polythene or plastic bags; Use of compost and vermicompost in place of fertilisers; Harvesting rainwater.

Question 5.

  1. What is biological magnification? Will the levels of this magnification be different at different levels of the ecosystem? Where will the magnification be maximum?
  2. DDT that was sprayed in minute amount on food plants was detected in high concentration in man? How did it happen? Explain.

Answer:

  1. The accumulation or increasing concentration of a substance such as a toxic chemical in the body of living organisms at different trophic levels in a food chain is called biological magnification. Yes, the concentration of these harmful chemicals will be different at different levels of the ecosystem. It will be maximum at the last trophic level which is mostly occupied by the top carnivores (quaternary consumers),
  2. DDT that was sprayed in minute amount on food plants was detected in high concentration in man due to biological magnification. When pesticides like DDT are used to protect crops from diseases and pests. These non-biodegradable substances enter the soil. From soil these substances are absorbed by plants along with water and minerals. The food plants when consumed by organisms, they get accumulated at different trophic levels. As the human beings occupy the top position in any food chain, maximum concentration of such harmful chemicals get accumulated in the bodies of man.

Haryana Board Class 10 Biology Solutions For Chapter 4 Control And Coordination

Haryana Board Class 10 Biology Solutions For  Chapter 4 Control And Coordination Multiple Choice Questions And Answers

Question 1. The image shows the structure of a neuron.

HBSE Class 10 Biology Solutions For The Structure Of A Neuron

How will information travel within a neuron?

  1. Dendrite → cell body → axon → nerve ending
  2. Dendrite → axon → cell body → nerve ending
  3. Axon → dendrite → cell body → nerve ending
  4. Axon -> cell body -> dendrite -> nerve ending

Answer: 1. Dendrite → cell body → axon → nerve ending

Question 2. The image shows structure of a neuron.

HBSE Class 10 Biology Solutions For Structure Of A Neuron

After our nose senses a smell, which option shows the mechanism of the travelling of sense in our body?

  1. Olfactory receptors- dendritic tip of a nerve cell- – axon- nerve ending-release of signal -dendritic tip of other nerve cell
  2. Olfactory receptors- dendritic tip of a nerve cell- axon- cell body- release of signal-dendritic tip of other nerve cell
  3. Gustatory receptors- dendritic tip of a nerve cell- cell body- axon- release of signal-dendritic tip of other nerve cell
  4. Gustatory receptors- dendritic tip of a nerve cell- axon- cell body- release of signal-dendritic tip of other nerve cell

Answer: 1. Olfactory receptors- dendritic tip of a nerve cell- – axon- nerve ending-release of signal -dendritic tip of other nerve cell

Question 3. Which option correctly shows the sequence of events that occur when we touch a hot utensil?

HBSE Class 10 Biology Solutions For Sequence Of Events That Occur When We Touch A Hot Utensil

Answer: 2.

Question 4. Which option correctly shows the order of events when a bright light is focused on our eyes?

  1. Bright light- receptors in eyes – sensory neuron – spinal cord – motor neurons – eyelid closes
  2. Bright light – receptors in eyes – spinal cord -sensory neuron – motor neurons – eyelid closes
  3. Bright light – receptors in eyes – sensory neuron – motor neurons – spinal cord – eyelid closes
  4. Bright light – receptors in eyes – spinal cord – motor neurons – sensory neuron – eyelid closes

Answer: 1. Bright light- receptors in eyes – sensory neuron – spinal cord – motor neurons – eyelid closes

Question 5. The image shows the labelled structure of a brain.

HBSE Class 10 Biology Solutions For Labelled Structure Of A Brain

Which parts of the brain controls the blood pressure?

  1. Spinal cord, skull, hypothalamus
  2. Cord, skull, cerebrum
  3. Pons, medulla, cerebellum
  4. Pons, medulla, pituitary

Answer: 3. Pons, medulla, cerebellum

Question 6. Which option illustrates the location of centre that controls the feelings associated with hunger (M) and the centre that allows a person to walk in a straight line (N)?

HBSE Class 10 Biology Solutions For Location Of Centre That Controls The Feelings Associated With Hunger (M)

Answer: 4.

Question 7. When we touch the leaves of “touch-me-not” plant, they began to fold up and droop. How does the plant communicate the information of touch?

  1. The plant uses electrical signals to transfer information from external environment to cells.
  2. The plant uses electrical- chemical signals to transfer information from cell to cell.
  3. The plant uses electrical- chemical signals to transfer information from tissue to specialized cells.
  4. The plant uses electrical signals to transfer information from cell to specialized tissues.

Answer: 2. The plant uses electrical- chemical signals to transfer information from cell to cell.

Question 8. Aditya potted some germinated seeds in a pot. He put the pot in a cardboard box that was open from one side. He keeps the box in a way that the open side of box faces sunlight near his window. After 2-3 days he observes the shoot bends towards light as shown in image.

HBSE Class 10 Biology Solutions For Aditya Potted Some Germinated Seeds In A Pot

Which type of tropism he observes?

  1. Geotropism
  2. Phototropism
  3. Chemotropism
  4. Hydrotropism

Answer: 2. Phototropism

Question 9. What is a likely limitation of electric impulses?

  1. The electric impulses travel slowly between the neurons.
  2. The electric impulses allow signal transmission in multiple directions.
  3. The electric impulses are transmitted to only those body parts that are connected to neurons.
  4. The electric impulses once generated needs to be transmitted quickly within the body.

Answer: 3. The electric impulses are transmitted to only those body parts that are connected to neurons.

Question 10. Organisms depend on hormones as well as electric impulses for the transmission of signals from brain to rest of the body. What can be a likely advantage of hormones over electric impulses?

  1. It is secreted by all types of cells present in the body.
  2. It is secreted by stimulated cells and reaches all cells of the body.
  3. It is relayed to the target organ at a faster rate than electric impulses
  4. It does not depend on an external stimulus to be generated in the cells.

Answer: 2. It is secreted by stimulated cells and reaches all cells of the body.

Question 11. What is the function of pituitary gland?

  1. To develop sex organs in males
  2. To stimulate growth in all organs
  3. To regulate sugar and salt level in the body
  4. To initiate metabolism in the body

Answer: 2. To stimulate growth in all organs

Question 12. A female is suffering from irregular menstrual cycle. The doctor prescribed her some hormonal tablets. Which option shows the hormone she lacks in her body from the endocrine gland?

  1. Oestrogen
  2. Testosterone
  3. Adrenalin
  4. Thyroxin

Answer: 1. Oestrogen

Question 13. The movement of a plant part in response to the force of attraction exerted by the earth is called:

  1. Hydrotropism
  2. Geotropism
  3. Chemotropism
  4. Phototropism

Answer: 2. Geotropism

Question 14. A big tree falls in a forest, but its roots are still in contact with the soil. The branches of this fallen tree grow straight up (vertically). This happens in response to:

  1. Water and light
  2. Water and minerals
  3. Gravity and water
  4. Light and gravity

Answer: 4. Light and gravity

Question 15. The main function of the plant hormone called abscisic acid is to:

  1. Increase the length of cells
  2. Promote cell division
  3. Inhibit growth
  4. Promote growth of stem and roots

Answer: 3. Inhibit growth

Question 16. The growth of tendrils in pea plants is due to the:

  1. Effect of sunlight on the tendril cells facing the sun
  2. Effect of gravity on the part of tendril hanging down towards the earth
  3. Rapid cell division and elongation in tendril cells that are away from the support
  4. Rapid cell division and elongation in tendril cells in contact with the support

Answer: 3. Rapid cell division and elongation in tendril cells that are away from the support

Question 17. The plant hormone which triggers the fall of mature leaves and fruits from the plant body is:

  1. Auxin
  2. Gibberellin
  3. Abscisic acid
  4. Cytokinin

Answer: 3. Abscisic acid

Chapter 4 Control And Coordination Very Short Questions And Answers

Question 1. Why is it advised to use iodised salt in our diet?

Answer: Iodine stimulates the thyroid gland to produce thyroxin hormone. Deficiency of this hormone results in the enlargement of the thyroid gland. This can lead to goitre.

Question 2. Give an example of a plant hormone that promotes its growth. Where it is synthesized?

Answer: Plant hormone that promotes growth is auxin. It is synthesized at the tip of the plant stem.

Question 3. State the function of:

  1. Gustatory receptors, and
  2. Olfactory receptors.

Answer:

  1. Gustatory receptors – these are sensitive to taste;
  2. Olfactory receptors – these are sensitive to smell.

Question 4. Name the part of the brain which controls posture and balance of the body.

Answer: Cerebellum

Question 5. Mention the part of the body where gustatory and olfactory receptors are located.

Answer: Gustatory receptors are located in Cerebrum of fore-brain. Olfactory receptors are located in Olfactory lobe of fore-brain.

Question 6. How is the spinal cord protected in the human body?

Answer: Spinal cord is enclosed in a bony cage called vertebral column.

Question 7. A potted plant is made to lie horizontally on the ground. Which part of the plant will show

  1. Positive geotropism?
  2. Negative geotropism?

Answer:

  1. Root
  2. Shoot.

Question 8. Mention the function of the hind-brain in humans.

Answer: Hind brain controls respiration, cardio-vascular reflexes and gastric secretions. It also modulates the motor commands initiated by the cerebrum.

Question 9. A young green plant receives sunlight from one direction only. What will happen to its shoots?

Answer: Shoots will bend towards the light and roots away from the light.

Question 10. What is the function of thyroxine hormone in our body?

Answer: Thyroxine hormone regulates the carbohydrate, protein and fat metabolism in the body so as to provide the best growth balance.

Question 11. Name two tissues that provide control and coordination in multicellular animals.

Answer: Nervous tissue and Muscular tissue

Chapter 4 Control And Coordination Short Questions And Answers

Question 1.

  1. Name the hormones that are released in human males and females when they reach puberty.
  2. Name a gland associated with brain. Which problem is caused due to the deficiency of the hormone released by this gland?

Answer:

  1. Testes in males produces hormone testosterone. Ovaries in females produces hormone
    oestrogen.
  2. Pituitary gland present in the brain is responsible for body growth, development of bones and muscles (if excess-gigantism) (if less-dwarfism).

Question 2. Which part of the brain controls involuntary actions? Write the function of any two regions of it.

Answer: Hind-brain controls the involuntary actions. Cerebellum controls the coordination of body movement and posture. Medulla oblongata regulates center for swallowing, coughing, sneezing and vomiting.

Question 3. Mention the functions of adrenaline hormone.

Answer: Adrenaline hormone is released into the blood from the adrenal gland during stimulation of the nervous system on seeing any adverse situation of fight or fright, it: Increases the blood pressure; Increases heart beat rate; Increases breathing rate; Diverts blood to essential organs including the heart, brain and skeletal muscles by dilating their blood vessels and constricting those of less essential organs, such as the skin and digestive system.

Question 4. Name, the two main organs of our central nervous system. Which one of them plays a major role in sending command to muscles to act without involving thinking process? Name the phenomenon involved.

Answer: Brain and Spinal cord. Spinal cord plays a major role in sending command to muscles to act without involving thinking process. This phenomenon is called reflex action.

Question 5. What is the need for a system of control and coordination in human beings?

Answer: A living being does not live in isolation. It has to constantly interact with its external environment and has to respond properly for its survival, e.g. when a hungry lion spots a deer, the lion has to quickly make a move so that it can have its food. On the other hand, the deer needs to quickly make a move to run for its life. The responses which a living being makes in relation to external stimuli are controlled and coordinated by a system; especially in complex animals. So, control and coordination is essential in maintaining a state of stability and a steady state between the internal conditions of an organism and the external environment.

Chapter 4 Control And Coordination Short Questions And Answers

Question 1. State how concentration of auxin stimulates the cells to grow longer on the side of the shoot which is away from light?

Answer: When light falls on the side of the shoot auxin diffuses towards the shady side of the shoot. This concentration of the auxin stimulates the cell to grow longer on the side of the shoot which is away from light. Thus plant appears to bend towards light

Question 2. Write one example each of the following tropic movements :

  1. Positive phototropism
  2. Negative phototropism
  3. Positive geotropism
  4. Negative geotropism
  5. Hydrotropism
  6. Chemotropism

Answer:

  1. Positive phototropism: shoots growing towards light.
  2. Negative phototropism: roots growing away from light towards ground,
  3. Positive geotropism: growth of roots towards earth due to the pull of the
    earth.
  4. Negative geotropism: shoots growing away from the earth,
  5. Hydrotropism: roots growing towards the source of water.
  6. Chemotropism: growth of pollen tubes towards the ovules.

Question 3. List the components of reflex arc in correct sequence. State in brief the role of brain in reflex action.

Answer: The reflex arc pathway is shown in the flow chart as follows: The Reflex arc does not involve brain. It minimises the overloading of brain.

HBSE Class 10 Biology Solutions For Reflex Arc Pathway

Question 4. Smita’s father has been advised by a doctor to reduce his sugar intake.

  1. Name the disease he is suffering from and name the hormone whose deficiency is?
  2. Identify the gland that secretes it and mention the function of this hormone.
  3. Explain how the time and amount of secretion of this hormone is regulated in human system.

Answer:

  1. He is suffering from diabetes. Deficiency of insulin causes diabetes,
  2. Pancreas secretes insulin. Insulin helps in regulating blood sugar,
  3. When the sugar level in blood increases, it is detected by the acells of the pancreas which responds by producing more insulin. As the blood sugar level falls, insulin secretion is reduced.

Question 5.

  1. Which plant hormone is present in greater concentration in the areas of rapid cell division?
  2. Give one example of a plant growth promoter and a plant growth inhibitor.

Answer:

  1. Cytokinin is present in greater concentration in the areas of rapid cell division,
  2. An example of a plant growth promoter is gibberellins and example of a plant growth inhibitor is abscisic acid.

Question 6. Which organ secretes a hormone when blood sugar rises in our body? Name the hormone and name one enzyme released by this organ.

Answer: Pancreas. Insulin is the hormone released by this organ and the name of the enzyme is pancreatic juice.

Question 7. What causes a tendril to encircle or coil around the object in contact with it is? Explain the process involved.

Answer: When a tendril comes in contact with any support, the part of the tendril in contact with the object does not grow as rapidly as the part away of the tendril away from the object. This causes the tendril to circle around the object and cling to it.

Chapter 4 Control And Coordination Long Questions And Answers

Question 1.

  1. Name the hormone which is released into the blood when its sugar level rises.
  2. Explain the need of Chemical communication in multicellular organisms.
  3. Name the organ which produces this hormone and its effect on blood sugar level. Also mention the digestive enzymes secreted by this organ with one function of each.

Answer:

  1. Insulin
  2. Since nerves cannot reach each and every part of the body of multicellular organisms, chemical communication is important for communication of messages between the central nervous system (brain and spinal cord) and other parts of the body
  3. Pancreas.

Question 2.

  1. How is brain protected from injury and shock?
  2. Name two main parts of hind brain and state the functions of each.

Answer:

  1. Skull and cerebrospinal fluid protect the brain from injury and shock.
  2. Two main parts of hind-brain are- Medulla and Cerebellum. Their functions are: medulla: Invounatary actions such as blood pressure, salivation and vomiting. Cerebellum: It is responsible for precision of voluntary actions and maintaining the posture and balance of the body.

Haryana Board Class 10 Biology Solutions For Sample Question Paper

Haryana Board Class 10 Biology Solutions For Sample Question Paper

Question 1. When sodium bicarbonate reacts with dilute hydrochloric acid, the gas evolved is:

  1. Hydrogen; it gives pop sound with burning match stick.
  2. Hydrogen; it turns lime water milky.
  3. Carbon dioxide; it turns lime water milky.
  4. Carbon dioxide; it blows off a burning match stick with a pop sound.

Answer: 3. Carbon dioxide; it turns lime water milky.

Question 2. When aqueous solutions of potassium iodide and lead nitrate are mixed, an insoluble substance separates out. The chemical equation for the reaction involved is:

  1. \(\mathrm{KI}+\mathrm{PbNO}_3 \rightarrow \mathrm{Pbl}+\mathrm{KNO}_3\)
  2. \(2 \mathrm{KI}+\mathrm{Pb}\left(\mathrm{NO}_3\right)_2 \rightarrow \mathrm{Pbl}_2+2 \mathrm{KNO}_3\)
  3. \(\mathrm{KI}+\mathrm{Pb}\left(\mathrm{NO}_3\right)_2 \rightarrow \mathrm{Pbl}+\mathrm{KNO}_3\)
  4. \(\mathrm{KI}+\mathrm{PbNO}_3 \rightarrow \mathrm{Pbl}_2+\mathrm{KNO}_3\)

Answer: 2. \(2 \mathrm{KI}+\mathrm{Pb}\left(\mathrm{NO}_3\right)_2 \rightarrow \mathrm{Pbl}_2+2 \mathrm{KNO}_3\)

Question 3. A metal ribbon ‘X’ burns in oxygen with a dazzling white flame forming a white ash Y. The correct description of X, Y and the type of reaction is:

  1. X = Ca; Y = CaO; Type of reaction = Decomposition
  2. \(2 \mathrm{KI}+\mathrm{Pb}\left(\mathrm{NO}_3\right)_2 \rightarrow \mathrm{Pbl}_2+2 \mathrm{KNO}_3\)
  3. X = Al; Y= Al203; Type of reaction = Thermal decomposition
  4. X = Zn; Y = ZnO; Type of reaction = Endothermic

Answer: 2. \(2 \mathrm{KI}+\mathrm{Pb}\left(\mathrm{NO}_3\right)_2 \rightarrow \mathrm{Pbl}_2+2 \mathrm{KNO}_3\)

Question 4. Acid present in tomato is:

  1. Methanoic acid
  2. Acetic acid
  3. Lactic acid
  4. Oxalic acid

Answer: 4. Oxalic acid

Question 5. Sodium hydroxide is termed an alkali while Ferric hydroxide is not because:

  1. Sodium hydroxide is a strong base, while Ferric hydroxide is a weak base.
  2. Sodium hydroxide is a base which is soluble in water while Ferric hydroxide is also a base but it is not soluble in water.
  3. Sodium hydroxide is a strong base while Ferric hydroxide is a strong acid.
  4. Sodium hydroxide and Ferric hydroxide both are strong base but the solubility of Sodium hydroxide in water is comparatively higher than that of Ferric hydroxide.

Answer: 2. Sodium hydroxide is a strong base, while Ferric hydroxide is a weak base.

Question 6. The name of the salt used to remove permanent hardness of water is:

  1. Sodium hydrogen carbonate (NaHC03)
  2. Sodium chloride (NaCI)
  3. Sodium carbonate decahydrate (Na2C03.10H20)
  4. Calcium sulphate hemihydrate (CaS04. 12H2O)

Answer: 3. Sodium carbonate decahydrate (Na2C03.10H20)

Question 7. The electron dot structure of chlorine molecule is:

HBSE Class 10 Biology Solutions For The Electron Dot Structure Of Chlorine Molecule

Answer: 3.

Question 8. Observe the following diagram and identify the process and its significance from the following
options:

HBSE Class 10 Biology Solutions For Transpiration

  1. Evaporation: maintains water contents in leaf cells.
  2. Transpiration: creates a suction force which pulls water inside the plant.
  3. Excretion: helps in excreting out waste water from the plant.
  4. Translocation: helps in transporting materials from one cell to another.

Answer: 2. Transpiration: creates a suction force which pulls water inside the plant.

Question 9. Opening and closing of stomata is due to:

  1. High pressure of gases inside the cells.
  2. Movement of water in and out of the guard cells.
  3. Stimulus of light in the guard cells.
  4. Diffusion of C02 in and out of the guard cells.

Answer: 2. Movement of water in and out of the guard cells.

Question 10. A cross between pea plant with white flowers (vv) and pea plant with violet flowers (W) resulted in F2 progeny in which ratio of violet (VV) and white (w) flowers will be:

  1. 1 : 1
  2. 2 : 1
  3. 3 : 1
  4. 1 : 3

Answer: 1. 1 : 1

Question 11. In plants the role of cytokinin is:

  1. Promote cell division
  2. Wilting of leaves
  3. Promote the opening of stomatal pore
  4. Help in the growth of stem

Answer: 1. Promote cell division

Question 12. The number of chromosomes in parents and offsprings of a particular species undergoing sexual reproduction remain constant due to:

  1. Doubling of chromosomes after zygote formation.
  2. Halving of chromosomes after zygote formation.
  3. Doubling of chromosomes before gamete formation.
  4. Halving of chromosomes at the time of gamete formation.

Answer: 4. Halving of chromosomes at the time of gamete formation.

Question 13. Two LED bulbs of 12W and 6W are connected in series, if the current through 12W bulb is 0.06A the current through 6W bulb will be:

  1. 0.04A
  2. 0.06A
  3. 0.08A
  4. 0.12A

Answer: 2. 0.06A

Question 14. The correct pattern of magnetic field lines of the field produced by a current carrying circular
loop is:

HBSE Class 10 Biology Solutions For Pattern Of Magnetic Field Lines Of The Field

Answer: 3.

Question 15. The resistance of a resistor is reduced to half of its initial value. If other parameters of the electrical cirenit remain unaltered, the amount of heat produced in the resistor will become:

  1. Four times
  2. Two times
  3. Half
  4. One fourth

Answer: 2. Two times

Question 16. An alpha particle enters a uniform magnetic field as shown. The direction of force experienced by the alpha particle is:

HBSE Class 10 Biology Solutions For Alpha Particle Enters A Uniform Magnetic Field

  1. Towards right
  2. Towards left
  3. Into the page
  4. Out of the page

Answer: 4. Out of the page

Assertion – Reasoning based questions. These consist of two statements Assertion (A) and Reason (R). Answer these questions selecting the appropriate option given below.

  1. Both (A) and (R) are true and (R) is the correct explanation of (A).
  2. Both (A) and (R) are true but (R) is not the correct explanation of (A).
  3. (A) is true but (R) is false.
  4. (A) is false but (R) is true.

Question 1. Assertion (A): Reaction of quicklime with water is an exothermic reaction.

Reason (R): Quicklime reacts vigorously with water releasing a large amount of heat.

Answer: 1. Both (A) and (R) are true and (R) is the correct explanation of (A).

Question 2. Assertion (A): In humans, if gene (B) is responsible for black eyes and gene (b) is responsible for brown eyes, then the colour of eyes of the progeny having gene combination Bb, bb or BB will be black only.

Reason (R): The black colour of the eyes is a dominant trait.

Answer: 4. (A) is false but (R) is true.

Question 3. Assertion (A): The inner walls of the small intestine have finger like projections called villi which are rich in blood.

Reason (R): These villi have a large surface area to help the small intestine in completing the digestion of food.

Answer: 1. Both (A) and (R) are true and (R) is the correct explanation of (A).

Question 4. Assertion (A): A current carrying straight conductor experiences a force when placed perpendicular to the direction of magnetic field.

Reason (R): The net charge on a current carrying conductor is always zero.

Answer: 2. Both (A) and (R) are true but (R) is not the correct explanation of (A)

Very Short Answer Questions

Question 1.

1. A student took a small amount of copper oxide in a conical flask and added dilute hydrochloric acid to it with constant stirring. He observed a change in colour of the solution.

    1. Write the name of the compound formed and its colour.
    2. Write a balanced chemical equation for the reaction involved.

OR

2. The industrial process used for the manufacture of caustic soda involves electrolysis of an aqueous solution of compound ‘X’, In this process, two gases Y and ‘Z’ are liberated. Y is liberated at cathode and which is liberated at anode, on treatment with dry slaked lime forms a compound ‘B’. Name X, Y, Z and B.

Answer:

The compound formed is Copper Chloride or cupric chloride (CuCI2). The colour of CuCI2 is blue-green.

The balanced chemical equation is: CuO(s) + 2HCI(aq) CuCI2(aq) + H20(l)
OR
Compound X is NaCI(aq), which is a concentrated aqueous solution of sodium chloride in water.

Compound Y is H2 (Hydrogen gas), liberated at cathode.

Compound Z is Cl2, (Chlorine gas), liberated at anode.

Treatment of gas Z (Cl2) with dry slaked lime, Ca(OH)2, gives bleaching powder (CaOCI2).

\(\mathrm{Ca}(\mathrm{OH})_2+\mathrm{Cl}_2 \rightarrow \mathrm{CaOCl}_2+\mathrm{H}_2 \mathrm{O}\)

Hence, Compound B is CaOCI2

The overall balanced chemical equation is: \(2 \mathrm{NaCl}(\mathrm{aq})+2 \mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightarrow 2 \mathrm{NaOH}(\mathrm{aq})+\mathrm{Cl}_2(\mathrm{~g})+\mathrm{H}_2(\mathrm{~g})\)

Question 2.

1. Name the part of brain which is responsible for the following actions:

  1. Maintaining posture and balance
  2. Beating of heart
  3. Thinking
  4. Blood pressure

OR

2. Where are auxins synthesized in a plant? Which organ of the plant shows:

  1. Positive phototropism
  2. Negative geotropism
  3. Positive hydrotropism

Answer:

  1. Cerebellum
  2. Medulla
  3. Cerebrum
  4. Medulla

OR
Auxins are synthesized in the growing regions at the tip of the shoot and root in a plant.

  1. Positive phototropism -Shoot
  2. Negative geotropism- Shoot
  3. Positive hydrotropism- Root

Question 3. Write one specific function each of the following organs in relation with excretion in human beings:

  1. Renal Artery
  2. Urethra
  3. Glomerulus
  4. Tubular part of nephron

Answer:

  1. Renal artery carries nitrogenous excretory wastes towards the kidneys.
  2. Urethra releases urine out of the body.
  3. Glomerulus helps infiltration of blood passing through it and passes it on to the Bowman’s capsule.
  4. The tubular part of nephron reabsorbs useful products like glucose, amino acids, ions and a large amount of water into blood capillaries.

Question 4. Two green plants are kept separately in oxygen free containers, one in the dark and other in sunlight. It was observed that plant kept in dark could not survive longer. Give reason for this observation.

Answer:

Plants release oxygen during photosynthesis, which can be utilized by the plant for respiration. The plant kept in dark was not able to perform photosynthesis to generate some oxygen for respiration.

The carbon dioxide released by the plant after respiration was utilized by the plants to photosynthesize food, hence, the plant was able to survive.

Question 5.

Observe the following diagram and answer the questions following it:

HBSE Class 10 Biology Solutions For Myopia Before The Retina For Far Away Objects

  1. Identify the defect of vision shown,
  2. List its two causes,
  3. Name the type of lens used for the correction of this defect.

OR

The colour of clear sky from the earth appears blue but from the space it appears black. Why?

Answer:

  1. Myopia, as image is formed before the retina for far away objects.
  2. Possible causes are:
    1. Elongation of eyeballs.
    2. Excessive curvature of eye lens.
  3. Correction: Using concave lens of suitable focal length.

OR

2. The sky appears blue because of the presence of atmosphere, which scatters maximum blue light. Since, Earth has atmosphere, so particles present in it, make scattering of light possible for blue colour.

Space has no atmosphere so there is no intervening particle to obstruct light so as to scatter light. Hence, it appear dark from space.

Question 6. Use of several pesticides which results in excessive accumulation of pesticides in rivers or ponds, is a matter of deep concern. Justify this statement.

Answer:

Pesticides are toxic, non-biodegradable substances that often run off into water bodies. Once, the toxin enters the body of living organisms, it is not metabolized or excreted, leading to its accumulation inside the body.

The accumulation increases the concentration of toxin with each successive trophic level of the food chain, leading to a phenomenon called biological magnification. The topmost trophic level accumulates the maximum amount of toxin, leading to severe health issues and sometimes the death of the individual.

Short Answer Questions

Question 1.

While electrolysing water before passing the current some drops of an acid are added.

  1. Why ? Name the gases liberated at cathode and anode. Write the relationship between the volume of gas collected at anode and the volume of gas collected at cathode.
  2. What is observed when silver chloride is exposed to sunlight ? Give the type of reaction involved.

Answer:

1. Water is a non-electrolyte so, the addition of acid will make it a good conductor of electricity. By this action, ions will be produced due to dissociation of acidified water that acts as a charge carriers for the faster conduction of electricity.

The gas liberated at the cathode is Hydrogen gas. The gas liberated at the anode is Oxygen gas. In the electrolysis of water, the ratio of volumes of hydrogen and oxygen obtained is 2:1 i.e. 2 volumes of hydrogen and 1 volume of oxygen combines to form water.

2. When silver chloride is exposed to light, it decomposes to form silver metal and chlorine gas. The type of reaction is photolytic decomposition.

Photolytic Decomposition

Question 2.

  1. Suggest a safe procedure of diluting a strong concentrated acid.
  2. Name the salt formed when sulphuric acid is added to sodium hydroxide and write its pH.
  3. Dry HCI gas does not change the colour of dry blue litmus paper. Why?

Answer:

The acid is always diluted by adding the acid to water slowly and with constant stirring. In addition to it, it is advised to wear gloves and goggles and maintain safe distance.

The salt formed when sulphuric acid is added to sodium hydroxide is sodium sulphate. The balanced equation is:

\(2 \mathrm{NaOH}+\mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{Na}_2 \mathrm{SO}_4+2 \mathrm{H}_2 \mathrm{O}\)

Its pH is approximately equal to 7.

The colour of litmus paper changes only in the presence of ions like hydrogen (H+) or hydronium (H30+) ions. HCI can produce these ions only in the form of aqueous solution. Hence, dry HCI gas does not change the colour of dry litmus paper.

Question 3.

1. (1) How does Paramecium obtain its food ?

(2) List the role of each of the following in our digestive system:

  • Hydrochloric acid
  • Trypsin
  • Muscular walls of stomach
  • Salivary amylase

OR

2. (1) What is double circulation ?

(2) Why is the separation of the right side and the left side of the heart useful? How does it help birds and mammals?

Answer:

1. (1) Paramecium takes its food at a specific spot by endocytosis. Food is moved to this spot by the movement of cilia which cover the entire surface of the cell.

(2) Hydrochloric acid: The hydrochloric acid creates an acidic medium which facilitates the action of enzyme pepsin and it also kills the germs in the food.

  • Trypsin: Trypsin helps in digestion of proteins into smaller peptides.
  • Muscular wall of stomach: The muscular wall of stomach help in mixing the food thoroughly with more digestive juices.
  • Salivary amylase: Saliva contains an enzyme called salivary enzyme which breaks down starch to give simple sugar.

or

2. (1) Double circulation is the process during which blood passes twice through the heart during one complete cycle. Blood is circulated to the body tissues through systemic circulation and to the lungs through pulmonary circulation.

(2) The separation of right side and left side of the heart helps in separation of oxygenated and deoxygenated blood, which allows a more efficient supply of oxygen to the body cells. 1 Birds and mammals are warm blooded animals. Since, they require more energy to maintain a constant body temperature; hence, the separation provides availability of oxygen during respiration to generate more energy for thermoregulation.

Question 4.

1. Define the following terms in the context of a diverging mirror:

  1. Principal focus
  2. Focal length Draw a labelled ray diagram to illustrate your answer.

OR

2. An object of height 10 cm is placed 25 cm away from the optical centre of a converging lens of focal length 15 cm. Calculate the image-distance and height of the image formed.

Answer:

  1. Principal focus: It is the point on the axis of a mirror to which parallel rays of light appear to diverge after reflection from diverging mirror.
  2. Focal length: The distance between pole of convex mirror and principal focus.

HBSE Class 10 Biology Solutions For The Distance Between Pole Of Convex Mirror And Principal Focus

2. Given, Object’s height h0 = 10 cm; Object distance, u = -25 cm; Focal length of convex lens; f = 15 cm

Image distance v = ? Height of image hi = ?

Using the lens formula:

\(\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\)

or, \(\frac{1}{v}=\frac{1}{f}+\frac{1}{u}=\frac{1}{15}+\frac{1}{-25}\)

or, \(v=\frac{75}{2}=37.5 \mathrm{~cm}\)

From the magnification formula

\(\frac{h_i}{h_{\mathrm{o}}}=\frac{\mathrm{v}}{\mathrm{u}}\)

or, \(h_i=\frac{v}{u} \times h_o=\frac{37.5 \times 10}{-25.0}\)

or, hi = —15 cm

An inverted image of 15 cm will be formed at a distance of 37.5 cm from the given lens.

Question 5. The power of a lens is +4D. Find the focal length of this lens. An object is placed at a distance of 50 cm from the optical centre of this lens. State the nature and magnification of the image formed by the lens and also draw a ray diagram to justify your answer.

Answer:

Power of lens, P = +4D

Since \(\)

Or, f = 0.25 m = 25 cm

Since power is positive, so the lens is convex lens.

Now using lens formula,

\(\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\)

or, \(\frac{1}{v}=\frac{1}{f}+\frac{1}{u}=\frac{1}{25}+\frac{1}{-50}\)

or, v = 50 cm

Magnification, \(m=\frac{v}{u}=\frac{50}{-50}=-1\)

Since, image distance is positive and its magnification is -1, it indicates a real, inverted and image of same size.

Question 6.

Why is an alternating current (A.C.) considered to be advantageous over direct current (D.C.) for the long distance transmission of electric power?

How is the type of current used in household supply different from the one given by a battery of dry cells?

How does an electric fuse prevent the electric circuit and the appliances from a possible damage due to short circuiting or overloading.

OR

For the current carrying solenoid as shown, draw magnetic field lines and give reason to explain that out of the three points A, B and C, at which point the field strength is maximum and at which point it is minimum?

HBSE Class 10 Biology Solutions For The Current Carrying Solenoid

Answer:

  1. Alternating current can be transmitted over long distances at very high voltages so as to ensure that current value is very low and as a result very low amount of energy is lost during transmission. When it reaches near the household, the voltage can be stepped down to restore required value of current whereas the voltage of direct current cannot be stepped up and down.
  2. The type of current used in domestic household supply is alternating while a battery of dry cells supplies direct current.
  3. Electric fuse is made of wires of very low melting point. If there is a current larger than the specified value, the temperature of fuse wire increases and it melts to break the electric circuit and stop the flow of unduly high electric current. This saves appliances from possible damage.

or

  1. The field strength will be maximum at point A as A is inside the solenoid where the field lines have highest density.
  2. The field strength will be minimum at point B as field lines have least density.

HBSE Class 10 Biology Solutions For The Field Strength Will Be Minimum At point B as Field Lines have Least Density

Question 7. Write one difference between biodegradable and non-biodegradable wastes. List two impacts of each type of the accumulated waste on environment if not disposed off properly.

Answer:

HBSE Class 10 Biology Solutions For Difference Between Diodegradable And Non-biodegradable Wastes

Effects of biodegradable substances on environment:

  1. Decomposition of biodegradable wastes is accompanied by foul smell which spreads in the environment and affects the people of nearby areas.
  2. It act as breeding grounds for houseflies, etc. which act as vectors of various diseases.

Effects of non-biodegradable substances on environment:

  1. Excessive use of non-biodegradable pesticides increases the soil pollution and also affects the soil fertility.
  2. Certain non-biodegradable wastes enter the food chains and affects the various biotic components of the environment.

Long Answer Questions And Answers

Question 1.

1. Draw the structure of the following components?

  • Butanoic acid
  • Chloropentane

2. How are structure

  • And structure
  • Given below related to one another ? Give reason to justify your answer.

image

Draw one more possible structure for above case.

3. Differentiate between saturated and unsaturated compounds on the basis of their general formula.

OR

2. What happens when a small piece of sodium is dropped on ethanol? Write the equation for this reaction.

  • Why is glacial acetic acid called so?
  • What happens when ethanol is heated at 443 K in the presence of cone. H2S04? Write the role of cone. H2SO4 in this case.
  • Write an equation showing saponification.

Answer:

HBSE Class 10 Biology Solutions For Strucutre Of Butanoic Acid And Chloropentane

2. Structures (1) and (2) are structural isomers of each other as they have different structures but same molecular formula (C6H14). Another structural isomer is n-hexane as shown below:

Structure Isomer Is n-hexane

3. Difference between saturated and unsaturated compounds are:

HBSE Class 10 Biology Solutions For Difference Between Saturated And Unsaturated Compounds

or

2. Ethanol reacts with sodium to produce hydrogen gas and sodium ethoxide.

2CH3CH2OH + 2Na -> 2CH3CH20-Na++ H2
Ethanol                        Sodium Ethoxide

  • The freezing point of pure ethanoic acid is 290 K and hence, it freezes during winter in cold climates. That is why; it is called as glacial acetic acid.
  • When ethanol is heated with concentrated sulphuric acid at 443 K, it loses one water molecule and forms ethene.

Ethanol Is Heated With Concentrated Sulphuric Acid

Role of H7SO4: It acts as dehydrating agent and removes water molecules from ethanol.

  • Equation for saponification reaction:

HBSE Class 10 Biology Solutions For Dehydrating Agents And Remove Water Moleules From Ethanol

Question 2.

  1. Name and explain the two modes of asexual reproduction observed in hydra.
  2. What is vegetative propagation? List two advantages of using this technique.

Answer:

1. The two modes of asexual reproduction observed in Hydra are : Budding and Regeneration.

Budding in Hydra: In Budding, a small part of the body of the parents grows out as a ‘bud’ which then detaches and becomes a new organism. Hydra reproduces by budding using the regenerative cells.

A bud develops as an outgrowth in Hydra due to repeated cell division at one specific site. When fully mature, the bud detaches itself from the parent body and develops into new independent individuals.

HBSE Class 10 Biology Solutions For Budding In Hydra

Regeneration in Hydra: In this method, small cut or broken parts of the organisms’ body grow or regenerate into separate individuals. Regeneration of Hydra from the body parts is carried out by specialized cells, which proliferate and make a large number of cells.

HBSE Class 10 Biology Solutions For Regeneration In Hydra

2. Vegetative propagation is the development of a new plant from the vegetative parts (roots, stem and leaves) of a plant.

Advantages:

  1. Such plants can bear flowers and fruits earlier than those produced from seeds.
  2. Allows propagation of plants (banana, orange etc) that have lost capacity to produce seeds.
  3. All plants produced are genetically similar to the parent plant and hence have all its characters.

Question 3.

  1. How is electric current related to the potential difference across the terminals of a conductor? Draw a labelled circuit diagram to verify this relationship.
  2. Why should an ammeter have low resistance?
  3. Two V – 1 graphs A and B for series and parallel combinations of two resistors are as shown. Giving reason state which graph shows (a) series, (b) parallel combination of the resistors.

HBSE Class 10 Biology Solutions For Series And Parallell Combinanations Of Two Resistors

Answer:

1. The ratio of potential difference to the flow of current in conductor is constant which is called the resistance of the conductor provided the physical conditions of the conductor remains constant. This is called Ohm’s Law. Mathematically, V/l = R (constant)

Circuit Diagram:

HBSE Class 10 Biology Solutions For Circuit

2. An ammeter is a device used to measure the amount of current flowing in a circuit. The resistance of an ideal ammeter should be zero as it is always connected in series in a circuit.

Hence, its resistance adds to the total resistance of the circuit. If the resistance of the ammeter would be high, then the total resistance would be too high. This in turn would decrease the amount of current flowing through the circuit.

3. The slope of l-V graph gives the reciprocal of resistance. So, the graph with a lesser slope will indicate a larger resistance.

The series combination of resistances has a large net resistance than the corresponding parallel combination. So, Graph A indicates series combination and graph B indicates parallel combination.

Source Based Or Case Based Questions And Answers

Question 1. The melting points and boiling points of some ionic compounds are given below:

HBSE Class 10 Biology Solutions For The Melting Points Of Some Iconic Compounds

These compounds are termed ionic because they are formed by the transfer of electrons from a metal to a non-metal. The electron transfer in such compounds is controlled by the electronic configuration of the elements involved. Every element tends to attain a completely filled valence shell of its nearest noble gas or a stable octet.

  1. Show the electron transfer in the formation of magnesium chloride.
  2. List two properties of ionic compounds other than their high melting and boiling points.
  3. While forming an ionic compound say sodium chloride how does sodium atom attain its stable configuration ?

OR

2. Give reasons:

  1. Why do ionic compounds in the solid state not conduct electricity?
  2. What happens at the cathode when electricity is passed through an aqueous solution of sodium chloride ?

Answer:

1. Electron transfer in the formation of magnesium chloride:

HBSE Class 10 Biology Solutions For Electron Transfer In The Formation Of Magnesium Chloride

2. Two properties of ionic compounds are:

  1. They conduct electricity in solution or molten state.
  2. They are hard and brittle crystalline solids. ‘A + ‘A

3. Sodium chloride is formed by the combination of sodium and chloride ions. Sodium loses one electron from valence shell to attain stable noble gas configuration.

HBSE Class 10 Biology Solutions For Sodium Chloride Is Formed By The Combination Of Sodium And Chloride Ions

or

  1. Ionic compounds do not conduct electricity in solid state due to the absence of free ions but they conduct electricity in molten and aqueous state due to presence of free ions.
  2. Reduction takes place at cathode. Therefore, during electrolysis of aqueous solution of sodium chloride, hydrogen gas is evolved at cathode due to reduction of H+ ions.
\(2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow \mathrm{H}_2 \uparrow\)

Question 2. The most obvious outcome of the reproductive process is the generation of individuals of similar design, but in sexual reproduction they may not be exactly alike. The resemblances as well as differences are marked. The rules of heredity determine the process by which traits and characteristics are reliably inherited. Many experiments have been done to study the rules of inheritance.

  1. Why an offspring of human being is not a true copy of his parents in sexual reproduction?
  2. While performing experiments on inheritance in plants, what is the difference between FI and F2 generation?
  3. Why do we say that variations are useful for the survival of a species over time ?

OR

Study Mendel’s cross between two plants with a pair of contrasting characters.

RRYY            X                rrYY

Round Yellow   Wrinkiled Green

He observed 4 types of combinations in F2 generation. Which of these were new combinations? Why do new features which are not present in the parents, appear in F2 generation?

Answer:

  1. An offspring is not a true copy of his parents in sexual reproduction because of variations due to recombination of genetic material from two different parents.
  2. All the FI offspring are genetically identical and exhibit the dominant trait. F2 offspring can exhibit a range of dominant to recessive traits.
  3. Variations help the organism to become more adapted to the changing environmental conditions. This helps the organisms to overcome the adverse conditions.

or

Round Green and Wrinkled Yellow are the new combinations in F2 generation. New features which are not present in parents appear in F2 generation due to the segregation of different combinations of alleles of different characters independently during gamete formation and fertilisation.

Question 3. The ability of a medium to refract light is expressed in terms of its optical density. Optical density has a definite connotation. It is not the same as mass density. On comparing two media, the one with the large refractive index is optically denser medium than the other. The other medium with a lower refractive index is optically rarer. Also the speed of light through a given medium is inversely proportional to its optical density.

Determine the speed of light in diamond if the refractive index of diamond with respect to vacuum is 2.42. Speed of light in vacuum is 3 x 108 m/s.

Refractive indices of glass, water and carbon disulphide are 1.5, 1.33 and 1.62 respectively. If a ray of light is incident in these media at the same angle (say q), then write the increasing order of the angle of refraction in these media.

The speed of light in glass is 2 x 108 m/s and in water is 2.25 x 108 m/s.

Which one of the two is optically denser and why ?

A ray of light is incident normally at the water-glass interface when it enters a thick glass container filled with water. What will happen to the path of the ray after entering the glass? Give reason.

OR

The absolute refractive indices of water and glass are 4/3 and 3/2 respectively. If the speed of light in glass is 2 x 108 m/s, find the speed of light in (i) vacuum and (ii) water.

Answer:

1. \(\mu=\frac{\text { Speed of light in vacuum (c) }}{\text { Speed of light in any medium (v) }}\)

Given, p = 2.42, c = 3 x 108 m/s

Therefore, speed of light in medium (diamond) = \(\frac{3 \times 10^8}{2.42}=1.24 \times 10^8 \mathrm{~m} / \mathrm{s}\)

2. Carbon disulphide is most dense and water is least dense.

From Snell’s Law, \(\mu=\frac{\sin i}{\sin r}\)

So, \(\frac{\sin i}{1.33}>\frac{\sin i}{1.5}>\frac{\sin i}{1.62}\)

Hence, for angle of refraction, increasing order will be rcs > rg > rw

3. Speed of light is more in rarer medium than denser medium. So, glass is denser.

When light is incident normally at waterglass interface, it goes straight through the medium with no bending. It is because, by bending light prefers to go shortest path. In case when normally incident, it is already along shortest path.
Or
3. \(\mu=\frac{\text { Speed of light in vacuum (c) }}{\text { Speed of light in any medium (v) }}\)

Given, \(\mu_w=\frac{4}{3} ; \mu_{\mathrm{g}}=\frac{3}{2}\)

Speed of light in glass, vg= 2 x 108 m/s

\(\mu_{\mathrm{g}}=\frac{\mathrm{c}}{v_g}\)

or, \(\mathrm{c}=\frac{3}{2} \times 2 \times 10^8=3 \times 10^8 \mathrm{~m} / \mathrm{s}\)

\(\mu_w=\frac{c}{v_w}\)

or, \(\frac{c}{v_w}=\frac{3 \times 10^8}{4 / 3}=2.25 \times 10^8 \mathrm{~m} / \mathrm{s}\).

Haryana Board Class 10 Biology Solutions For Chapter 2 How Do Organisms Reproduce?

Haryana Board Class 10 Biology Solutions For Chapter 2 How Do Organisms Reproduce? Multiple Choice Questions And Answers

 

Question 1. The image shows the model of a family of dogs.

HBSE Solutions Class 10 Biology Chapter 2 The Model Of A family Of Dogs

It can be observed that the offspring is similar to the parent but not identical. What is the likely reason for this?

  1. Variation in the genetic material
  2. Fast multiplication of body cells
  3. Asexual mode of reproduction
  4. Effect of environment on the offspring.

Answer: 1. Variation in the genetic material

Question 2. A population of thermophilic archaebacteria are generally found in hot springs. Any change to the temperature of the water affects the survival of the archaeabacteria. If the temperature of hot springs gets reduced, change in which component can allow survival of few members of these archaeabacteria?

  1. Cell wall
  2. Cytoplasm
  3. DNA
  4. Ribosomes

Answer: 3. DNA

Question 3. The image shows the process of division in plasmodium.

HBSE Solutions Class 10 Biology Chapter 2 The Process Of Division In Plasmodium

What can be concluded about the division in plasmodium?

  1. The cyst divides repeatedly to form many daughter cells.
  2. The cell divides multiple times giving rise to many daughter cells.
  3. The nucleus divides repeatedly inside the cell to form new daughter cells.
  4. The cyst enlarges in size and then bursts producing many new daughter cells.

Answer: 2. The cell divides multiple times giving rise to many daughter cells.

Question 4. The image shows the process of binary fission in amoeba.

HBSE Solutions Class 10 Biology Chapter 2 The Process Of Binary Fission In Amoeba

Which option correctly predicts about the daughter cells?

  1. The parent cell will lead to the formation of four daughter cells of equal sizes.
  2. The parent cell will lead to the formation of two daughter cells of equal sizes.
  3. The parent cell will lead to the formation of four daughter cells of different sizes.
  4. The parent cell will lead to the formation of two daughter cells of different sizes.

Answer: 2. The Parent cell will lead to the formation of two daughter cells of equal sizes.

Question 5. The image shows the formation of spores in Rhizopus.

HBSE Solutions Class 10 Biology Chapter 2 The Formation Of Spores In Rhizopus

How spores develop into Rhizopus?

  1. Spores divide and grow into new individual
  2. Spores combine with other spores and grow
  3. Spores enlarge in size for the growth of new individual
  4. Spores land on other organisms and increase with their growth in size

Answer: 1. Spores divide and grow into new individual

Question 6. The image shows the division in Spirogyra

HBSE Solutions Class 10 Biology Chapter 2 The Division In Spirogyra

What can be concluded about the Spirogyra from this division?

  1. It is a multicellular organism gives rise to two new equal sized individuals.
  2. It is a unicellular organism that gives rise to two new equal sized individuals.
  3. It is a unicellular organism that breaks into pieces that grows into new individuals.
  4. It is a multicellular organism that breaks into pieces that grows into new individuals.

Answer: 4. It is a multicellular organism that breaks into pieces that grows into new individuals.

Question 7. A student takes a planaria in the lab and cuts into three parts as shown.

HBSE Solutions Class 10 Biology Chapter 2 A Planaria In The Lab And Cuts Into Three Parts

What will likely happen?

  1. The cells around the cut start to divide to form lost part
  2. The cells around the cut enlarge to take the shape of lost part
  3. The cells around the cut start to divide to form a complete organism
  4. The cells around the cut attracts other planarians to fuse with the separated part.

Answer: 3. The cells around the cut start to divide to form a complete organism

Question 8. A student observes the process of regeneration in Planaria.

HBSE Solutions Class 10 Biology Chapter 2 The Process Of Regeneration In Planaria

The student claimed that the newly formed planarians have identical genome. Which statement support the claim?

  1. It is a single celled organism
  2. All planarians share the same genome.
  3. Division in Planaria involves a single parent.
  4. Planaria divides only under unfavourable condition

Answer: 3. Division in Planaria involves a single parent.

Question 9. The image shows a bud developing on a Hydra.

HBSE Solutions Class 10 Biology Chapter 2 HBSE Solutions Class 10 Biology Chapter 2 A Bud Developing On A Hydra

How does the bud develop in the Hydra?

  1. Bud develops due to separation of body parts of hydra
  2. Bud develops due to repetitive cell division at a specific site
  3. Bud develops due to change in the environmental conditions
  4. Develops due to attachment of another hydra at a specific site

Answer: 2. Bud develops due to repetitive cell division at a specific site

Question 10. The model shows the process of budding in Hydra.

HBSE Solutions Class 10 Biology Chapter 2 The Process Of Budding In Hydra

What is the likely purpose of this division in Hydra?

  1. To increase the body size
  2. To recover lost body parts
  3. To induce variation in body
  4. To develop new independent individual

Answer: 4. To develop new independent individual

Question 11. The image shows the process of vegetative propagation in a plant

HBSE Solutions Class 10 Biology Chapter 2 The Process Of Vegetative Propagation In A Plant

The shoot of the parent plant is pushed below the soil that results in growth of a new plant. What is the advantage of this process?

  1. This results in plant of different flowers
  2. This helps grow plants without adding extra manure
  3. This eliminates the need of producing plant using seeds
  4. This allows growth of plants with new genetic composition

Answer: 3. This eliminates the need of producing plant using seeds

Question 12. The image shows the production of a new sugarcane from an existing sugarcane plant.

HBSE Solutions Class 10 Biology Chapter 2 The Production Of A New Sugarcane From An Existing Sugarcane Plant

The method is called vegetative propagation. Which option supports the name of this process?

  1. It is a sexual method of producing new plants.
  2. It is an asexual method of producing new plants.
  3. It does not require a parent plant for reproduction.
  4. It involves fusion two parts of a single parent for reproduction.

Answer: 2. It is an asexual method of producing new plants.

Question 13. The image shows the different parts of a flower.

HBSE Solutions Class 10 Biology Chapter 2 The Different Parts Of A Flower

Which part of the pistil is responsible for receiving pollen from stamen in order to perform reproduction?

  1. Anther
  2. Ovary
  3. Petal
  4. Stigma

Answer: 4. Stigma

Question 14. The image shows the structure of a flower.

HBSE Solutions Class 10 Biology Chapter 2 The Structure Of A Flower

Which process will likely be disturbed or not occur, if labelled part is removed from the flower?

  1. Formation of fruit
  2. Transport of pollen
  3. Formation of pollen
  4. Development of pollen tube

Answer: 1. Formation of fruit

Question 15. Which option correctly lists the changes that occur in males during puberty?

HBSE Solutions Class 10 Biology Chapter 2 Lists The Changes That Occur In Males During Puberty

Answer: 1. HBSE Solutions Class 10 Biology Chapter 2 During Puberty

Question 16. The table shows the changes that occur in girls during puberty.

HBSE Solutions Class 10 Biology Chapter 2 The Changes That Occur In Girls During Puberty

What is the likely significance of these changes?

  1. Aging of the body
  2. Sexual maturation
  3. Production of germ cells
  4. Abnormal division of the cells
  5. Answer: 2. Sexual maturation

Question 17. The image shows the male reproductive system.

HBSE Solutions Class 10 Biology Chapter 2 The Male Reproductive System

Which option correctly shows the path that the sperms take when they are released from the body?

  1. Testis → ureter → urethra → penis
  2. Testis → vas deferens → ureter → penis
  3. Testis → ureter → vas deferens → penis
  4. Testis → vas deferens → urethra -→ penis

Answer: 4. Testis → vas deferens → urethra → penis

Question 18. The image shows the male reproductive system outside abdominal cavity.

HBSE Solutions Class 10 Biology Chapter 2 The Male Reproductive System Outside Abdominal Cavity

What will likely happen if testes are located inside the abdominal cavity?

  1. Delayed puberty
  2. Sperm formation
  3. Increase in body temperature
  4. Change in genetic composition of sperms

Answer: 1. Delayed puberty

Question 19. The image shows the reproductive organ in females

HBSE Solutions Class 10 Biology Chapter 2 The Reproductive Organ In Females

Which event will likely occur in the ovaries of females after attaining puberty?

  1. Fertilisation
  2. Synthesis of eggs
  3. Production of eggs
  4. Growth and development of embryo

Answer: 3. Production of eggs

Question 20. The image shows the female reproductive system.

HBSE Solutions Class 10 Biology Chapter 2 The Female Reproductive System

Which event will be likely affected, if a female’s uterus is implanted with intrauterine device?

  1. Release of eggs
  2. Entering of sperms
  3. Maturation of eggs
  4. Implantation of embryo

Answer: 4. Implantation of embryo

Question 21. The table lists some changes that occur inside the female body after fertilization of egg with sperm.

HBSE Solutions Class 10 Biology Chapter 2 The Female Body After Fertilization Of Egg With Sperm

Which option correctly sequences these events?

  1. C→B E  A→D
  2. E→C→D→B→A
  3. E→C→B →D→A
  4. C→E→A→B→D

Answer: 3. E→C→B→D→A

Question 22. After a female attains puberty, the body undergoes many changes for maturation. Which event will likely happen in the female body when there is no fertilization of egg?

  1. Disintegration of uterine wall
  2. Development of egg into zygote
  3. Increase in the production of eggs in the ovaries
  4. Of a mature egg into an immature egg

Answer: 1. Disintegration of uterine wall

Question 23. Which contraceptive can be used to prevent the entry of sperm inside the female reproductive organ?

  1. Inserting copper-t inside the uterus
  2. Wearing condoms on the penis
  3. Consuming oral pills containing hormones
  4. Undergoing surgery for blocking fallopian tube

Answer: 2. Wearing condoms on the penis

Question 24. The image shows a surgical method in females to prevent pregnancy.

HBSE Solutions Class 10 Biology Chapter 2 A Surgical Method In Females To Prevent Pregnancy

Which event will be likely prevented from this method?

  1. Maturation of eggs
  2. Production of eggs
  3. Entry of eggs into the uterus
  4. Entry of sperm into the uterus

Answer: 3. Entry of eggs into the uterus

Chapter 2 How Do Organisms Reproduce Very Short Questions And Answers

Question 1. What is the main advantage of sexual reproduction over asexual reproduction?

Answer: Sexual reproduction provides diversity in the genetic makeup of the offspring. This is due to the combination of genes from two different individuals.

Question 2. Why is DNA copying an essential part of the process of reproduction?

Answer: DNA copying is essential for inheritance of features from parents to the next generation.

Question 3. How does the embryo get nourishment inside the mother’s body?

Answer: The embryo gets nourishment through the placenta.

Question 4. Why does bread mould grow profusely on a moist slice of bread rather than on a dry slice of bread?

Answer: The spores of bread mould need favourable conditions like moist surface to germinate. Since, moistened bread slice offers both moisture and nutrients to the bread mould, hence it grows profusely. Dry slice of bread offers nutrients, but not moisture hence, hyphae fail to grow.

Question 5. Why cannot fertilisation take place in flowers if pollination does not occur?

Answer: In a flower fertilization requires both male and female gametes. If pollination does not occur, male gamete is not available hence fertilization cannot take place.

Question 6. What is the basic difference between male and female germ cell?

Answer: The male germ cell is motile while the female germ cell is non-motile.

Question 7. State the condition necessary for germination of pollen grain.

Answer: Pollen grain germinates only if it has fallen on the stigma of the same plant species, or else it decomposes.

Question 8. Which organ of female reproductive system is also known as birth canal?

Answer: Vagina

Question 9. What is the effect of DNA copying which is not perfectly accurate in the reproductive process?

Answer: It results in variations in populations

Question 10. Which artificial propagation method is used to propagate roses?

Answer: Grafting.

Question 11. Specify the first event which occurs in the reproductive system of a human female if egg is fertilised.

Answer: If egg is fertilized, it gets implanted in the uterine lining resulting in pregnancy.

Question 12. Specify the events which occur in the reproductive system of a human female if egg is not fertilised.

Answer: If egg is not fertilized, it is shed out of the body along with blood causing menstruation.

Question 13. In a tobacco plant, the male gametes have 24 chromosomes. What is the number of chromosomes in the female gamete? What is the number of chromosomes in the zygote?

Answer: Number of chromosomes in female gamete is 24. Number of chromosomes in zygote is 48.

Question 14. Neha took two bread slices and kept them in the following conditions:

  1. Slice 1 in a dried and dark place
  2. Slice 2 in moist and dark place

What would she observe in each of the above conditions?

Answer:

  1. No spores will be formed in Slice 1 due to lack of moisture,
  2. In Slice 2, fungal hyphae develop due to the presence of favourable condition, i.e, moisture.

Question 15. Kashyap noticed that an organism by mistake was cut into parts. After sometime, both parts developed into new individuals.

  1. Name the mode of reproduction used by the organism.
  2. State the type of cells which carry out this process.

Answer:

  1. Regeneration
  2. Regenerative cells.

Question 16. Name the site of implantation and development of baby in human female.

Answer: Uterus

Question 17. Mention any two benefits of using barrier method during sexual act.

Answer:

  1. Protection from sexually transmitted diseases,
  2. Prevention of pregnancy

Question 18. Trace the path of sperms from where they are produced in human body to the exterior.

Answer: Testes-Epididymis-Vas deferens-Ejaculatory duct-Urinogenital duct-Penis.

Question 19. Why is DNA copying an essential part of the process of reproduction?

Answer: DNA copying is essential for ensuring that exact blue print of body design of parents is inherited by the offsprings

Question 20. Why are asexually reproducing organisms capable of showing hereditary features?

Answer: The asexually reproducing organisms require only single parent, so when they divide the cell itself divide into two bearing no variation in DNA. So there is more chance of hereditary features,

In asexual reproduction, organisms raised are the exact copy of their parents, They exhibit very little variation due to some environmental factors or mutations which are due to sudden changes in genes out of these two factors, only mutations are inheritable.

Question 21. Offspring formed due to sexual reproduction have better chances of survival. Why? Is this statement always true?

Answer: Offspring formed due to sexual reproduction have better chances of survival because Sexual reproduction introduces genetic variation in the offspring which is essential for evolution and survival of the species; Offspring exhibit hybrid vigour which enables them to adapt better to the changing environment; Enhanced traits allow the offspring to be tolerant and survive under adverse conditions.

It is not always true that the offspring formed by sexual reproduction have better chances of survival because during sexual reproduction, there is crossing over and random selection due to which the offspring produced may have traits which are inferior to the parents.

Question 22. Protozoans reproduce by binary fission as well as multiple fission. Which process is better and why? Give your opinion.

Answer: Multiple fission is better than binary fission because it produces several daughter cells inside a protective structure called cyst. This ensures their survival under adverse conditions.

Question 23. List four ways of preventing pregnancy. State two advantages of using such preventive methods.

Answer: Ways of preventing pregnancy: Natural method: The sexual act is avoided from day 10 to day 17 of the menstrual cycle, i.e. during the period ovulation is expected. Barrier methods: The fertilisation of ovum and sperm is prevented with the help of physical devices (condom and diaphragm). Oral contraceptives: Tablets or drugs are taken orally. These contain small doses of hormones which prevent the release of eggs and prevent fertilisation. Intrauterine contraceptive devices: Contraceptive devices such as CopperT rods are placed in the uterus to prevent pregnancy. Two advantages of using preventive methods are:

  1. These helps in prevention of sexually transmitted diseases such as syphilis, AIDS etc.
  2. These helps in maintaining control over human population by avoiding the unwanted pregnancies.

Question 24. Basis the understanding of size of the organism and chromosome number, answer the following
questions:

  1. Do larger organisms have more number of chromosomes/cells?
  2. More the number of chromosomes/cells greater in the DNA content. Justify.

Answer:

  1. No, there is no relationship between size of organism and its chromosome number,
  2. Yes, since the major component of chromosome is DNA, if there are more chromosomes in a cell, the quantity of DNA will also be more.

Question 25. A married woman used a device made of common metal for preventing pregnancy. This device was put into her uterus by some trained medical professional. Unfortunately she got pregnant after two months of insertion of device. She was in shock to learn that her birth control device has failed.

  1. How does device X prevent pregnancy?
  2. Why do you think the woman got pregnant even after using device X?

Answer:

  1. Device X (Copper-T) suppresses fertilizing capacity of sperms and thus, prevents pregnancy,
  2. Intra-uterine devices can be expelled anytime without the knowledge of the woman. The couple continued active sexual life thinking that their birth control device is still in action.

Question 26. Compare the vegetative propagation in Bryophyllum and money plant.

Answer:  Bryophyllum reproduces by the adventitious buds present on the margins of its leaves. Money plant reproduces by stem cutting which can produce the whole plant when placed in moist soil.

Question 27. Sujata’s parents received a proposal for her marriage from a boy living in London. Before everything could get finalised, Sujata asked her parents to ask the boy to get his blood test report.

  1. Do you think it was right on the part of Sujata to do so?
  2. Name two STDs along with their causative organism.

Answer:

  1. Yes, it was right on Sujata’s part to do so because blood reports would indicate whether the person has any disease, any sexually transmitted disease or is free of them,
  2. Two sexually transmitted diseases are:
    1. AIDS caused by HIV
    2. Gonorrhoea caused by Neisseria gonorrhoea

Question 28. Explain how, offspring and parents of organisms reproducing sexually have the same number of chromosomes.

Answer:

The process of cell division called meiosis halves the number of chromosomes present in the cell of an organism. The gametes produced as a result of meiosis, fuse during sexual reproduction and restore the number of chromosomes in the offspring of the individual.

Question 29. What substances pass

  1. From the maternal to the foetal blood?
  2. From the foetal to the maternal blood?

Answer:

  1. Oxygen, nutrients (food and water) pass from maternal to foetal blood,
  2. Carbon dioxide and other waste substances like nitrogenous waste (urea) pass from foetal to maternal blood.

Question 30. The sperms are tiny bodies that consist of mainly genetic material and a long tail,

  1. Where are the sperms produced? What is the role of the long tail?
  2. How are the sperms delivered from the site of their production?

Answer:

  1. Sperms are produced in seminiferous tubules present in testers of male reproductive system by process known as spermatogenesis, ii. Sperm consists of 2 main parts head and tail, both of which are connected by neck. The long tail also known as flagellum of sperms helps in quick movement or motility of sperms through the female reproductive tract through wave like motion,
  2. Sperms are delivered from the site of their production by vas deferens, also known as sperm duct to the urethra in order to allow the passage of semen outside the body. Movement of sperm is facilitated by prostatic fluid.

Question 31. If a woman is using Copper-T, will it help in protecting her from sexually transmitted diseases? Why?

Answer:

If a woman is using a copper-T,it will not help in protecting her from sexually transmitted diseases. Copper-T prevents only implantation in the uterus.

Question 32. A potato is cut into a number of small pieces. These potato pieces are placed on wet cotton kept in a tray. After a few days, green shoots and roots appear only from some potatoes and not from all potato pieces. Why?

Answer:

The pieces of potato that bear nodes only can give rise to new plants by producing shoots and roots. But the pieces not bearing nodes cannot produce new plants. Thus, from a whole potato only some pieces that bear nodes give rise to roots and shoots. This is an example of vegetative propagation which is an asexual mode of reproduction in plants.

Question 33. It is a well-known fact that a pregnant woman’s health is the backbone of every family, society and thus nation.

  1. Which tissue is responsible for providing nutrition from the mother to the growing embryo?
  2. According to you, what can be the likely measures to maintain a woman’s health during pregnancy?

Answer:

  1. The placenta is responsible for providing nutrition from the mother to the growing embryo. Exchange of nutrients, oxygen and waste products between the embryo and the mother takes place through the placenta,
  2. Measures a woman can take to maintain health during pregnancy: Consume a balanced diet; Take food or medicinal supplements as required; Be careful about diet and hygiene; Exercise regularly; Stop bad habits like smoking and drinking alcohol; Take adequate rest.

Question 34.

  1. What is the role of the seminal vesicles and the prostate gland?
  2. If a woman is using a copper-T, will it help in protecting her from sexually transmitted disease?

Answer:

  1. The prostate gland makes the medium of semen alkaline as sperms require an alkaline medium to thrive. The seminal vesicle adds fluid content to the semen,
  2. No, because the copper-T just hampers the process of implantation of embryo in the uterine wall but not provides protection against sexually transmitted diseases.

Question 35. Write the full form of AIDS. Name the causative agent of AIDS. How is this disease transmitted?

Answer:

‘AIDS’ stands for “Acquired Immuno Deficiency Syndrome”. Its causative agent is Human Immunodeficiency Virus (HIV). It is transmitted from an infected person to a healthy person in the following ways – Through sexual intercourse, if one of the partner is infected; Through contaminated blood transfusion; The virus from infected mother may cross through the placenta and reach the embryo in the womb.

Question 36. How will an organism be benefited if it reproduces through spores?

Answer:

There are several advantages if the organism reproduces by spores: There are large numbers of spores formed from one sporangium; The spores can remain in dormant state till the favourable conditions are available; The spores have thick covering called cyst which helps them to survive in unfavourable conditions; The spores are very light and can spread easily through water, air or through the animals.

Question 37. Mrs. Kumar is pregnant for the last three months and she goes for regular check-up to her doctor. During her last visit, the doctor asked Mrs. Kumar to get an ultrasound done. Both Mr. and Mrs. Kumar went to a radiologist and got the ultrasound done. Once the ultrasound is over, Mr. Kumar asked the doctor whether the baby in the womb is a boy or a girl.

  1. What should the doctor reply to the couple?
  2. Is it ethical to determine the sex of a child?
  3. What should government do to discourage sex determination?

Answer:

  1. The doctor should not tell the sex of the foetus,
  2. No, it is not ethical to determine the sex of a child,
  3. The government should ban the process of sex determination and should punish or fine whosoever does so.

Haryana Board Class 10 Biology Solutions For Chapter 3 Heredity And Evolution

Haryana Board Class 10 Biology Solutions For Chapter 3 Heredity And Evolution Multiple Choice Questions And Answers

Question 1. Which statement explains the Mendel’s law of segregation?

  1. A trait in an offspring is due to the combination of an allele each from both the parent.
  2. A trait in an offspring is due to the combination of two alleles each from both the parent.
  3. A trait in an offspring is due to the combination of two alleles each from either of the parent.
  4. A trait in an offspring is due to the combination of one allele each from either of the parent.

Answer: 1. A trait in an offspring is due to the combination of an allele each from both the parent.

Question 2. The inheritance of color trait in flower is as shown.

HBSE Solutions Class 10 Biology Chapter 3 Mendel's Law Of Segregation

R and r denote two different genes for color. Which law of Mendel can be explained using the image?

  1. Only Law of segregation
  2. Only Law of independent assortment
  3. Law of segregation and Law of dominance
  4. Law of segregation and Law of independent assortment

Answer: 3. Law of segregation and Law of dominance

Question 3. Humans have two different sex chromosomes, X and Y. Based on the Mendel’s laws, a male offspring Will inherit which combination of chromosomes?

  1. Both the x chromosomes from one of its parents
  2. Both the y chromosomes from one of its parents
  3. Combination of x chromosomes from either of its parents
  4. Combination of x and y chromosome from either of its parents.

Answer: 4. Combination of x and y chromosome from either of its parents.

Question 4. Two individuals are as shown using geometric shapes

HBSE Solutions Class 10 Biology Chapter 3 Geometric Shapes

Their sex chromosomes are respectively denoted by Xf Xm, and Y. What are the possible combinations of sex chromosomes for their male and female offspring respectively?

  1. Xf Xm and XmXm
  2. XmY and XmXm
  3. Xf Y and XmY
  4. XmY and XmXf

Answer: 4. XmY and XmXf

Question 5. An individual is tall with black hair, and free earlobes. The individual learnt to play football from his father and got a scar on his forearm in a match. Which table shows the correct classification of his traits into acquired traits and inherited traits.

HBSE Solutions Class 10 Biology Chapter 3 Classification Of His Traits Into Acquired Traits And Inherited Traits

Answer: 4. HBSE Solutions Class 10 Biology Chapter 3 Acquired Traits And Inherited Traits

Question 6. The image shows the traits present across generations of a family.

HBSE Solutions Class 10 Biology Chapter 3 The Traits Present Across Generations Of A Family

Based on the image, what can be inferred about the eye color trait?

  1. Acquired trait because both male and females have it.
  2. Acquired trait because it is expressed in all the generations.
  3. Inherited trait because it is expressed in two different colors.
  4. Inherited trait because it depends on the traits of preceding generation.

Answer: 4. Inherited trait because it depends on the traits of preceding generation.

Question 7. In which case does the change in DNA contributes to speciation?

  1. Changes in the DNA of zygote
  2. Changes in the DNA of brain cells
  3. Changes in the DNA of bone cells
  4. Changes in the DNA of sperm cells.

Answer: 4. Changes in the DNA of sperm cells.

Question 8. The image shows the extent of similarity in the DNA of humans and the organisms.

HBSE Solutions Class 10 Biology Chapter 3 The Extent Of Similarity In The Dna Of Humans And The Organisms

Based on the image, what can be inferred about DNA changes in context of speciation?

  1. Newly evolved species have inactive ancestral genes.
  2. Species retain their DNA and evolve new proteins with time.
  3. Some of the genes remain conserved during the evolution of species.
  4. Species undergo a complete change of DNA sequences as they evolve.

Answer: 3. Species undergo a complete change of DNA sequences as they evolve.

Chapter 3 Heredity And Evolution Very Short Questions And Answers

Question 1. A normal pea plant bearing coloured flowers suddenly start producing white flowers. What could be the possible cause?

Answer: The appearance of white flowers is due to mutation

Question 2. How does the creation of variations in a species promote survival?

Answer: Variations enable a species to adapt according to the changes and new needs and thus provide better chances for the survival of the species.

Question 3. Two pea plants one with round green seeds (RRyy) and another with wrinkled yellow (rrYY) seeds produce FI progeny that have round, yellow (RrYy) seeds. When FI plants are selfed, what will be the ratio of the F2 progeny?

Answer: 9:3:3:1.

Question 4. List any two characters used by Mendel to conduct experiments on pea plants.

Answer: Seed colour and Flower colour.

Question 5. Do genetic combination of mothers play a significant role in determining the sex of new born?

Answer: No, mothers have no role in determining the sex of the new born. Mothers have a pair of X chromosome. And all children will inherit an ‘X’ chromosome from their mother regardless of whether they are boys or girls.

Question 6. Give reason for the appearance of new combinations of characters in the F2 progeny.

Answer: Since an organism can inherit each character independently. So, in the F2 progeny new combination of character appears, e.g. Tall/Short and round/wrinkled seed trait are independently inherited.

Question 7. Newly formed DNA copies may not be identical at times. Give reason.

Answer: Newly formed DNA copies may not be identical at times if there are errors or inaccuracies in DNA copying.

Question 8. If a tall pea plant is crossed with a dwarf pea plant, what is the ratio of dwarf plants in the F2 generation?

Answer: 3 Tall : 1 Dwarf

Question 9. No two individuals are absolutely alike in a population. why?

Answer: No two individuals are absolutely alike in a population as there are variation in the DNA due to the crossing-over and recombination during the DNA copying process.

Chapter 3 Heredity And Evolution Short Questions And Answers

Question 1. What are the two possible sources of variation?

Answer: Two possible sources of variation are Errors in DNA copying and Random fertilisation.

Question 2. Why did Mendel choose pea plants for his experiments?

Answer: Mendel choose pea plants for his experiments because of the following reasons: Easy to grow; Flowers have bisexual characteristics; Easy to obtain pure breed plant through self-fertilization; Generation time is less.

Question 3. An animal (guinea pig) having black colour is crossed with guinea pig having the same colour. They produced 100 offsprings, of which 75 were black and 25 were white. Find out

  1. What is the possible genotype?
  2. Which trait is dominant and which is recessive? How is this determined?

Answer:

Let BB/ Bb be the dominant colour (black) of guinea pig and bb be the recessive trait which codes for white colour of guinea pig.

  1. The possible genotype of the parent pigs is Bb i.e. the animals are heterozygous dominant as they produced offspring in the ratio of 3: 1. The 25 of the black pigs have the gene BB whereas the rest 50 of the black pigs have genes Bb and the white pigs have bb.
  2. The trait coding for black colour is dominant while the one coding for white is recessive. Traits are defined to be dominant when they are present in more individuals than the recessive trait-bearing individuals in the wild (i.e. not in human regulated breeding).

Question 4. ‘Different species use different strategies to determine the sex of a new born individual. It can be environmental cues or genetically determined’. Explain the statement by giving example for each strategy.

Answer: Environmental cue: In some animals, the temperature at which fertilised eggs are incubated determines whether the developing animal in egg is a male or a female. In some animals like snail, individuals can change sex.

Genetic cue: A child who inherits an X chromosome from the father will be a girl and the one who inherits a Y chromosome from the father will be a boy

Question 5. “It is possible that a trait is inherited but may not be expressed.” Give a suitable example to justify this statement.

Answer: Yes, it is possible that a trait is inherited but may not be expressed, e.g. when pure tall pea plants are crossed with pure dwarf pea plants, only tall pea plants are obtained in FI generation. The trait for dwarfness was present but could not express itself in the presence of the dominant trait for tallness. On selfing the tall plants of FI generation, both tall and dwarf plants are obtained in F2 generation in the ratio 3 : 1. Reappearance of the dwarf character, a recessive trait in F2 generation shows that the dwarf trait was present in individuals of FI, but it did not express. It got suppressed in the presence of the dominant trait.

Question 6. “A brother and sister are more related to each other as compared to the case when any one of them is related with his/her cousin”. Through this statement what will we get to know about their ancestors?

Answer: A brother and sister are more closely related to each other which mean that they have common ancestors more recently as compared to the case when any one of them (brother or sister) is related to the cousin. A brother and sister have their “parents” in common while a brother or sister and cousin have “grandparents” in common.

Question 7. How are variant genotypes produced?

Answer: Variant genotypes are produced by the following mechanisms: Mutation in genes and chromosomes; Recombination of genes; Hybridization of genes

Question 8. A woman with blonde curly hair married a man with black soft hair. All of their children in the first generation had black soft hair but in next generation children had different combinations in the ratio of 9:3:3:1. State the law that governs this expression.

Answer: The law of independent inheritance governs the given situation. The law states that the factors of different pairs of contrasting characters do not influence each other. They are independent of one another in the inheritance.

Question 9. The genotype of green stemmed tomato plants is denoted as GG and that of purple stemmed tomato plants as gg. When these two are crossed:

  1. What is the percentage of purple stemmed plants in F2 progeny if FI plants are self-pollinated?
  2. In what ratio would you find the genotype of GG and Gg in the F2 progeny?

Answer:

  1. Purple stemmed plants in F2 progeny – 25%
  2. Ratio of GG and Gg in the F2 progeny- 1:2:1

Question 10. A child questioned his teacher that why do organisms resemble their parents more as compared to grandparents. In which way will the teacher explain to the child?

Answer: The two parents involved in sexual reproduction produce gametes which fuse together forming a zygote. It gradually develops into a young child showing certain similarities with the parents. Since, a child inherits its characters from both the parents the resemblance with them is very close. The grandparents and the child resemble less closely because a gap of gene pool is created by the parents of the child. Since the child is immediate generation next to his parent thus to carry more similar genes as that of parents. Variations are more with grandparents.

Question 11. Mustard was growing in two fields- A and B, while field A produced brown coloured seeds, field B produced yellow coloured seeds. It was observed that in field A, the offsprings showed only the parental trait for consecutive generations, whereas in field B, majority of the offsprings showed a variation in the progeny. What are the probable reasons for these?

Answer: Self-pollination in field A; Cross pollination in field B.

Question 12. In a cross between plants with homozygous pink flowers and plants with homozygous white flowers, all the offsprings ofFI generation had pink flowers. When the FI generation was self-crossed, it was observed in the F2 generation that out of 100, 75 flowers were pink. Make a cross and answer the following questions:

  1. What are the genotypes of the F2 progeny?
  2. What is the ratio of Pink: White flowers in the F2 generation?

Answer:

1. Genotypes of F2 progeny are PP, Pp and pp.

2. HBSE Solutions Class 10 Biology Chapter 3 The Ratio Of Pink White Flowers In The F2 Generation

Question 13.

  1. If a purple-flowered pea plant (PP) is crossed with a white-flowered pea plant (pp), will we have white-flowered pea plants in the FI generation? Explain.
  2. What do you mean by dominant and recessive traits

Answer:

  1. No. We will not have white-flowered pea plants in the FI generation. This is because all the FI progeny plants show genetic makeup Pp. Since P is a trait dominant over p, all the plants in the FI generation have purple flowers.
  2. A dominant trait is a genetic trait which is expressed in a person who has only one copy of that gene. A recessive trait is a genetic trait which is expressed only when two copies of the gene are present.

Question 14. A pea plant with blue flowers denoted by BB is cross-bred with a pea plant with white flowers denoted by ww.

  1. What is the expected flower colour in the FI progeny?
  2. What will be the percentage of plants bearing white flowers in the F2 generation when the flowers of FI plants are self-pollinated?
  3. State the expected ratio of the genotypes BB and Bw in the F2 progeny.

Answer:

HBSE Solutions Class 10 Biology Chapter 3 A Pea Plant With Blue Flowers Is cross-bred With A Pea Plant With White Flowers

  1. The F1 progeny is expected to have plants with blue flowers.
  2. % of the F2 generation bears white flowers. So, 25% of the F2 progeny bears white flowers in the F2 generation when the flowers of F1 plants are self-pollinated.
  3. The ratio of the genotype BB and Bw in the F2 progeny is 1 (BB):2 (Bw).

Question 15. A cross was carried out between a pure-bred pea plant with axial flowers and a purebred pea plant with terminal flowers, and the FI progeny was obtained. This progeny was selfed to obtain the F2 progeny. Answer the following questions:

  1. What is the phenotype of the FI progeny and why?
  2. Give the phenotypic ratio of the F2 progeny.
  3. Why is the F2 progeny different from the FI progeny?

Answer:

A pea plant with axial flowers (AA; dominant) was crossed with a pea plant with terminal flowers (aa, recessive).

1. All the FI progeny would bear axial flowers because the trait for axial flowers is dominant over the trait for terminal flowers.

2. In the F2 generation. Parents → Aa x Aa Gametes → A, a A, a

HBSE Solutions Class 10 Biology Chapter 3 A Pea Plant With Axial Flowers

3. Phenotypic ratio → Axial : Terminal = 3:1

FI plants are heterozygous (Aa), and hence, only the dominant trait is visible in the FI generation. In the F2 generation, factors responsible for the two traits are segregated and recombined to form a homozygous recessive trait for terminal flowers (aa).

Question 16. A blue-flowered plant denoted by BB is crossbred with a white-flowered plant denoted by bb.

  1. State the colour of the flowers you would expect in the FI generation plants.
  2. What must be the percentage of white-flowered plants in the F2 generation if flowers of FI plants are self-pollinated?
  3. State the expected ratio of the genotypes BB and Bb in the F2 progeny.

Answer:

The cross between blue-flowered plant (BB) and white-flowered plant (bb) is a monohybrid cross which involves a single trait, i.e. colour of the flower under study.

  1. All the FI generation plants would be blue.
  2. If flowers of FI plants are self-pollinated, then we would have 75% plants with blue flowers and 25% plants with white flowers in the F2 generation.
  3. The expected ratio of the genotypes BB and Bb in the F2 progeny is 1:2.

Question 17. Basis the understanding of size of the organism and chromosome number, answer the following
questions:

  1. Do larger organisms have more number of chromosomes/cells?
  2. More the number of chromosomes/cells greater in the DNA content. Justify.

Answer:

  1. No, there is no relationship between size of organism and its chromosome number.
  2. Yes, since the major component of chromosome is DNA, if there are more chromosomes in a cell, the quantity of DNA will also be more.

Question 18. A potato is cut into a number of small pieces. These potato pieces are placed on wet cotton kept in a tray. After a few days, green shoots and roots appear only from some potatoes and not from all potato pieces. Why?

Answer: The pieces of potato that bear nodes only can give rise to new plants by producing shoots and roots. But the pieces not bearing nodes cannot produce new plants. Thus from a whole potato only some pieces that bear nodes give rise to roots and shoots. This is an example of vegetative propagation which is an asexual mode of reproduction in plants.

Question 19. In a tobacco plant, the male gametes have 24 chromosomes. What is the number of chromosomes in the female gamete? What is the number of chromosomes in the zygote?

Answer: Number of chromosomes in female gamete is 24. Number of chromosomes in zygote is 48.

Question 20. A man with blood group A marries a woman with blood group 0 and their daughter has blood group 0. Is this information enough to tell you which of the traits-blood group A or O- is dominant? Why or why
not?

Answer: No. This information is not sufficient to determine which of the traits – blood group A or 0 – is dominant. This is because we do not know about the blood group of all the progeny. Blood group A can be genotypically AA or AO. Hence, the information is incomplete to draw any such conclusion.

Chapter 3 Heredity And Evolution Long Questions And Answers

Question 1. Sex determination is the method by which distinction between males and females is established in a species. The sex of an individual is determined by specific chromosomes. These chromosomes are called sex chromosomes or allosomes. X and Y chromosomes are called sex chromosomes. The normal chromosomes other than the sex chromosomes of an individual are known as autosomes.

In XX-XY type of sex determination who produces two different types of gametes?

A couple has six daughters. What is the possibility of their having a girl next time?

What is the number of autosomes present in the liver cells of a human female?

Answer:

  1. Males produce two different types of gametes.
  2. 50% The possibility of having a girl or boy child is equal i.e., 50%, as 50% male gametes are Y type and 50% are X type.
  3. 22 pairs In humans, the number of autosomes are 2n = 44 or 22 pairs regardless of the sex.

Question 2.

  1. How many characters are transmitted in the following cross? Name them.
  2. Define dominant trait and recessive trait.

HBSE Solutions Class 10 Biology Chapter 3 Dominant Trait And Recessive Trait

Answer:

  1. Two characters are transmitted. They are shape of seed and colour of seed.
  2. A dominant trait is a genetic trait which is considered dominant if it is expressed in a person who has only one copy of that gene. A recessive trait is a genetic trait which is expressed only when two copies of the same gene are present.

Question 3. Explain Mendel’s experiment on inheritance of characters considering only one visible contrasting character in pea plant.

Answer:

Mendel’s experiment can be studied in the following ways:

  1. Mendel first crossed pure-bred, tall pea plants with pure-bred, short pea plants and found that tall pea plants were produced in the FI generation.
  2. Mendel crossed tall pea plants of the FI generation and found that tall plants and dwarf plants are in the ratio 3:1. Mendel observed that the dwarf trait of the pea plant which had disappeared in the FI generation progeny reappeared in the F2 generation. He concluded by saying that traits are inherited independently.

Question 4. After self-pollination in pea plants with round, yellow seeds, following types of seeds were obtained by Mendel:

HBSE Solutions Class 10 Biology Chapter 3 Types Of Seeds Were Obtained By Mendel

Analyse the result and describe the mechanism of inheritance which explains these results.

Answer: The ratio obtained is 9:3:3:1 in which parental as well as new combinations are observed. This indicates that progeny plants have not inherited a single whole gene set from each parent. Every germ cell takes one chromosome from the pair of maternal and paternal chromosomes. When two germ cells combine, segregation of one pair of characters is independent of other pair of characters.

Question 5. In humans, there is a 50% probability of the birth of a boy and 50 % probability that a girl will be born. Justify the statement on the basis of the mechanism of sex-determination in human beings.

Answer: In human beings, the genes inherited from the parents decide whether the child born will be a boy or a girl. Women have a perfect pair of sex chromosomes (XX). But, men have a mismatched pair (XY) of sex chromosomes. All children will inherit an X chromosome from their mother regardless of whether they are boys or girls. Thus, the sex of the children will be determined by what they inherit from their father. A child who inherits an X chromosome from her father will be a girl, and one who inherits a Y chromosome from him will be a boy.

Question 6. A blue colour flower plant denoted by BB is cross bred with that of white colour flower plant denoted by bb.

  1. State the colour of flower you would expect in their FI generation plants.
  2. What must be the percentage of white flower plants in F2 generation if flowers of F1 plants are selfpollinated?
  3. State the expected ratio of the genotypes BB and Bb in the F2 progeny.
  4. State the type of plants which are not found in FI generation but reappeared in F2 generation. Write the reason for the same.

Answer:

  1. All flowers will be blue.
  2. White: 25%
  3. Genotype ratio in F2: BB: Bb :: 1:2
  4. White flowering plants; White flower being a recessive trait can only be expressed in the recessive homozygous condition.

Question 7. Fertilization and implantation are the most critical events in the reproduction process. In this process, both egg and sperm are fused together to form a zygote. Later, it gets implanted into the uterus and the development of an organism starts.

The given flowchart explains the process. Study the chart and answer the questions given below:

HBSE Solutions Class 10 Biology Chapter 3 Fertilization And Implantation Are The Most Critical Events In The Reproduction Process

  1. Identify the labels X, Y and Z.
  2. Where does the process of fertilisation takes place in female body?

Answer:

  1. W: Sperm; X: Ovary; Y: Zygote; Z: Foetus;
  2. Fallopian Tubes

Question 8. A horticulturist took stems of two different plants, plant X with roots and plant Y without roots. He fixed the cut stem X in soil and fitted and bound tightly the other cut stem Y over the surface of X. He fastened the joint properly with the help of polythene. The cut soon healed and the two plant stems (X and Y) grew together as one plant.

  1. What are stems X and Y respectively known as?
  2. Why the area where two stems are joined is covered with polythene?

Answer:

  1. The cut stem of a plant having roots (X) and is fixed in soil is called stock. The cut stem of another plant (without roots) (Y) is called scion.
  2. The area where two stems are joined is covered with polythene to prevent the loss of water and cell sap from the cut and joined ends of the stems. It also helps to prevent harmful infection by bacteria or fungi.

Haryana Board Class 10 Physics Solutions For Chapter 4 Magnetic Effects Of Current

Haryana Board Class 10 Physics Solutions For Chapter 4 Magnetic Effects Of Current Multiple Choice Questions And Answers

Question 1. A student learns that the magnetic field strength around a bar magnet is different at every point. Which diagram shows the correct magnetic field lines around a bar magnet?

HBSE Class 10 Biology Solutions For Magnetic Field Strength Around A Bar Magnet Is Different At Every Point

Answer: 3.

Question 2. A student places some iron fillings around a magnet. The iron fillings arrange themselves as shown in image.

HBSE Class 10 Biology Solutions For A Student Place Some Iron Fillings Around A Magnet

The student labelled four different regions around the magnet. Where would be the magnetic be the strongest?

  1. P
  2. Q
  3. R
  4. S

Answer: 3. R

Question 3. A student placed a magnetic compass around a straight current carrying wire. The student noticed when he moved the compass away from the wire, the deflection in compass decreases. How would be the magnetic field lines around the conductor?

HBSE Class 10 Biology Solutions For A Magnetic Compass Around A Straight Current Carrying Wire

Answer: 2.

Question 4. The image shows the magnetic field lines around a straight current-carrying conductor.

HBSE Class 10 Biology Solutions For The Magnetic Field Lines Around A Straight Current Carrying Conductor

If the direction of the current in the straight wire is changed, what change in the magnetic field line will be observed?

HBSE Class 10 Biology Solutions For The Direction Of Current In The Straight Wire Is Changed

Answer: 1.

Question 5. Which diagram shows the magnetic field lines around a current-carrying circular loop?

HBSE Class 10 Biology Solutions For The Magnetic Field Lines Around A Current Carrying Circular Loop

Answer: 2.

Question 6. Which diagram shows the correct direction of the magnetic field lines at point P and Q in current carrying circular loop?

HBSE Class 10 Biology Solutions For Direction Of Magnetic Field Line At point P nd Q In Current Carrying Circular Loop

Answer: 2.

Question 7. The magnetic field lines of solenoid are similar to the magnetic field lines of bar magnet. Which image correctly shows the solenoid as a bar magnet?

HBSE Class 10 Biology Solutions For The Magnetic Field Lines Of Solenoid Are Similar To The Magnetic Field Lines Of Bar Magnet

Answer: 3.

Question 8. Where should the magnetic compass be placed in the solenoid to get maximum deflection in the magnetic compass?

HBSE Class 10 Biology Solutions For The Magnetic Compass Be Placed In Solenoid To Get Maximum Deflection In The Magnetic Compass

Answer: 3.

Question 9. The image shows Fleming’s left-hand rule.

HBSE Class 10 Biology Solutions For The Fleming's Left-Hand Rule

Which option explains the rule to understand the working of the motor?

  1. When a current-carrying conductor is moved with a force, it creates a magnetic field.
  2. When a conductor is moved inside a magnetic field, current is produced in the conductor.
  3. When a magnetic field is moved relative to the conductor, current is produced in the conductor.
  4. When a current carrying conductor placed in a magnetic field, it experiences a force by magnetic field.

Answer: 4. When a current carrying conductor placed in a magnetic field, it experiences a force by magnetic field.

Question 10. A metal rod PQ. is placed in the magnetic field. The ends of the rod are connected with a battery using wires.

HBSE Class 10 Biology Solutions For A Metal Rod PQ Is Placed In The Magnetic Field

Where will the rod move?

  1. Upward
  2. Downwards
  3. Into the field
  4. Out of the field

Answer: 4. Out of the field

Question 11. A student inserts a bar magnet in the coil. The student observes deflection in the galvanometer connected to the coil. What will happen if the magnet is continuously getting in and out of the coil?

  1. The current induced in the coil will increase
  2. The current will change its direction continuously
  3. The magnetic field will create a motion in the coil
  4. The magnetic field of the bar magnet would keep decreasing

Answer: 2. The current will change its direction continuously

Question 12. A student makes an arrangement to study electromagnetic induction, as shown.

HBSE Class 10 Biology Solutions For Electromagnetic Induction

She changes the arrangement in four different ways.

HBSE Class 10 Biology Solutions For Electromagnetic Induction In Four Different Ways

In which trial would the galvanometer remain undeflected?

  1. Trial 1
  2. Trial 2
  3. Trial 3
  4. Trial 4

Answer: 4. Trial 4

Question 13. The image shows the components of an electric generator.

HBSE Class 10 Biology Solutions For Electric Generator

When the coil PQRS is rotated as shown. What is the direction of electric current when coil completes half cycle of the rotation?

HBSE Class 10 Biology Solutions For The Direction Of Electric Current When Coil Completes Half Cycle Of The Rotation

Answer: 2.

Question 14. Appliances that have a metal body are generally connected to the earthing wire. What is the reason to earth these wires?

  1. To prevent an excess of current
  2. To prevent the leakage of current
  3. To provide extra current to appliance
  4. To provide high resistance to the appliance

Answer: 2. To prevent the leakage of current

Question 15. Which diagram shows the domestic electric circuit?

HBSE Class 10 Biology Solutions For The Domestic Electric Circuit

Answer: 4.

Chapter 4 Magnetic Effects Of Current Very Short Questions And Answers

Question 1. Why does a compass needle show deflection when brought near a current carrying conductor?

Answer: The magnetic field is produced around the current carrying conductor.

Question 2. Positively charged particles moving towards the west are deflected towards the north by a magnetic field. What will be the direction of the magnetic field?

Answer: Upward direction in accordance with Fleming’s left hand rule.

Question 3. If you hold a coil of wire next to magnet, no current will flow in the coil. What else is needed to induce a current?

Answer: Relative motion between the coil and the magnet.

Question 4. What is the qualitative effect of inserting an iron core into a current-carrying solenoid?

Answer: Magnetic field becomes very strong.

Question 5. How is the magnetic field inside the long solenoid carrying current?

Answer: The Magnetic field is uniform inside the long solenoid carrying current.

Question 6. Name the rule for finding the direction of the magnetic field produced by a straight current-carrying conductor.

Answer: Maxwell’s right-hand thumb rule.

Question 7. What is the shape of a current-carrying conductor whose magnetic field pattern resembles that of a bar magnet?

Answer: Solenoid

Question 8. A circular conducting coil has ‘n’ turns. If the magnitude of the magnetic field produced by a single coil is B, then what is the magnetic field of the entire coil?

Answer: Since the magnetic field produced by current-carrying wire at a given point depends directly on the current passing through it. Therefore, if there is a circular coil having “n” turns, the field produced is “n” times as large as that produced by a single turn. This is because the current in each circular turn has the same direction, and the field due to each turn then just adds up.

Question 9. A soft iron bar is enclosed by a coil of insulated copper wire as shown in the figure. When the plug of the key is closed, then where would the face B of the iron bar be marked?

Answer: N-Pole

Chapter 4 Magnetic Effects Of Current Short Questions And Answers

Question 1. For the coil in the diagram below, when the switch is pressed:

HBSE Class 10 Biology Solutions For The Coil When The Switch Is Pressed

  1. What is the polarity of end A?
  2. Which way will the compass point then?

Answer:

  1. End A becomes the S-pole because current flows in a clockwise direction at A.
  2. When A becomes S-pole, the other end becomes N-pole. So, the tip of the compass (which also has North polarity) moves away from this end i.e., the tip moves towards the right.

Question 2. What, according to Maxwell’s Right Hand Thumb rule, will be the direction of the current when lines of the magnetic field are in the anti-clockwise direction?

Answer:

The current will be flowing in the upward direction when the direction of the magnetic field is in the anti-clockwise direction.

Question 3. The figure given below shows the magnetic field between two magnets:

  1. Label the other poles of the magnets.
  2. Which is the weaker magnet?

Answer:

HBSE Class 10 Biology Solutions For The Other Poles Of The Magnets

Magnet 2 is weaker.

Question 4. What are the two ways in which you can trace the magnetic field pattern of a bar magnet?

Answer:

  1. By using iron filings
  2. By using compassion

Question 5. Copy the figure below which shows a plotting compass and a magnet. Label the North pole of the magnet and draw the field line on which the compass lies.

HBSE Class 10 Biology Solutions For A Plotting Compass And A Magnet

Answer:

HBSE Class 10 Biology Solutions For The North Pole Of The Magnet

Question 6.

  1. Mention the factors on which the direction of force experienced by a current-carrying conductor placed in a magnetic field depends.
  2. A proton beam is moving along the direction of a magnetic field. What force is acting on the proton beam?

Answer:

  1. The direction of the force exerted on experienced by a current carrying conductor mainly depends on the two factors:
    • The direction of the current , the direction of flow of positive charge determine the direction of the force as described by the Flemings left hand rule,
    • The direction of the magnetic field; the force exerted on the conductor depends on the orientation of the magnetic field. This is also defined according to the Flemings left hand rule,
  2. As the proton beam is moving along the direction of the magnetic field it will experience no force. The strength of the magnetic field depends on the orientation of the conductor in the magnetic field.

It experiences maximum force when the direction of flow of current and the magnetic field is perpendicular to each other i.e. the angle between the two is, let’s say θ = 90° as shown in the figure (2). If the direction of the current and the magnetic field is parallel i.e. θ = 0° as shown in the figure (1), the conductor will experience no force.

HBSE Class 10 Biology Solutions For The Direction Of The Current And The Magnetic Field Is Parallel

Question 7. What are the three factors which cause a change in magnetic flux of the coil?

Answer:

Changes in the magnetic flux of a coil occur due to:

  1. Relative motion between a coil and a magnet placed near it.
  2. Relative motion between a coil and a current-carrying conductor placed near it.
  3. Change of current in the conductor placed near the coil.

Question 8. The diagram below shows a bar magnet surrounded by two compasses numbered 2 and 4. What directions will these compasses show?

HBSE Class 10 Biology Solutions For A Bar Magnet Surrounded By Two Compasses Numbered 2 And 4

Answer:

Directions shown by the needle of compasses 2 and 4 would be as shown below.

HBSE Class 10 Biology Solutions For The Direction Needle Of Compasses

This is because the magnetic field lines arise from the North Pole and end at the South Pole.

Question 9. A simple motor is made in a school laboratory. A coil of wire is mounted on an axle between the poles of a horseshoe magnet, as illustrated.

HBSE Class 10 Biology Solutions For A Simple Motor Is Made In A School Laboratory

In the example above, coil ABCD is horizontal and the battery is connected as shown.

  1. For this position, state the direction of the force on the arm AB.
  2. Why does the current in the arm BC not contribute to the turning force on the coil?

Answer:

  1. Downwards
  2. Because BC is in the same direction as the direction of field lines. Force is minimum when the direction of current in the conductor is the same as that of the magnetic field. BC will not contribute as the force on this part of the coil will be cancelled by the force on DA.

Question 10. Geeta performed an activity in the school laboratory to observe the effects of magnetic field lines. Based on her observation, she has drawn the magnetic field lines of two magnets as shown in fig. 1 and fig. 2.

HBSE Class 10 Biology Solutions For The Magnetic Field Lines OfTwo Magnets

  1. Select the figure that represents the correct pattern of field lines. Give reasons for your answer.
  2. Also, name the poles of the magnets facing each other.

Answer:

  1. Figure B represents the correct pattern of field lines. In figure A, field lines cross each other which is not possible because if they cross each other, at the point of intersection, there would be two directions of field lines,
  2. In figure B, field lines are emerging in nature, so poles of magnet facing each other are north poles while opposite faces will have south polarity.

Question 11. A student drew three magnetic field lines 1, 2 and 3 of a bar magnet with the help of a compass needle as shown in the figure.

HBSE Class 10 Biology Solutions For Three Magnetic Field Lines 1, 2 And 3 Of A Bar Magnet

Is this configuration possible? If not, then what is wrong in the given figure and why?

Answer:

No, this type of configuration is not possible as:

Two field lines cannot intersect each other,

Direction of field line as it emerged from south pole and merged at north pole. ‘3’ is wrong as, it emerged from south pole and merged at north pole.

Question 12.

  1. What are the patterns formed by the circular loop carrying current?
  2. Which rule is used to find the direction of the magnetic field produced due to the electric current in a circular loop?
  3. On which factors does the strength of a magnetic field depend?

Answer:

  1. The circular loops carrying current forms concentric circular patterns of the magnetic field due to electric current.
  2. The direction of the magnetic field of the loop carrying current can be determined by the clock face rule.
  3. The strength of the magnetic field due to current depends on the
    1. Number of turns of wire in the coil
    2. Radius of the coil
    3. Current is flowing in the coil.

Question 13. How does the strength of the magnetic field at the centre of a circular coil of a wire depend on?

  1. Radius of the coil
  2. Number of turns of wire in the coil
  3. Draw the magnetic lines of force in case of a circular coil of wire

Answer:

  1. The strength of the magnetic field (B) is inversely proportional to the radius of the circular loop (r).
  2. The strength of the magnetic field (B) is directly proportional to the number of turns in the coil (N).
  3. The magnetic field lines will be as shown below.

HBSE Class 10 Biology Solutions For The Strength Of The Magnetic Field

Question 14. A coil of insulated copper wire is connected to a galvanometer. What will happen if a bar magnet is held stationary inside the coil?

Answer:

When a bar magnet is held stationary inside the coil, there will be no deflection in the galvanometer indicating that no current is produced in the coil.

Question 15.

  1. If a current carrying straight conductor is placed in east-west direction, then find the direction of the force experienced by the conductor due to earth’s magnetic field.
  2. Mona draws magnetic field lines of field close to the axis of a current carrying circular loop. As she moves away from the centre of the circular loop she observes that the lines keep on diverging. How will you explain her observation?

Answer:

  1. The force will act in upward direction perpendicular to both, the direction of current as well as to the field. The direction of force experienced by the conductor gets reversed, i.e., in the downward direction.
  2. Strength of the magnetic field decreases as distance increases. This is indicated by the decrease in degree of closeness of the lines of field.

Question 16. A magnetic field is described by drawing the magnetic field lines. When a small north magnetic pole is placed in the magnetic field created by a magnet, it will experience a force. And if the North Pole is free, it will move under the influence of the magnetic field. The path traced by a north magnetic pole free to move under the influence of a magnetic field is called a magnetic field line.

HBSE Class 10 Biology Solutions For Magnetic Field Lines

Since the direction of magnetic field line is the direction of force on a north pole, so the magnetic field lines always begin from the N-pole of a magnet and end on the S-pole of the magnet. Inside the magnet, however the direction of magnetic field lines is from the S-pole of the magnet to the N-pole of the magnet. Thus, the magnetic field lines are closed curves. When a small compass is moved along a magnetic field line, the compass needle always sets itself along the line tangential to it. So, a line drawn from the south pole of the compass needle to its north pole indicates the direction of the magnetic field at that point.

  1. Do the magnetic field lines intersect? If not, why?
  2. A strong bar magnet is placed vertically above a horizontal wooden board. What would be the magnetic lines of force?
  3. Where do we use Magnetic field lines?

Answer:

  1. No two magnetic field lines are found to cross each other. If two field lines crossed each other, it would mean that at the point of intersection, the compass needle would point in two directions at the same time, which is not possible.
  2. The magnetic field and hence the magnetic line of force exist in all the planes all around the magnet.
  3. The relative strength of the magnetic field is shown by the degree of closeness of the field lines and the direction of the magnetic field is obtained by the tangent to the field lines at the point of intersection.

Question 17. Diagram below shows a circuit containing a coil wound over a long and thin hollow cardboard tube.

HBSE Class 10 Biology Solutions For Circuit Containing A Coil Wound Over A long And thin Hollow Cardboard tube

  1. Show the polarity acquired by each face of the solenoid.
  2. Draw the magnetic field lines of force inside the coil and also show their direction.

Answer:

  1. The polarity acquired by the two ends is as shown below. (A shows North and South polarity).

HBSE Class 10 Biology Solutions For The Magnetic Field Lines Of Force Inside the Coil And Their Directions

Question 18. Identify the poles of the magnet in the given figures:

HBSE Class 10 Biology Solutions For The Poles Of The Magnet

Answer:

1. Field lines emerge from north pole (N) and merge at the south pole (S) in figure (1), So the poles of the magnet are:

HBSE Class 10 Biology Solutions For Field Lines Emerge From The North Pole

2. Field lines emerge from north pole (N) and merge at the south pole (S) in figure (2), So the poles of the magnet are:

HBSE Class 10 Biology Solutions For The Pole Of The Magnet

Question 19. Richa is performing an experiment on magnetism for which she has used two magnets A and B. She felt that there was an attraction between the magnets. So, she drew the magnetic field lines as below.

HBSE Class 10 Biology Solutions For Two Magnets A And B The Magnetic Field Lines

But her teacher said that the diagram was wrong.

  1. Why was the diagram wrong?
  2. Draw the correct diagram.
  3. What does the direction of thumb indicate in the right-hand thumb rule. In what way this rule is different from Fleming’s left-hand rule?

Answer:

1. From the drawing of Richa it is clear that the magnetic north poles are facing each other. But there was an attraction between the magnets. So, north pole of one magnet was facing the south pole of the other magnet.

2. The correct diagram is either of the two diagrams shown below:

HBSE Class 10 Biology Solutions For Attraction Between The Magnets

3. In the right-hand thumb rule, the r thumb indicates the direction of current in the straight conductor held by curled fingers, whereas in Fleming’s left-hand rul,e the thumb gives the direction of force experienced by a current-carrying conductor placed in an external magnetic field.

Chapter 4 Magnetic Effects Of Current Long Questions And Answers

Question 1. Is the magnetic field same all around a bar magnet? Explain with reasons.

Answer:

No, magnetic field strength varies at every point around it. Magnetic field strength depends on the number of field lines per unit area.

If the field lines per area is more, then the magnetic strength in that area is strong, and if the field lines per area is less, then the magnetic strength is weak.

As the magnetic field lines per unit area is maximum at the poles, the magnetic strength is also maximum in that region.

Question 2. PQ is a current-carrying conductor in the paper plane, as shown in the figure below.

HBSE Class 10 Biology Solutions For PQ Is A Current Carrying Conductor In The Plane Of The Paper

  1. Find the directions of the magnetic fields produced by it at points R and S.
  2. Given r1 > r2, where will the strength of the magnetic field be larger? Give reasons.
  3. If the polarity of the battery connected to the wire is reversed, how would the direction of the magnetic field be changed?
  4. Explain the rule that is used to find the direction of the magnetic field for a straight current-carrying conductor.

Answer:

  1. The magnetic field lines are produced is into the plane of the paper at R and out of it at S.
  2. Field at S > Field at P The Magnetic field strength for a straight current-carrying conductor is inversely proportional to the distance from the wire.
  3. If polarity is reversed, the current will go from top to bottom in the wire and the magnetic field lines goes in the clockwise direction on the plane which is perpendicular to the wire carrying current,
  4. According to the Right-Hand thumb rule, the thumb is aligned to the direction of the current, and the direction in which the fingers are wrapped around the wire will give the direction of the magnetic field.

Question 3. An insulated copper wire wound on a cylindrical cardboard tube such that its length is greater than its diameter is called a solenoid. When an electric current is passed through the solenoid, it produces a magnetic field around it. The magnetic field produced by a current-carrying solenoid is similar to the magnetic field produced by a bar magnet. The field lines inside the solenoid are in the form of parallel straight lines. The strong magnetic field produced inside a current-carrying solenoid can be used to magnetise a piece of magnetic material like soft iron when placed inside the solenoid. The strength of the magnetic field produced by a current-carrying solenoid is directly proportional to the number of turns and the strength of the current in the solenoid.

  1. What would be the strength of the magnetic field inside a long current-carrying straight solenoid?
  2. By using which rule we can find the north-south polarities of an electromagnet?
  3. A long solenoid carrying a current produces a magnetic field B along its axis. If the current is double and the number of turns per cm is halved, then what will be the new value of magnetic field?
  4. A soft iron bar is enclosed by a coil of insulated copper wire as shown in figure.

HBSE Class 10 Biology Solutions For A Soft Iron Bar Is Enclosed By A Coil Of Insulated Copper Wire

When the plug of the key is closed, then where would the face B of the iron bar be marked?

Answer:

  1. Magnetic field inside an infinite solenoid is uniform. Hence it is same at all points.
  2. Clock face rule
  3. For a long solenoid, magnetic field B oc In; where I is the flowing current and n is number of turns per unit length in the solenoid. Therefore, in the given case magnetic field will remain unchanged.
  4. N-pole.

Question 4.

1. An induced emf is produced when a magnet is plunged into a coil. On which factor or factors does the magnitude of induced emf depend?

2. The wire in the figure below is being moved downwards through the magnetic field, so as to produce an induced current.

HBSE Class 10 Biology Solutions For The Wire Moved Downwards Through The Magnetic Field

  • What would be the effect of using a stronger magnet?
  • What would be the effect of holding the wire still in the magnetic field?

Answer:

1. The number of turns of the coil, the speed with which the magnet is moved, and the strength of the magnet.

  • The magnitude of the induced current increases.
  • The induced current is zero.

Question 5. When a current is passed through the circular loop of wire, the magnetic field lines near the coil are nearly circular and concentric. At the centre of the circular loop, the magnetic field lines are straight. The strength of the magnetic field produced by a current-carrying circular coil (or circular wire) depends on:

  1. Radius of the circular coil.
  2. Number of turns of wire in the circular coil. The direction of the field lines can be found by applying the Right-Hand Thumb Rule.

HBSE Class 10 Biology Solutions For The Direction Of Field Lines Can Be Found By Applying Right-Hand Thumb Rule

  • A long horizontal power line is carrying a current of 100 A in the east-west direction. What is the direction of the magnetic field at a point 1.0 m below it?
  • What type of curve we get, between magnetic field and distance along the axis of a current-carrying circular coil?

Answer:

  1. The current flows in the east-west direction. From the right-hand rule, we get the direction of the magnetic field as from north to south. The direction of the magnetic field will be the same at every point below the power line.
  2. At smaller distances, the magnetic field will be described by concentric circles around the wire. As the distance increases, the circles become larger and larger. At the centre of the loop or coil, the magnetic field will appear as a straight line.

Haryana Board Class 10 Physics Solutions For Chapter 3 Electricity

Haryana Board Class 10 Physics Solutions For Chapter 3 Electricity Multiple Choice Questions And Answers

Question 1. A charge of 100 C flows through a bulb in 5 minutes. How much current is flowing through the bulb?

  1. 500 A
  2. 100 A
  3. 20 A
  4. 0.3 A

Answer: 3. 20 A

Question 2. A conducting wire carries 1021 electrons in 4 minutes. What is the current flowing through the wire?

  1. 40 A
  2. 7 A
  3. 4 A
  4. 0.7 A

Answer: 4. 0.7 A

Question 3. Work of 14 J is done to move 2 C charge between two points on a conducting wire. What is the potential difference between the two points?

  1. 28 V
  2. 14 V
  3. 7 V
  4. 3.5 V

Answer: 3. 7 V

Question 4. In order to move a charge of 3 C between two points on a conducting wire, 12 J of work is done. How much increase or decrease in the voltage will increase the work done on the same amount of charge to 36 J?

  1. -12 V
  2. -8 V
  3. +8 V
  4. +12 V

Answer: 3. +8 V

Question 5. A circuit has a charge of 2C moving through it in 3 s. Which electrical component in the circuit, if present, will show the current?

  1. Voltmeter will show a current of 6 A.
  2. Ammeter will show a current of 0.7 A.
  3. Rheostat will show a current of 0.7 A.
  4. Resistor will show a current of 0.35 A

Answer: 2. Ammeter will show a current of 0.7 A.

Question 6. The image shows a circuit diagram.

HBSE Class 10 Biology Solutions For A Circuit Diagram Using The Voltmeter

What is being measured using the voltmeter?

  1. Current in the circuit
  2. Voltage in the circuit
  3. Voltage across the resistor
  4. Resistance offered by the resistor

Answer: 3. Voltage across the resistor

Question 7. What is the relationship between resistance and current?

  1. They are directly related to each other.
  2. They are inversely related to each other.
  3. The resistance has a greater magnitude than current.
  4. The current has a greater magnitude than resistance.

Answer: 2. They are inversely related to each other.

Question 8. Which plot shows the change in voltage when the current is gradually decreased across a resistor?

HBSE Class 10 Biology Solutions For The Change In Voltage When The Current Is Gradually Decreased Across A Resistor

Answer: 1.

Question 9. Which one among a bar of an alloy of mass 2 kg and a 3 kg iron bar of same dimension has greater resistivity?

  1. An iron bar because it has a higher mass,
  2. Alloy bar because it has lower mass.
  3. Iron bar because it has the same types of atoms,
  4. Alloy bar because it has different types of atoms.

Answer: 4. Alloy bar because it has different types of atoms.

Question 10. A piece of wire is measured to have resistivity in the order of 1019 Ω m. What should its material be classified into?

  1. Alloys
  2. Insulators
  3. Good conductors
  4. Poor conductors

Answer: 2. Insulators

Question 11. Which combination of a 2 Ω resistor and 4 Ω resistor offers the least resistance to current in the circuit?

  1. Series combination, which results in a net resistance of 2Ω.
  2. Parallel combination, which results in a net resistance of 2Ω.
  3. Series combination, which results in a net resistance of 1.5Ω.
  4. Parallel combination, which results in a net resistance of 0.5Ω.

Answer: 4. Parallel combination, which results in a net resistance of 0.5.

Question 12. The image shows a combination of 4 resistors

HBSE Class 10 Biology Solutions For A Combination Of 4 Resistors

What is the net resistance between the two points in the circuit?

  1. 0.5 Ω
  2. 1.0 Ω
  3. 1.5 Ω
  4. 2.0 Ω

Answer: 2. 1.0 Ω

Question 13. How much more heat is produced if the urrent is doubled?

  1. Twice the original amount
  2. Thrice the original amount
  3. Four times the original amount.
  4. Five times the original amount.

Answer: 2. Thrice the original amount

Question 14. A bulb has a resistance of 5. If 2 A of current at 200 V flows through the bulb, how much heat is produced by the bulb in 10 minutes?

  1. 2.0 x 105 J
  2. 4.0 x 105 J
  3. 1.2 x 106 J
  4. 2.4 x 106 J

Answer: 4. 2.4 x 106 J

Question 15. In order to reduce electricity consumption at home, what kind of appliance should one purchase?

  1. One which draws low power
  2. One which produces less heat
  3. One which operates at a higher voltage
  4. One which draws a high amount of current

Answer: 1. One that draws low power

Question 16. An electric toaster has a power rating of 200 W. It operates for 1 hour in the morning and 1 hour in the evening. How much does it cost to operate the toaster for 10 days at Rs 5 per kW h?

  1. Rs 20
  2. Rs 400
  3. Rs 5000
  4. Rs 10000

Answer: 1. Rs 20

Chapter 3 Electricity Very Short Answer Questions

Question 1. What is meant by saying that the electric potential at a point is 1 volt?

Answer: Electric potential at a point is 1 volt means 1 joule of work is done in moving lunit positive charge from infinity to that point.

Question 2. The potential difference across the wire having fixed resistance is tripled. By how much does the electric power increase?

Answer: P = V2/R = (3V)2/R; P = 9P

Question 3. What is the commercial unit of energy?

Answer: Kilowatt-hour

Question 4. What is the relation between the resistance (R) of the filament of a bulb, its power (P), and a constant Voltage (V) applied across it?

Answer: P = V2/R

Question 5. What is the general name of the substances having infinitely high electrical resistance?

Answer: Insulators

Question 6. How is an ammeter connected in a circuit to measure the current flowing through it?

Answer: An Ammeter is connected in series in an electric circuit.

Question 7. Calculate the current in the circuit if 40 C of charge is passed through a point in 10 seconds.

Answer: I = Q/t = 40/10 = 4A

Question 8. State whether a voltmeter has a high resistance or a low resistance. Give a reason for your answer.

Answer: Voltmeter has a high resistance so that it takes a negligible current from the circuit.

Question 9. If the current in a wire is flowing in the vertically downward direction and a magnetic field is applied from west to east, what is the direction of the force on the wire?

Answer: According to Fleming’s left-hand rule direction of the force will be in the South direction.

Question 10. State whether an electric heater will consume more electrical energy or less electrical energy per second when the length of its heating element is reduced.

Answer: The electric heater will consume more electrical energy when the length of its heating element is reduced because resistance is directly proportional to length (R = ρ/ I x A ). Hence, resistance will decrease. Power is inversely proportional to resistance (P = V2R ) . Hence, power will increase.

Question 11. Define one coulomb charge.

Answer: One coulomb of charge is that quantity of charge which exerts a force of 9 x 109 Newton on an equal charge is placed at a distance of 1 m from it.

Question 12. A car headlight bulb working on a 12 V car battery draws a current of 0.5 A. The resistance of the light bulb is:

  1. 0.5 Ω
  2. 6 Ω
  3. 12 Ω
  4. 24 Ω

Answer: 3. 12

Question 13. If 20 C of charge passes a point in a circuit in 1 s, what current is flowing?

Answer: Q = 20 C, t = Is; I = Q/t; Thus, I = 20/1 = 20 A

Question 14. What is an electric current?

Answer: An electric current is a flow of electric charges (electrons) through a conductor.

Question 15. Which effect of current is utilized in an electric light bulb?

Answer: The Heating effect of electric current.

Question 16. State the expression for electrical power in terms of potential difference and resistance.

Answer: Electrical power, P = V2/R

Question 17. Name a device that helps to measure the potential difference across a conductor.

Answer: Voltmeter

Question 18. If a charge of 420 C is flowing from the wire for 5 minutes. Find the value of current.

Answer: 1 = q/t; q = 420 C; t = 5 minutes = 5 x 60 = 300 s ; I = 420/300 = 1.4 A

Question 19. How does the resistance of a pure metal change if its temperature decreases?

Answer: On decreasing the temperature, the resistance decreases.

Question 20. How many milliamperes are there in 1 ampere?

Answer: 1A = 102mA

Question 21. Why should the resistance of an ammeter be very small?

Answer: The resistance of an ammeter should be very small so that it may not change the value of the current flowing in the circuit.

Question 22. An electron enters a magnetic field at right angles to it, as shown in the figure. What is the direction of force acting on the electron?

HBSE Class 10 Biology Solutions For An Electron Enters A Magnetic Field At Right Angles

Answer: The direction of force is perpendicular to the direction of magnetic field and current as given by Fleming’s left hand rule. We know that the direction of current is taken opposite to the direction of motion of electrons. The force is therefore directed into the page.

Chapter 3 Electricity Short Questions And Answers

Question 1. Why it necessary to connect an earth wire to electric appliances having metallic bodies?

Answer:

An electric appliance with having metallic body can have current leak and current may flow through its body. It can be fatal if a person touches its current-carrying metallic body. So, it must be connected to an earth wire to pass the extra current, flowing through its body, to the earth plate.

Question 2. Two resistors each of 10 Ω are connected in

  1. Series
  2. And then in parallel to a battery of 6 V. Calculate the ratio of power consumed in the combination of resistor in two case.

Answer:

V = 6V; Total resistance in series, Rs = 20 Ω

Power, Ps = V2/R = 1.8 W; Total resistance in parallel, Rp = 5 Ω; Power, Pp = V2/R = 7.2 W

Ps/Pp = 1.8/7.2 = 4

Question 3. An electrical appliance is rated 240 V, 500 W. How much current will this appliance draw? It is planned to spend Rs 200 for running this appliance in a month. If the per unit cost is Rs 4.60, how many hours can this appliance be operated in a month of 30 days?

Answer:

Let the running time per day be “n” hours.

Running cost for 30 days = n x 0.5 x 4.60 x 30 = 200

Hence, n = 200/ (0.5 x 4.60 x 30) = 2 hours 54 minutes.

Question 4. Observe the given table, in which the values of current flowing through a conductor for corresponding values of potential difference across the conductor.

HBSE Class 10 Biology Solutions For The Values Of Current Flowing Through A Conductor For Corresponding Values Of Potential

  1. Plot a graph between current and voltage.
  2. Calculate the resistance of the conductor.

Answer:

HBSE Class 10 Biology Solutions For Plot A Graph Between Current And Voltage

Resistance, R = V/l = 10/0.4 = 25 Ω

Question 5. Draw a schematic labelled diagram of a closed circuit which connects all the given components in series and is connected across a 12 V battery:

  1. 20 W lamp
  2. An ammeter
  3. A switch
  4. 10 Ω/100 W resistor

Answer:

HBSE Class 10 Biology Solutions For A Closed Circuit Which Connects All The Given Components In Series And Connected Across A 12V Battery

Question 6. The circuit diagram given below shows the combination of three resistors R1, R2 and R3:

HBSE Class 10 Biology Solutions For The Combination Of 3 Resistors

Find:

  1. Total resistance of the circuit.
  2. Total current flowing in the circuit.

Answer:

The total resistance of two resistors that are connected in parallel is

1/R’ = 1/3+1/6 1/R’ = 3/6 R’ = 2 Ω

Total resistance of the circuit = 2+4 = 6 Ω

Total current flowing through the circuit = V/total resistance

I = 12/6 = 2 A

Question 7. Study the circuit and find:

  1. Total resistance in arm CE
  2. Current in arm AB

HBSE Class 10 Biology Solutions For Total Resistance In Arm CE

Answer:

Total resistance in arm CE Rs = 6 + 6 = 12 Ω

Current in arm AB, I = 5V/17 0 = 0.29A

Question 8. Observe the table given below and answer the questions:

HBSE Class 10 Biology Solutions For The Symbols Used In Circuit Diagram

  1. What does the symbol (1) in the table given above represents?
  2. Which electrical component is represented when symbol (2) is used in circuit diagram?

Answer:

  1. Ammeter
  2. Rheostat

Question 9. Draw a circuit diagram to show how 3 bulbs can be lit from a battery so that 2 bulbs are controlled by the same switch while the third bulb has its own switch.

Answer:

HBSE Class 10 Biology Solutions For 3 Bulbs Can It Form A Battery

Question 10. An electric heater connected to the 240 V main supply draws a current of 10 A. Find:

  1. Power of the heater
  2. Energy consumed in 2 minutes

Answer:

P = VI P = 240 x 10 = 2400 W

Energy consumed = Vlt = 2400 x 120 = 28800 Joules

Question 11. When a 6Ω resistor is connected across the terminals of an 18 V battery, Calculate the coulomb passing through the resistor per second.

Answer:

V = IR; 18 =I x 6; I = 3A = 3C/s

Question 12. Name the black substance of pencil. Will the current flow through the electrical circuit when we use the sharpened ends of the pencil to complete the circuit?

Answer:

Graphite. Yes, current will flow through the circuit since graphite is a good conductor of electricity.

Question 13.

  1. Which gas is filled in an electric bulb and why?
  2. State Ohm’s law.

Answer:

  1. Argon or neon gas is filled in electric bulbs. These gases are used because they are inactive or inert. This prolongs the life of the filament,
  2. Ohm’s law: At a constant temperature, the current flowing through a conductor is directly proportional to the potential difference across its ends.

Question 14. The values of potential difference V applied across a resistor and the corresponding values of current I flowing in the resistor are given below:

HBSE Class 10 Biology Solutions For The Values Of Potential Difference IIn Volts

  1. What is the nature of the V-l graph plotted for the above values of potential difference and current?
  2. Which law is illustrated by such a type of graph?

Answer:

  1. Straight line graph
  2. Ohm’s law is illustrated when the nature of V-1 graph is a straight line.

Question 15. From the following electric circuit:

HBSE Class 10 Biology Solutions For Resultant Resistance And Current

Calculate:

  1. Resultant resistance and current
  2. Heat energy evolved when the circuit is switched on for 30 minutes

Answer:

For the given circuit,

  1. Resultant resistance, Req = 10.33 Ω; Total current, I = V/Req = 6/ 10.33 = 0.58 A
  2. Heat energy evolved in this circuit if it is switched on for 30 min. By Joule’s law of heating, H = l2Rt = (0.58)2 x 10.33 x 1800 = 6255J

Question 16.

  1. Define power,
  2. What is the SI unit of power?
  3. What is the commercial unit of electrical energy?

Answer:

  1. Work done per unit time is called power,
  2. The SI unit of power is Watt (W).
  3. The commercial unit of electrical energy is kilowatt-hour (kWh)

Question 17. Three resistors of 5Ω, 10Ω, and 20Ω, respectively, are connected to a battery of 12 V as shown in the circuit given below.

HBSE Class 10 Biology Solutions For The Three Resistors

Calculate:

  1. Current through each resistor
  2. Total current in the circuit
  3. Total resistance of the circuit

Answer:

Effective resistance for the parallel combination of 5 Ω and 20 Ω is given by 1/R = 20+5

20×5; R=4 . Hence, total resistance in the circuit 12 + 4 = 16 Ω.

Current drawn from the battery = 16/12 = 4/3 = 1.33 A

  1. Potential difference across the parallel combination of resistors is (4/3) x 4 = (16/3) V. Hence, current through 5 Ω resistor is (16/15) A and current through 20 Ω resistor is (16/60) =(4/15) A.
  2. Total current drawn from the battery (4/3) = 1.33 A
  3. Total resistance in the circuit 12 + 4 = 16

Question 18. An electric appliance of 1.5 kW power rating operates on a 220V main supply and has a current rating of 5 A. Is this fuse suitable for this electrical appliance? Explain.

Answer:

Power, P = 1.5 kW = 1500 W; V = 220 V; Current drawn, P = V x I

Therefore, I = P/V = 1500/220 = 6.81 A ≈ 7 A

The current drawn by the electrical appliance is 7 A, which is beyond the fuse rating capacity in the circuit, therefore, it will melt and break the circuit.

Hence, the fuse wire of 5 A rating would not be suitable for this electrical appliance.

Question 19.

  1. Write the expression for the relationship between resistance and resistivity.
  2. A 1.25 kW electric oven is connected to a 230 V power source. Calculate
    • Electric current passing through the oven
    • Resistance of the oven
    • Electric energy consumed in a 4-hour use of the oven

Answer:

  1. ρ = R x A/ I, where ρ = resistivity of material, R = Resistance of conductor, I = Length of conductor
  2. Power rating of the oven, P = 1.25 kW = 1250 W ; Potential difference of the power supply, V = 230 V
    • Power, P = V x I; Therefore, I = P/V = 1250/230 = 5.43 A
    • Resistance and power are related as P = V2/R; Therefore, R = V2/P = (230)2/1250 = 42.32 Ω
    • Energy consumed by the oven is E = P x t = 1.25 kW x 4 h = 5 kWh

Question 20. If three resistors of 6 Ω, 7 Ω, and 10 Ω are connected in series. Calculate the equivalent resistance in the circuit.

  1. What is the potential difference across the three resistors R1, R2, and R3 in a series combination if the potential across the circuit is V?
  2. What is the current along each of the resistors in a series combination if the current flowing in the circuit is I?
  3. Is the equivalent resistance in the parallel combination of resistors more than the value of each individual resistor?

Answer:

Equivalent resistance when the resistors are connected in series is Rs = R1 + R2 + R3

Therefore, Rs = 6 + 7 + 10 = 23 ohm

The potential difference (PD) across each resistor will be different in a series combination.

Across resistor R1, PD will be V1; for R2, PD will be V2, and for R3, PD will be V3.

The current across the three resistors in a series combination will be the same, i.e., T.

No, when the resistors are connected in parallel, the equivalent resistance is less than each of the individual resistors.

Question 21. An electric heater of resistance 10 Ω and a resistance wire of 8 Ω are connected in series with a 6 V battery. Find:

  1. Current through the circuit
  2. Potential difference across the electric heater
  3. Potential difference across an electric wire.

Answer:

1. V = IR

6 = I x (8+10)

I = 6/18 = 1/3A

2. Velectric heater= 1/3 x 10 = 10/3 V

3. Velectric wire=1/3×8 = 8/3 V

Question 22. In the given circuit, calculate

HBSE Class 10 Biology Solutions For Total Resistance Of The Circuit

Total resistance of the circuit

Current flowing through the circuit

Potential difference across the lamp and the resistor

  1. Total resistance R = R1 + R2 = 18 Ω + 6Ω = 24 Ω;
  2. Current flowing through the circuit, I = V/R = 6/24 = 0.25 A;
  3. Potential difference across the lamp, V2 = IR1 = 0.25 x 18 = 4; Potential difference across the resistor R2, V2 = IR2 = 0.25 x 6 = 1.5 V

Question 23. An electrical appliance is rated 200 V, 100 W. What is the resistance of the appliance? Five such appliances run simultaneously for 4 hours. What is the energy consumed? Calculate the cost of running these appliances if the per unit cost is Rs 4.60.

Answer:

Rating of the electrical appliance is 200 V-100 W. Since P = V2/R.

Therefore, R = V2/P = (200)2/100 = 400 Ω.

Total power consumed by 5 bulbs will be 500 W.

Hence, for a four-hour operation, the energy consumed is E = P x t

E = 500×4 = 2000 Wh = 2kWh

Cost of electricity per unit is Rs 4.60. Hence, the total cost is Cost = 2 x 4.60 = Rs 9.2

Question 24.

  1. What is the function of an earth wire in electrical instruments? Why is it necessary to earth the metallic electric appliances?
  2. Explain what is short-circuiting and overloading in an electric supply.

Answer:

  1. The main function of earth wire is to absorb excess electricity passing through the appliance and to prevent overloading of live wire. The metallic appliances are earthed because whenever there is leakage of current, the leaked current passes thro ugh the earth wire to earth.
  2. Short Circuit: When live wire comes in contact with Neutral wire then the resistance becomes zero and with Ohm’s law current becomes infinite. Overloading connected sWhen a large number of home appliances are simultaneously to the same socket and they withdraw large amounts of current than mentioned on the switch.

Chapter 3 Electricity Long Questions And Answers

Question 1.

  1. The domestic electric supply of a house is through a 15 A fuse. Along with a 2000 W heater, how many 100 W bulbs can be used simultaneously in the house without causing overload? The domestic supply is 220 V all appliances are rated for 220 V.
  2. What is meant by one cycle of alternating current? and
  3. In a typical lightning strike, 2.5 C flows from the cloud to the ground in 0.20 milliseconds. What is the current during the strike?

Answer:

The current through the heater can be calculated as

I = P/V = 2000/ 220 = 100/11

Now the current rating through one bulb can be calculated in a similar way

i = P/V = 100/220 = 5/11

Now if there are n bulbs, then for the fuse the total current should be

15 = I + ni

15 = 100/11 + n x 5/11

15 x 11 -100 = 5n

n = 165-100/5 =13

Therefore, a total of 13 bulbs can be used simultaneously.

Question 2.

  1. Derive an expression for the heat produced in a conductor of resistance R when a current I flows through it for time t.
  2. Two identical resistors of resistance R are connected in series with a battery of potential difference V for time t. The resistors are then connected in parallel with the same battery for the same time t. Compare the heat produced in the two cases.

Answer:

A conductor offers resistance to the flow of current. Hence, work must be continuously done by the current to keep itself flowing.

When an electric charge Q moves against a potential difference V, the work done is

W = QV ….(1)

From the definition of current, Q = I x t ….(2)

From Ohm’s law, V = I x R …(3)

Therefore, W = I x t x I x R

Assuming that all this work goes in producing heat energy (H).

Therefore, the heat produced (H) in a conductor of resistance ‘R’ when current T is flowing for time ‘t’ is

H = I2Rt

2. When resistors are connected in series:

Rs + R + R = 2R

Therefore, Hs = V2/Rs = V2/2R …(1)

When resistors are connected in parallel:

∴ \(\frac{1}{R p}=\frac{1}{R}+\frac{1}{R}=\frac{2}{R}\)

Therefore, Rp = R/2

Hs = V2/ Rs = 2V2/R ….(2)

From (1) and (2)

∴ \(\frac{\mathrm{Hs}}{\mathrm{Hp}}=\frac{1}{4}\)

Therefore, Hp = 4Hs

Question 3. Derive the expression for equivalent resistance if three resistors R1, R2 and R3 are connected in a parallel combination.

Answer:

Derivation for equivalent resistance if three resistors are connected in parallel:

HBSE Class 10 Biology Solutions For Three Resistors Are Connected In Parallel

Potential difference in a parallel circuit remains the same across all resistors.

Current is the sum of the currents across all the individual resistors.

I = l1 + I2 + I3 ….(1)

Let Rp be the resultant resistance of the circuit. On applying Ohm’s law to the entire circuit,

l = V/Rp ….(2)

Now applying Ohm’s law to individual resistances,

⇒ \(\left.\begin{array}{l}
\mathrm{I}_1=\mathrm{V} / \mathrm{R}_1 \\
\mathrm{I}_2=\mathrm{V} / \mathrm{R}_2 \\
\mathrm{I}_3=\mathrm{V} / \mathrm{R}_3
\end{array}\right]…..(3)\)

From equations (1), (2) and (3),

V/ Rp = V/R1 + V/R2 +V/R3

Cancelling “V” on both sides, we get

1/ Rp =1/R1 + 1/R2 + 1/R3

Question 4. A circuit has a fuse of 5 A. What is the maximum number of 110 W (220 V) bulbs that can be safely used in the circuit?

Answer:

Suppose ‘n’ bulbs can be used safely.

Power of ‘n’ bulbs, P = 110 x n Watt

Potential difference, V = 220 V; Current, I = 5 A

Power, P = V x I

110 xn =220 x 5/110

Thus, the maximum number of bulbs that can be connected safely in the circuit is 10.

Question 5.

  1. List the factors on which the resistance of a conductor depends.
  2. A 4-kW heater is connected to a 220-V power source. Calculate
    • Electric current passing through the heater
    • Resistance of the heater
    • Electric energy consumed in a 2-hour use of the heater

Answer:

  1. Power, P = VI
    Therefore, I = P/V = 4000/220 = 18.18 A
  2. P = V2/R
    Therefore, R = V2/P = (220)2/4000 = 12.1 Ω
  3. E = P x t = 4 kW x 2 h = 8kWh

Question 6. What is meant by the statement ‘the potential difference between two points is 1 V? Study the circuit shown in which three identical bulbs B1, B2, and B3 are connected in parallel with a battery of 4.5 V.

HBSE Class 10 Biology Solutions For The Potential Difference Between Two points Three identical Bulbs

  1. What will happen to the glow of the other two bulbs if bulb B3 gets fused?
  2. If the wattage of each bulb is 1.5 W, what readings will the ammeter A show when all the three bulbs glow simultaneously?
  3. Find the total resistance of the circuit.

Answer:

Potential difference of 1 volt means that one joule of work is done to move a charge of one coulomb from one point to another.

  1. If bulb B3 gets fused, then the other two bulbs will continue glowing with the same brightness.
  2. When the bulbs are in parallel, wattage will be added (4.5 W) and the ammeter reading would be 45/45 = 1.0 ampere.
  3. Since the ammeter reading is 1.0 ampere, the resistance of the combination = 4.5V/1.0A = 4.5 Ω

Question 7. In the given circuit, A, B, C and D are four lamps connected with a battery of 60V. Analyse the circuit to answer the following questions.

HBSE Class 10 Biology Solutions For Four Lamps Connected With A Battery Of 60 V

  1. What kind of combination are the lamps arranged in (series or parallel)?
  2. Explain with reference to your above answer, what are the advantages (any two) of this combination of lamps?
  3. Explain with proper calculations which lamp glows the brightest?
  4. Find out the total resistance of the circuit.

Answer:

1. The lamps are connected in parallel.

2. Advantages of parallel combination: If one lamp stops working, it will not affect the working of the other lamps. – Potential difference across each lamp will be same.

3. The lamp with the highest power will glow the brightest.

P=VI

Even if the potential difference across all the lamps are same, lamp C has the highest current.

Hence, for Lamp C, P=5 x 60 Watt = 300 W.

Thus, lamp C glows the brightest.

4. Total current, I = 3+4+5+3 A = 15A

Voltage = 60 V

Since V = IR,

Hence R = V/I = 60/15 A = 4 A.

Haryana Board Class 10 Physics Solutions For Sample Question Paper

Haryana Board Class 10 Physics Solutions For Sample Question Paper

Question 1. The change in colour of the moist litmus paper in the given set-up is due to 

HBSE Class 10 Biology Solutions For The Change In Colour Of The Moist litmus Paper

  • Presence of acid
  • Presence of base
  • Presence of H+(aq) in the solution
  • Presence of Litmus which acts as an indicator
  1. 1 and 2
  2. Only 2
  3. Only 3
  4. Only 4.

Answer: 3. Only 3

Question 2. In the redox reaction

MnO2 + 4HCI → MnCI2 + 2H2O + Cl2

  1. MnO2 is reduced to MnCI2 & HCI is oxidized to H2O
  2. MnO2 is reduced to MnCI2 & HCI is oxidized to Cl2
  3. MnO2 is oxidized to MnCI2 & HCI is reduced to Cl2
  4. MnO2 is oxidized to MnCI2 & HCI is reduced to H2O

Answer: 2. MnO2 is reduced to MnCI2 & HCI is oxidized to Cl2

Question 3. Which of the following is the correct observation of the reaction shown in the setup?

HBSE Class 10 Biology Solutions For Magnesium Ribbon Burns With Brilliant White Light

  1. Brown powder of Magnesium oxide is formed.
  2. Colourless gas, which turns lime water milky, is evolved.
  3. Magnesium ribbon burns with brilliant white light.
  4. Reddish brown gas with a smell of burning Sulphur has evolved.

Answer: 3. Magnesium ribbon burns with brilliant white light.

Question 4. With reference to the four gases CO2, CO, Cl2 and O2, which one of the options in the table is correct?

HBSE Class 10 Biology Solutions For Four Gases CO2

Answer: 2. HBSE Class 10 Biology Solutions For Four Gases Acidic Oxide

Question 5. On placing a copper coin in a test tube containing green ferrous sulphate solution, it will be observed that the ferrous sulphate solution

  1. Turns blue, and a grey substance is deposited on the copper coin.
  2. Turns colourless, and a grey substance is deposited on the copper coin.
  3. Turns colourless, and a reddish-brown substance is deposited on the copper coin.
  4. Remains green with no change in the copper coin.

Answer: 4. Remains green with no change in the copper coin.

Question 6. Anita added a drop each of diluted acetic acid and diluted hydrochloric acid on pH paper and compared the colours. Which of the following is the correct conclusion?

  1. pH of acetic acid is more than that of hydrochloric acid.
  2. pH of acetic acid is less than that of hydrochloric acid.
  3. Acetic acid dissociates completely in aqueous solution.
  4. Acetic acid is a strong acid

Answer: 1. pH of acetic acid is more than that of hydrochloric acid.

Question 7. The formulae of four organic compounds are shown below. Choose the correct option

HBSE Class 10 Biology Solutions For The Formule Of Four Organic Compounds

  1. A and B are unsaturated hydrocarbons
  2. C and D are saturated hydrocarbons
  3. Addition of hydrogen in the presence of a catalyst changes A to C
  4. The addition of potassium permanganate changes B to D

Answer: 3. Addition of hydrogen in the presence of a catalyst changes A to C

Question 8. In the given transverse section of the leaf, identify the layer of cells where maximum photosynthesis occurs.

HBSE Class 10 Biology Solutions For Transverse Section Of The Leaf Identify the Layer Of Cells where Maximum Photosynthesis Occure

  1. 1, 2
  2. 2, 3
  3. 3, 4
  4. 1, 4

Answer: 2. 2, 3

Question 9. Observe the experimental setup shown below. Name the chemical indicated as ‘X’ that can absorb the gas which is evolved as a byproduct of respiration.

HBSE Class 10 Biology Solutions For A Byproduct Of Respiration

  1. NaOH
  2. KOH
  3. Ca (OH)2
  4. K2C03

Answer: 2. KOH

Question 10. If a tall pea plant is crossed with a pure dwarf pea plant, then what percentage of F1 and F2 generations, respectively, will be tall?

  1. 25%, 25%
  2. 50%, 50%
  3. 75%,100%
  4. 100%, 75%

Answer: 4. 100%, 75%

Question 11. Observe the three figures given below. Which of the following depicts tropic movements appropriately?

HBSE Class 10 Biology Solutions For Depict Tropic Movements Appropriately

  1. B and C
  2. A and C
  3. B only
  4. C only

Answer: 4. C only

Question 12. The diagram shown below depicts pollination. Choose the options that will show a maximum variation in the offspring.

HBSE Class 10 Biology Solutions For Depict Pollination A Maximum Variation In The Offspring

  1. A, B and C
  2. B and D
  3. B, C and D
  4. A and C

Answer: 2. B, C and D

Question 13. A complete circuit is left on for several minutes, causing the connecting copper wire to become hot. As the temperature of the wire increases, the electrical resistance of the wire

  1. Decreases
  2. Remains the same
  3. Increases
  4. Increases for some time and then decreases.

Answer: 3. Increases

Question 14. A copper wire is held between the poles of a magnet.

HBSE Class 10 Biology Solutions For A Copper Wire Is Held Between The Poles Of A Magnet

The current in the wire can be reversed. The pole of the magnet can also be changed over. In how many of the four directions shown can the force act on the wire?

  1. 1
  2. 2
  3. 3
  4. 4

Answer: 2. 2

Question 15. Plastic insulation surrounds a wire having a diameter d and a length I as shown above. A decrease in the resistance of the wire would be produced by an increase in the

HBSE Class 10 Biology Solutions For Plastic Insulation Surrounds A Wire Having Diameter

  1. Length I of the wire
  2. Diameter d of the wire
  3. Temperature of the wire
  4. Thickness of the plastic insulation

Answer: 2. Diameter d of the wire

Question 16. Which of the following patterns correctly describes the magnetic field around a long straight wire carrying current?

  1. Straight lines perpendicular to the wire,
  2. Straight lines parallel to the wire,
  3. Radial lines originating from the wire,
  4. Concentric circles centred around the wire.

Answer: 4. Concentric circles centred around the wire.

Assertion – Reasoning Based Questions And Answers

These consist of two statements: Assertion (A) and Reason (R). Answer these questions by  selecting the appropriate option given below:

  1. Both A and R are true, and R is the correct explanation of A
  2. Both A and R are true, and R is not the correct explanation of A
  3. A is true, but R is false
  4. A is False, but R is true

Question 1. Assertion: Silver bromide decomposition is used in black and white photography.

Reason: Light provides energy for this exothermic reaction.

Answer: 3. A is true but R is false

Question 2. Assertion: Height in pea plants is controlled by efficiency of enzymes and is thus genetically controlled.

Reason: Cellular DNA is the information source for making proteins in the cell.

Answer: 1. Both A and R are true, and R is the correct explanation of A.

Question 3. Assertion: Amphibians can tolerate mixing of oxygenated and deoxygenated blood.

Reason: Amphibians are animals with a chambered heart.

Answer: 3. A is true but R is false.

Question 4. Assertion: On freely suspending a current-carrying solenoid, it comes to rest in the Geographical N-S direction.

Reason: One end of the current-carrying straight solenoid behaves as a North pole and the other end as a South pole, just like a bar magnet.

Answer: 1. Both A and R are true, and R is the correct explanation of A.

Very Short Answer Questions And Answers

Question 1. A clear solution of slaked lime is made by dissolving Ca(OH)2 in an excess of water. This solution is left exposed to the air. The solution slowly goes milky as a faint white precipitate forms. Explain why a faint white precipitate forms, and support your response with the help of a chemical equation.
OR
Keerti added dilute Hydrochloric acid to four metals and recorded her observations as shown in the table given below:

HBSE Class 10 Biology Solutions For A Clear Solution Of Slaked Lime Is Made By Dissolving In An Excess Of Water

Select the correct observation(s) and give chemical equation(s) of the reaction involved.

Answer:

Calcium hydroxide reacts with Carbon dioxide present in the atmosphere to form Calcium carbonate which results in milkiness or white ppt or Formation of Calcium carbonate \(\mathrm{Ca}(\mathrm{OH})_2+\mathrm{CO}_2 \rightarrow \mathrm{CaCO}_3+\mathrm{H}_2 \mathrm{O}\)

or

⇒ \(\mathrm{Fe}+\mathrm{HCl} \rightarrow \mathrm{FeCl}_2 / \mathrm{FeCl}_3+\mathrm{H}_2\) (No deduction for balancing/ states)

∴ \(\mathrm{Zn}+\mathrm{HCl} \rightarrow \mathrm{ZnCl}_2+\mathrm{H}_2\)

Question 2. How is the mode of action in the beating of the heart different from reflex actions? Give four examples.

Answer:

HBSE Class 10 Biology Solutions For The Mode Of Action In Beating Of the Heart Different From Reflex Actions

Question 3. Patients whose gallbladders are removed are recommended to eat less oily food. Why?

Answer:

The gallbladder stores bile, which helps in the emulsification of lipids.

In the absence of stored bile, emulsification of fats will be negligible or affected or less and thus fat digestion will be slow. Hence, there are such diet restrictions.

Question 4. Name the substances, other than water, that are reabsorbed during urine formation. What are the two parameters that decide the amount of water that is reabsorbed in the kidney?

Answer:

Glucose, amino acids, salts and a major amount of water are selectively reabsorbed as the urine flows along the tube.

The amount of water reabsorbed depends on how much excess water there is in the body and on how much dissolved waste there is to be excreted.

Question 5. State the phenomena observed in the diagram. Explain with reference to the diagram, which of the two lights mentioned above will have the higher wavelength?

HBSE Class 10 Biology Solutions For the Phenomena The Lights Mentioned Will have Higher Wavelength

How will you use two identical prisms so that a narrow beam of white light incident on one prism emerges out of the second prism as white light? Draw the diagram.

Answer:

Dispersion- The splitting of white light into seven colours on passing through a prism.

Velocity is directly proportional to wavelength, given a constant frequency. So yellow will have a greater wavelength than blue as the velocity of yellow light is greater than blue.
or
The angles of deflection of the two prisms need to be equal and opposite. While the first prism splits the light in the seven colours due to different angles of deflection, the second prism combines the spectrum along a single ray and the colours again combine to give white light as the emergent light.

HBSE Class 10 Biology Solutions For Angle Of Deflections Of The Two Prisms Need to Be Equal And Opposite

Question 6. A lot of waste is generated in the neighborhood. However, almost all of it is biodegradable. What impact will it have on the environment or human health?

Answer:

Excess generation of biodegradable wastes can be harmful as – Its decomposition is a slow process leading to production of foul smell and gases.

It can be the breeding ground for germs that create unhygienic conditions.

Short Answer Questions And Answers

Question 1.

HBSE Class 10 Biology Solutions For The Types Of reaction The Type In the Form Of A Balanced Chemical Equations

Identify the types of reaction mentioned above in (1) and (2). Give one example for each type in the form of a balanced chemical equation.

Answer:

  1. Displacement
    • \(\mathrm{Fe}(\mathrm{~s})+\mathrm{CuSO}_4(\mathrm{aq}) \rightarrow \mathrm{FeSO}_4(\mathrm{aq})+\mathrm{Cu}(\mathrm{~s})\)
    • \(\mathrm{Zn}(\mathrm{~s})+\mathrm{CuSO}_4(\mathrm{aq}) \rightarrow \mathrm{ZnSO}_4(\mathrm{aq})+\mathrm{Cu}(\mathrm{~s})\)
    • \(\mathrm{Pb}(\mathrm{~s})+\mathrm{CuCl}_2(\mathrm{aq}) \rightarrow \mathrm{PbCl}_2(\mathrm{aq})+\mathrm{Cu}(\mathrm{~s})\)
  2. Double displacement
    • \(\mathrm{Na}_2 \mathrm{SO}_4(\mathrm{aq})+\mathrm{BaCl}_2(\mathrm{aq}) \rightarrow \mathrm{BaSO}_4(\mathrm{~s})+2 \mathrm{NaCl}(\mathrm{aq})\)

Question 2.

HBSE Class 10 Biology Solutions For The gasses Evolced At The Anode And Cathode In The Experimental Set Up

  1. Identify the gases evolved at the anode and cathode in the above experimental set-up.
  2. Name the process that occurs. Why is it called so?
  3. Illustrate the reaction of the process with the help of a chemical equation.

Answer:

  1. Anode: Chlorine; Cathode: Hydrogen
  2. Chlor alkali process as the products obtained are alkali, chlorine gas and hydrogen gas Electric current
  3. \(2 \mathrm{NaCl}(\mathrm{aq})+2 \mathrm{H}_2 \mathrm{O}(\mathrm{I}) \longrightarrow 2 \mathrm{NaOH}(\mathrm{aq})+\mathrm{Cl}_2(\mathrm{~g})+\mathrm{H}_2(\mathrm{~g})\)

Question 3. The leaves of a plant were covered with aluminium foil, how would it affect the physiology of the plant?
OR
How is lymph an important fluid involved in transportation? If lymphatic vessels get blocked, how would it affect the human body? Elaborate.

Answer:

No photosynthesis will occur, so no glucose will be made. Also, no respiration will take place as no Oxygen will be taken in.

No transpiration will occur, so there will be no upward movement of water or minerals from the soil as there will be no transpirational pull.

Temperature regulation of the leaf surface will be affected.
or
Lymph carries digested and absorbed fat from the intestine and drains excess fluid from the extracellular space back into the blood.

Blockage of the lymphatic system will lead to water retention and poor fat absorption in the body.

Question 4. Rohit wants to have an erect image of an object using a converging mirror of focal length 40 cm.

  1. Specify the range of distances where the object can be placed in front of the mirror. Justify.
  2. Draw a ray diagram to show image formation in this case.
  3. State one use of the mirror based on the above kind of image formation.

Answer:

1. The object has to be placed at a distance between 0 – 40 cm. This is because the image is virtual, erect and magnified when the object is placed between F and P.

HBSE Class 10 Biology Solutions For Image Is Virtual erect And Magnified When The Object Is Placed Between F and P

3. Used as a shaving mirror or used by dentists to get an enlarged image of teeth

Question 5.

  1. A lens of focal length 5 cm is being used by Debashree in the laboratory as a magnifying glass. Her least distance of distinct vision is 25 cm.
    • What is the magnification obtained by using the glass?
    • She keeps a book at a distance 10 cm from her eyes and tries to read. She is unable to read. What is the reason for this?
  2. Ravi kept a book at a distance of 10 cm from the eyes of his friend Hari. Hari is not able to read anything written in the book. Give reasons for this?

Answer:

1. Image distance, v = -25 cm

Focal length, f = 5 cm

Magnification, m =?

From the lens formula

⇒ \(\frac{1}{\mathrm{f}}=\frac{1}{v}-\frac{1}{u}\)

Therefore,

⇒ \(\frac{1}{u}=\frac{1}{-25}-\frac{1}{5}=-\frac{6}{25}\)

Object distance, \(u=-\frac{25}{6}\)

Since \(m=\frac{v}{u}=-\frac{25 x 6}{-25}=6 \mathrm{~cm}\)

2. This is because the least distance of distinct vision is 25 cm.

Question 6. A student fixes a white sheet of paper on a drawing board. He places a bar magnet in the centre and sprinkles some iron filings uniformly around the bar magnet. Then he taps gently and observes that iron filings arrange themselves in a certain pattern.

  1. Why do iron filings arrange themselves in a particular pattern?
  2. Which physical quantity is indicated by the pattern of field lines around the bar magnet?
  3. State any two properties of magnetic field lines.

OR

A compass needle is placed near a current-carrying wire. State your observations for the following cases and give reasons for the same in each case:

  1. The magnitude of the electric current in the wire is increased.
  2. The compass needle is displaced away from the wire.

Answer:

  1. When iron filings are placed in a magnetic field around a bar magnet, they behave like tiny magnets. The magnetic force experienced by these tiny magnets makes them rotate and align themselves along the direction of the field lines.
  2. The physical property indicated by this arrangement is the magnetic field produced by the bar magnet.
  3. Magnetic field lines never intersect; magnetic field lines are closed curves.

OR

  1. The deflection in the compass needle increases as the Magnetic field of the current-carrying conductor is directly proportional to the current flowing through it.
  2. The deflection in the needle decreases as the magnetic field is inversely proportional to the perpendicular distance from the wire.

Question 7. Why is damage to the ozone layer a cause for concern? What are its causes, and what steps are being taken to limit this damage?

Answer:

Damage to the ozone layer is a cause for concern because the ozone layer shields the surface of Earth from harmful UV radiations from the sun, which cause skin cancer in human beings.

Synthetic chemicals like chlorofluorocarbons (CFCs), which are used as refrigerants and in the fire

  • Extinguishers are the main reason for the depletion of the ozone layer.
  • Steps taken to limit this damage – Many developing and developed countries have signed and are obeying the directions of UNEP (United Nations Environment Programme) to freeze or limit the production and usage of CFCs at 1986 levels.

Long Answer Questions And Answers

Question 1. Shristi heated Ethanol with a compound A in presence of a few drops of concentrated sulphuric acid and observed a sweet smelling compound B is formed. When B is treated with sodium hydroxide it gives back Ethanol and a compound C.

  1. Identify A and C
  2. Give one use each of compounds A and B.
  3. Write the chemical reactions involved and name the reactions.

OR

  1. What is the role of concentrated Sulphuric acid when it is heated with Ethanol at 443 K. Give the reaction involved.
  2. Reshu by mistake, forgot to label the two test tubes containing Ethanol and Ethanoic acid.
  3. Suggest an experiment to identify the substances correctly? Illustrate the reactions with the help of chemical equations.

Answer:

1. A – Ethanoic acid Or any other carboxylic acid , C- Sodium salt of ethanoic acid or any other carboxylic acid or sodium ethanoate

2. Use of A- dil solution used as vinegar in cooking or preservative in pickles
Use of B-making perfumes, flavoring agent

3. \(\mathrm{CH}_3 \mathrm{COOH}+\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH} \longrightarrow \mathrm{CH}_3 \mathrm{COOC}_2 \mathrm{H}_5+\mathrm{H}_2 \mathrm{O}\)

⇒ \(\mathrm{CH}_3 \mathrm{COOC}_2 \mathrm{H}_5+\mathrm{NaOH} \longrightarrow \mathrm{CH}_3 \mathrm{COONa}+\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}\)

Or

1. Sulphuric acid acts as a dehydrating agent

Sulphuric Acid Acts As dehydrating Agent

2. By reaction with sodium carbonate or bicarbonate 1M with the samples, ethanol will not react whereas ethanoic acid gives brisk effervescence.

⇒ \(2 \mathrm{CH}_3 \mathrm{COOH}+\mathrm{Na}_2 \mathrm{CO}_3 \longrightarrow \mathrm{CH}_3 \mathrm{COONa}+\mathrm{H}_2 \mathrm{O}+\mathrm{CO}_2\)

or

∴ \(\mathrm{CH}_3 \mathrm{COOH}+\mathrm{NaHCO}_3 \longrightarrow \mathrm{CH}_3 \mathrm{COONa}+\mathrm{H}_2 \mathrm{O}+\mathrm{CO}_2\)

Question 2.

  1. Why is it not possible to reconstruct the whole organism from a fragment in complex multicellular organisms?
  2. Sexual maturation of reproductive tissues and organs is are necessary link for reproduction. Elucidate.

OR

  1. How are variations useful for species if there is a drastic alteration in the niches?
  2. Explain how the uterus and placenta provide necessary conditions for the proper growth and development of the embryo after implantation.

Answer:

  1. The reason is that many multi-cellular organisms are not simply a random collection of cells. Specialised cells are organised as tissues, and tissues are organised into organs, which then have to be placed at definite positions in the body. Therefore, cell-by-cell division would be impractical.
  2. Sexual maturation of reproductive tissues is a necessary link for reproduction because of the need for specialised cell called germ-cells to participate in sexual reproduction. The body of the individual organism has to grow to its adult size, the rate of general body growth begins to slow down, reproductive tissues begin to mature.

A whole new set of changes in the appearance of the body takes place like change in body proportions, new features appear. This period during adolescence is called puberty. There are also changes taking place that are different between boys and girls. In girls, breast size begins to increase, with darkening of the skin of the nipples at the tips of the breasts. Also, girls begin to menstruate at around this time. Boys begin to have new thick hair growth on the face and their voices begin to crack.

or

  1. If the niche were drastically altered, the population could be wiped out. However, if some variations were to be present in a few individuals in these populations, there would be some chance for them to survive. Variation is thus useful for the survival of species over time..
  2. The lining of the uterus thickens and is richly supplied with blood to nourish the growing embryo.
    • The embryo gets nutrition from the mother’s blood with the help of placenta. It is embedded in the uterine wall.
    • It contains villi on the embryo’s side of the tissue. On the mother’s side are blood spaces, which surround the villi.
    • This provides a large surface area for glucose and oxygen to pass from the mother to the embryo. The developing embryo will also generate waste substances which can be removed by
      transferring them into the mother’s blood through the placenta.
    • The child is born as a result of rhythmic contractions of the muscles in the uterus.

Question 3.

HBSE Class 10 Biology Solutions For A Schematic Diagram Of A Household Circuit

The diagram above is a schematic diagram of a household circuit. The house shown in the above diagram has 5 usable spaces where electrical connections are made. For this house, the mains have a voltage of 220 V and the net current coming from the mains is 22A.

  1. What is the mode of connection to all the spaces in the house from the mains?
  2. The spaces 5 and 4 have the same resistance and spaces 3 and 2 have respective resistances of 20Ω and 30Ω. Space 1 has a resistance double that of space 5. What is the net resistance for space 5.
  3. What is the current in space 3?
  4. What should be placed between the main connection and the rest of the house’s electrical appliances to save them from accidental high electric current?

Answer:

1. All spaces are connected in parallel.

2. Let Resistance of Space 5 and 4 be R ohms respectively

Resistance of Space 1 = 2 R ohms

Resistance of Space 2 = 30 ohms

Resistance of Space 3 = 20 ohms

Current = 22 A V= 220 V

Total Resistance= V/I

⇒ \(\frac{1}{R 1}+\frac{1}{R 2}+\frac{1}{R 3}+\frac{1}{R 4}+\frac{1}{R 5}+\frac{1}{R 6}=\frac{1}{R e q}\)

⇒ \(\frac{1}{2 R}+\frac{1}{30}+\frac{1}{20}+\frac{1}{R}+\frac{1}{R}=\frac{1}{R e q}\)

∴ \(\frac{150}{60 R}+\frac{5 R}{60 R}=\frac{1}{R e q}\)

R = 150.

Case-Based Or Data -Based Questions And Answers

Question 1. Two students decided to investigate the effect of water and air on the iron object under identical experimental conditions. They measured the mass of each object before placing it partially immersed in 10 ml of water. After a few days, the objects were removed, dried, and their masses were measured. The table shows their results.

HBSE Class 10 Biology Solutions For The Effect Of Water And Air On Iron Object under Identical Experimental Conditions

1. What might be the reason for the varied observations of the two students?

2. In another setup up the students coated iron nails with zinc metal and noted that iron nails coated with zinc prevent rusting. They also observed that zinc initially acts as a physical barrier, but an extra advantage of using zinc is that it continues to prevent rusting even if the layer of zinc is damaged. Name this process of rust prevention and give any two other methods to prevent rusting.

or

2. In which of the following applications of Iron will rusting occur most? Support your answer with a valid reason.

HBSE Class 10 Biology Solutions For Iron Bucket Electroplated With Zinc

A – Iron Bucket electroplated with Zinc B – Electricity cables having iron wires covered with aluminium C – Iron hinges on a gate D – Painted iron fence.

Answer:

1. Rusting occurs in both A and B, so there is an increase in mass.

As the surface area of B is greater, the extent of rusting is greater

2. Galvanisation

Oiling or greasing or painting or alloying or chromium plating or any other.

or

2. C – Iron hinges on a gate – Iron is in contact with both atmospheric oxygen and moisture or water vapour.

Question 2. Pooja has green eyes while her parents and brother have black eyes. Pooja’s husband Ravi has black eyes, while his mother has green eyes and his father has black eyes.

1. On the basis of the above given information, is the green eye colour a dominant or recessive trait? Justify your answer.

2. What is the possible genetic makeup of Pooja’s brother’s eye colour?

3. What is the probability that the offspring of Pooja and Ravi will have green eyes? Also, show the inheritance of eye colour in the offspring with the help of a suitable cross.

OR

4. 50% of the offspring of Pooja’s brother are green-eyed. With the help of cross show how this is possible.

Answer:

1. Yes, green eye colour is recessive as it will express only in a homozygous condition.

2. BB, Bb c. bb*Bb

∴ \(\begin{array}{|l|l|l|}
\hline & \mathrm{B} & \mathrm{~b} \\
\hline \mathrm{~b} & \mathrm{Bb} & \mathrm{bb} \\
\hline \mathrm{~b} & \mathrm{Bb} & \mathrm{bb} \\
\hline
\end{array}\)

Genetic cross – 50% of the offspring can have green eye colour

OR

3. Brother is heterozygous (Bb) and wife is green(bb)

Wife bb*Bb brother

∴ \(\begin{array}{|l|l|l|}
\hline & \mathrm{B} & \mathrm{~b} \\
\hline \mathrm{~b} & \mathrm{Bb} & \mathrm{bb} \\
\hline \mathrm{~b} & \mathrm{Bb} & \mathrm{bb} \\
\hline
\end{array}\)

50% of the offsprings can have green eye colour as per the cross shown.

Question 3.

HBSE Class 10 Biology Solutions For A Specialized Slide Projector

The above images are of a specialized slide projector. Slides are small transparencies mounted in sturdy frames, ideally suited to magnification and projection, since they have a very high resolution and a high image quality. There is a tray where the slides are to be put into a particular orientation so that the viewers can see the enlarged erect images of the transparent slides. This means that the slides will have to be inserted upside down in the projector tray. To show her students the images of insects that she investigated in the lab, Mrs. Iyer brought a slide projector. Her slide projector produced a 500 times enlarged and inverted image of a slide on a screen 10 m away.

1. Based on the text and data given in the above paragraph, what kind of lens must the slide projector have?

2. If v is the symbol used for image distance and u for object distance, then with one reason state what will be the sign for v/u in the given case?

3. A slide projector has a convex lens with a focal length of 20 cm. The slide is placed upside down 21 cm from the lens. How far away should the screen be placed from the slide projector’s lens so that the slide is in focus?

OR

3. When a slide is placed 15 cm behind the lens in the projector, an image is formed 3 m in front of the lens. If the focal length of the lens is 14 cm, draw a ray diagram to show image formation. (not to scale)

Answer:

1. Convex Lens

2. Negative as the image is real and inverted.

3. l/f = 1/v-1/u 1/20

= 1/v-1/-20 1/v

= 1/20 – 1/21

= (21 – 20)/420

= 1/420

v = 420 cm

or

HBSE Class 10 Biology Solutions For The Ray Diagram

Haryana Board Class 10 Physics Solutions For Chapter 2 The Human Eye And The Colourful World

Haryana Board Class 10 Physics  Chapter 2 The Human Eye And The Colourful World Multiple Choice Questions And Answers

Question 1. In which part of the human eye the image of an object is formed?

  1. Iris
  2. Pupil
  3. Retina
  4. Cornea

Answer: 3. Retina

Question 2. A person gets out into the sunlight from a dark room. How does his pupil regulate and control the light entering the eye?

  1. The size of the pupil will decrease, and less light will enter the eye
  2. The size of the pupil will decrease, and more light will enter the eye
  3. The size of the pupil will remain the same, but more light will enter the eye
  4. The size of the pupil will remain the same, but less light will enter the eye

Answer: 1. The size of the pupil will decrease, and less light will enter the eye

Question 3. A person is seeing an object closer to his eyes. What changes in his eyes will take place?

  1. The pupil size will expand
  2. The ciliary muscles will contract
  3. The focal length of the eye lens will increase
  4. The light entering the eye will be more

Answer: 2. The ciliary muscles will contract

Question 4. A person standing at point Y is watching a car coming from point X to 0 as shown.

HBSE Class 10 Physics Solutions For Chapter 2 A Person Standing At Point YIs Watching A Car Coming From A point X

The table shows the variation in the parts of the eye while seeing the car at X and O

  1. At X, the focal length is higher than that at O.
  2. At O, the focal length is higher than that at X.
  3. At X, the Ciliary muscle is thicker than at O.
  4. At 0, the Ciliary muscle is thicker than at X.

Which change in the person’s eye would likely to occur while watching the car?

  1. 1 and 3
  2. 1 and 4
  3. 2 and 3
  4. 2 and 4

Answer: 2. 1 and 4

Question 5. A person went for a medical check-up and found that the curvature of his eye lens is increasing. Which defects is he likely to suffer from?

  1. Myopia
  2. Cataract
  3. Presbyopia
  4. Hypermetropia

Answer: 1. Myopia

Question 6. The image shows the ray diagram of a defective eye.

HBSE Class 10 Physics Solutions For Chapter 2 The Ray Diagram Of A Defected Eye

Which option shows the correction of the defect of the eye?

HBSE Class 10 Physics Solutions For Chapter 2 The Defect Of he Eye

Answer: 4. HBSE Class 10 Physics Solutions For Chapter 2 Defect of the eye

Question 7. Which image shows the deviation of light in a prism?

HBSE Class 10 Physics Solutions For Chapter 2 The Deviation Of Light In A Prism

Answer:  HBSE Class 10 Physics Solutions For Chapter 2 Light In A Prism

Question 8. The image shows a light ray incident on a glass prism.

HBSE Class 10 Physics Solutions For Chapter 2 A Light Ray Incident On A Glass Prism

The various angles are labelled in the image. Which angle shows the angle of incidence and angle of refraction, respectively?

  1. A and D
  2. B and E
  3. C and F
  4. D and F

Answer: 1. A and D

Question 9. The image shows the dispersion of the white light in the prism.

HBSE Class 10 Physics Solutions For Chapter 2 The Dispersion Of The White Light In The Prism

What will be the colours of the X, Y and Z?

  1. X: red; Y: green; Z: violet
  2. X: violet; Y: green; Z: red
  3. X: green; Y: violet; Z: red
  4. X: red; Y: violet; Z: green

Answer: 2. X: violet; Y: green; Z: red

Question 10. A ray of light is incident on one face of the prism, as shown.

HBSE Class 10 Physics Solutions For Chapter 2 A Ray Of Light Is Incident On One Face Of The Prism

How will the ray of light disperse in the prism?

HBSE Class 10 Physics Solutions For Chapter 2 A Ray Light Dispire In The Prism

Answer: HBSE Class 10 Physics Solutions For Chapter 2 A Ray Of Light Dispire

Question 11. Why do stars appear to twinkle at night?

  1. Because the light of stars travels in a different medium
  2. Because the distance of the star varies when Earth rotates
  3. Because the star changes its position relative to Earth
  4. Because the atmosphere reflects the light at different angles

Answer: 1. Because the light of stars travels in a different medium

Question 12. A student learns that the scattering of sunlight depends on the wavelength of the light and the size of particles present in the atmosphere. The student collects the data about the wavelength of the visible lights and the size of the particle, as shown.

HBSE Class 10 Physics Solutions For Chapter 2 The Data About The Wavelength Of The Visible light And Size Of The Particle

Which particles will scatter blue light?

  1. P and R
  2. R and S
  3. P and Q
  4. Q and S

Answer: 3. P and Q

Question 13. A got his eye tested. The optician’s prescription for the spectacles was:

Left eye: -3 D

Right eye: -3.50 D

The person is having a defect of vision called:

  1. Presbyopia
  2. Myopia
  3. Astigmatism
  4. Hypermetropia

Answer: 3. Astigmatism

Question 14. A man finds it difficult to read the odometer on the dashboard of the car but is able to clearly read a distant road sign. Which of the following statements is correct about this man?

  1. The near point of his eyes has receded away.
  2. The near point of his eyes has come closer to him.
  3. The far point of his eyes has receded away.
  4. The far point of his eyes has come closer to him.

Answer: 1. The near point of his eyes has receded away.

Question 15. Figures a, b, c, respectively, indicate the point in case of:

HBSE Class 10 Physics Solutions For Chapter 2 The Hypermetropia Eye

  1. The Hypermetropia eye, the myopic eye and the normal eye
  2. The normal eye, the myopic eye and hypermetropic eye
  3. The normal eye, the Hypermetropia eye and the myopic eye
  4. The myopic eye, the normal eye and hypermetropic eye

Answer: 3. The normal eye, the Hypermetropia eye and the myopic eye

Question 16. In which of the following cases will no dispersion take place when sunlight passes through it?

HBSE Class 10 Physics Solutions For Chapter 2 Dispersion take Place When Sunlight Passes Throught It

Answer: HBSE Class 10 Physics Solutions For Chapter 2 Dispersion take Place

Chapter 2 The Human Eye And The Colourful World Very Short Questions And Answers

Question 1. By how much time the day would have been shorter if the Earth had no atmosphere?

Answer: 4 minutes

Question 2. What colour does the sky appear to an astronaut?

Answer: The Sky appears dark to astronauts.

Question 3. Which colour of light has the shorter wavelength- red or violet?

Answer: Violet

Question 4. Name the phenomenon due to which a swimming pool appears less deep than it really is.

Answer: Refraction of light

Question 5. Name the phenomenon due to which the stars seem higher in the sky than they actually are.

Answer: Atmospheric refraction of light.

Question 6. Why does the Sun appear white at noon?

Answer: The Sun is nearly overhead at noon. The sunlight has to pass through a much smaller portion of the Earth’s atmosphere. The scattering is much less, and thus, the Sun looks white.

Question 7. An object is moved closer to an eye. What changes must take place in the eye in order to keep the image in Sharp focus?

Answer: The shape of the eye-lens should must be changed by the ciliary muscles to make it thicker and increase its converging power in order to keep the image sharp and in focus.

Question 8. Name the part of our eyes that helps us to focus on near and distant objects in quick succession.

Answer: Ciliary muscles

Question 9. A person is advised to wear spectacles with concave lenses. What type of defect of vision is he suffering from?

Answer: Myopia or short-sightedness.

Question 10. A person can comfortably read a book but finds it difficult to read the number on a bus parked 5 m away from him. Name the type of defect of vision he is suffering from. Which type of lens should he use in his spectacles to correct his vision?

Answer: Myopia or short-sightedness. A concave lens should be used to correct his vision.

Question 11. Name the type of particles which act as a prism in the formation of the rainbow in the sky.

Answer: Water droplets present in the atmosphere.

Question 12. Give the condition required to achieve a larger magnification of a small object by a compound microscope?

Answer: To achieve a larger magnification of a small object, both the objective and the eyepiece of a compound microscope should have smaller focal lengths.

Chapter 2 The Human Eye And The Colourful World Short Questions And Answers

Question 1. A student observes the following phenomenon in the lab as a white light passes through a prism. Among many other colours, he observed the position of the two colours, Red and Violet.

HBSE Class 10 Physics Solutions For Chapter 2 A White Light Passes Through A Prism

What is the phenomenon called? What is the reason for the violet light to bend more than the red light?

Answer:

Dispersion. The speed of violet light inside the prism is the slowest, and that of red is the highest. Hence, the deviation of violet light is maximum and that of red is minimum.

Question 2.

  1. Name the colour of light which undergoes
    • More scattering and
    • Less scattering while passing through the atmosphere.
  2. Draw a ray diagram to show the formation of a rainbow.

Answer:

While passing through the atmosphere

  1. Blue light undergoes more scattering, while
  2. Red light undergoes less scattering.

Question 3. Observe the experimental setup given below and answer the following questions:

HBSE Class 10 Physics Solutions For Chapter 2 The Experimental Setup The Scattering Of Blue Component Of Sunlight In The Atmosphere

  1. Out of blue light and red light, which one is scattered more easily?
  2. What causes the scattering of the blue component of sunlight in the atmosphere?

Answer:

  1. Blue light scatters more easily due to its smaller wavelength
  2. Gas molecules are present in the air.

Question 4. Give reasons:

  1. The extent of deviation of a ray of light on passing through a glass prism depends on its colour.
  2. Lights of red colour are used for danger signals.

Answer:

  1. The refractive index of a medium is different for different colours of light,
  2. Due to a large wavelength, red colour is the least scattered and travels to longest distance.

Question 5. A student sitting at the back bench in a class is not able to see what is written on the blackboard. He, however, sees it clearly when sitting on the front seat at an approximate distance of 1.5 m from the blackboard. Draw ray diagrams to illustrate the image formation of the blackboard when he is seated at the

  1. Back seat
  2. Front seat.

Answer:

1. When a student is seated in the back seat

HBSE Class 10 Physics Solutions For Chapter 2 When A Student Is Seated At The Back Seat

2. When a student is seated in the front seat

HBSE Class 10 Physics Solutions For Chapter 2 When Student Is Seated At Front Seat

Question 6. How does the eye adjust itself to deal with light of varying intensity?

Answer:

The amount of light entering the eye is controlled by the pupil. If the intensity of the outside light is low, then the pupil expands to allow more light to enter the eye. If the outside intensity is high, then the pupil contracts so that less light enters the eye.

Question 7. How do we see colours?

Answer:

  1. The retina of our eyes has a large number of light-sensitive cells. The cells on the retina are of two shapes:
  2. Rod-shaped and cone-shaped. The rod-shaped cells of our retina respond to the intensity of light. The cone-shaped cells of our retina respond to colours. These cells make us see colours and distinguish between them. Thus cones make colour perception possible.

Question 8. When is a person said to have developed cataract in their eye? How is the vision of a person having cataracts restored?

Answer:

Cataract develops when the lens of a person becomes unclear due to the formation of a membrane over it. The vision of a person having cataract can be restored after getting cataract surgery done on the eye-lens having cataract. This defect cannot be corrected by any type of spectacle lenses.

Question 9. Why do we have two eyes? Binocular (or two-eyed) vision has several advantages, one of which is the ability to see the world in three dimensions.

Answer:

Binocular (or two-eyed) vision has several advantages, one of which is the ability to see the world in three dimensions.

Question 10.

  1. Which property of light is rainbow formation based on?
  2. Why is the rainbow not visible on polluted skies?

Answer:

  1. Refraction, Dispersion and Total Internal Reflection of Light;
  2. Pollution particles change how the air scatters and absorbs different colours of light.

Question 11. What material is the human lens made up of? How is that different from the artificial lenses which are installed in the eye during an eye surgery?

The human eye lens is made up of proteins. The main difference between an artificial eye and lens is that a human eye lens provides sight, whereas an artificial eye does not.

Answer:

The human eye lens is made up of proteins. The main difference between an artificial eye and lens is that a human eye lens provides sight, whereas an artificial eye does not.

Question 12. An old person is unable to see clearly nearby objects as well as distant objects. To correct the vision, what kind of lens will he require?

Answer:

The upper portion (concave lens) facilitates the distant vision, and the lower portion (convex lens) facilitates near vision. Hence, a bifocal lens should be used, whose upper portion is a concave lens and lower portion is a convex lens.

Question 13. Why are danger signal lights red in colour?

Answer:

Danger signal lights are red in colour because red colour is the least scattered by fog or smoke.

Question 14. How will you use two identical prisms so that a narrow beam of white light incident on one prism emerges out of the second prism as white light? Draw the diagram.

Answer:

When an inverted prism is kept a little distance away from the prism causing dispersion or basically in the path of the split beam, the spectrum recombines to form white light.

HBSE Class 10 Physics Solutions For Chapter 2 Recombination Of The Spectrum Of White Light

Question 15. Which part of human eye helps in the perception of colours?

Answer:

The retina of the human eye has a large number of cone-shaped and rod-shaped cells. The rods respond to the intensity of light and the cones to the colours. These help us to perceive colours.

Question 16.

  1. A person having a myopic eye used the concave lens of focal length 50cm. What is the power of the lens?
  2. What property of the eye is the principle of motion, pictures

Answer:

  1. The formula for the power of a lens is Power (P) = 1/f. P = 1/50 = 0.02 dioptre.
  2. Refraction of light through the eye lens is the basic principle for the visualization of pictures and motion.

Question 17. What happens when ciliary muscles get relaxed and then get contracted?

Answer:

When ciliary muscles get relaxed then the lens becomes thin, due to which the focal length increases and we can easily see distant objects clearly. When we see the objects which are near to our eyes, then the ciliary muscles get contracted, due to which the curvature of the eye lens increases and the eye lens becomes thicker. Hence, focal length decreases, which enables us to see objects which are placed near our eyes.

Question 18. A child, while playing with his father, threw the spectacles of his father. Now, father is not able to read the newspaper. Identify the defect of vision and how it was corrected. Draw a ray diagram of the defect and also a correcting diagram of the defect.

Answer:

As the father of the child is not able to read the newspaper means he can’t see the nearby objects clearly. This means that the father is suffering from the defect of hypermetropia. In this defect, the near point becomes farther away from the normal near point. His spectacles were made up of a convex lens of proper focal length to correct his defect.

Question 19. Give reason:

  1. Why do stars twinkle?
  2. Why do the planets not twinkle?
  3. Why does the sun appear reddish early in the morning?
  4. Why does the sky appear dark instead of blue to an astronaut?
  5. Why does the Sun appear white at noon?

Answer:

  1. Stars appear to twinkle due to atmospheric refraction. The light of a star after the entry of light into Earth’s atmosphere undergoes refraction continuously (due to the different density of the layers of the atmosphere) till it reaches the surface of the Earth. As the path of light coming from stars keeps on changing, the apparent position of stars keeps changing, and the amount of light from stars entering the eye keeps twinkling. Due to this, a star sometimes appears bright and
    sometimes dim (called the twinkling effect),
  2. The planets are much nearer to the Earth than stars, and because of this they can be considered as a large source of light. If a planet is considered to be a collection of a very large number of point sources of light, then the average value of change in the amount of light entering the eye from all point-sized light sources is zero. Due to this, the effect of twinkling is nullified.
  3. The light coming from the sun passes through various denser layers of air in the Earth’s atmosphere before reaching our eyes near the horizon. Most of the part of blue light and light of small wavelengths gets scattered by dust particles near the horizon. So, the light reaching our eyes is of a large wavelength. Due to this, the sun appears reddish at the time of sunrise and sunset.
  4. As an astronaut moves away from the atmosphere of Earth, the atmosphere becomes thin. Due to the absence of molecules (or dust particles) in the air, the scattering of light does not take place. Thus, the sky appears dark in the absence of scattering.
  5. At noon, the Sun appears white because the light from the Sun is directly overhead and travels a relatively shorter distance. Thus, only blue and violet colours are scattered.

Question 20. On a rainy day, Ram reached his grandfather’s place in the village. On the way to the house, he saw a beautiful rainbow in the sky. In the night, he saw lots of twinkling stars in the clear sky. He was very excited to see this beautiful natural phenomenon, which he was not able to see in the city, where he lived with his father. Explain the phenomenon on the basis of science. Do you think that pollution in the atmosphere affects the formation of rainbows and the twinkling of stars? Do you agree with the fact that a pollution-free environment will strengthen such natural phenomena in the cities as well. Elaborate.

Answer:

The twinkling of stars is due to atmospheric refraction. The formation of a rainbow is due to dispersion, refraction and internal reflection. Yes, pollution in the atmosphere affects the formation of rainbows and the twinkling of stars, and a pollution-free environment will strengthen such natural phenomena in the cities as well.