Class 11 Physics

Wave Motion Short Answer Type Questions

Question 1. The equation y= a(sinkxcosωt- coskxsinωt) represents

1. A progressive wave propagating along the positive x-axis with velocity ω/k
2. A progressive wave propagating along the negative x-axis with velocity ω/k
3. A stationary wave
4. Not a wave motion

Answer:

Here, y = asin(kx-ωt); a progressive wave of ω/k velocity, along +ve x direction.

The option 1 is correct.

Question 2. The wavelength of a traveling wave is 2 m. What is the phase difference between two particles 1 m apart in the path of the wave?
Answer:

Given

The wavelength of a traveling wave is 2 m.

For λ = 2m, the phase changes by 2π.

So, for two points separated by 1 m, phase difference = \(\frac{\pi}{2}\) = π

Question 3. Are sound waves longitudinal or transverse?
Answer:

Sound waves are longitudinal waves.

Question 4. The equation of a traveling wave is y- 20 sin (0.2πx-50πt), where x and y are in cm and t is in second. Find out the

1. Amplitude,
2. Wavelength,
3. Frequency and
4. Velocity of the wave.

Answer:

Equation of progressive wave, y = 20 sin (0.2πx-50πt)

1. Amplitude (a) = 20 cm

2. Here, k = \(0.2 \pi \quad \text { or, } \frac{2 \pi}{\lambda}=0.2 \pi\)

∴ \(\lambda=\frac{2 \pi}{0.2 \pi}=10 \mathrm{~cm}\)

So, wavelength = 10 cm

3. According to given equation, angular velocity (ω) = 50 π;

∴ n = \(\frac{\omega}{2 \pi}=\frac{50 \pi}{2 \pi}=25\)

So, frequency = 25 Hz

4. Velocity (V) =nλ = 25 x 10 = 250 cm · s^-1

Question 5. The percentage change of velocity of sound when temperature increases from 0°C to 20°C is

1. 1.5%
2. 2.5%
3. 1.75%
4. 2%

Answer:

The velocity of sound, c ∝ √T

Here, T₁ = 0 + 273 = 273 K

T₂ = 20 + 273 = 293 K

∴ \(\frac{c_{20}}{c_0}=\sqrt{\frac{293}{273}}=\left(1+\frac{20}{273}\right)^{1 / 2} \approx 1+\frac{1}{2} \cdot \frac{20}{273}=\frac{283}{273}\)

So, percentage change of velocity

= \(\left(\frac{283}{273}-1\right) \times 100=3.66 \%\)

None of the options is correct.

Question 6. The velocity of sound in oxygen gas at STP is v. What will be the velocity of sound in Helium gas under the same conditions?
Answer:

Given

The velocity of sound in oxygen gas at STP is v.

Velocity of sound, c = \(\sqrt{\frac{\nu R T}{M}}\)

At STP, the molar mass of oxygen, M = 32 g

The molar mass of helium, M’ = 4g

Now as oxygen is a diatomic gas, v = \(\frac{7}{5}\)

Again helium is a monoatomic gas, v’ = \(\frac{5}{3}\)

So, \(\frac{\nu^{\prime}}{\nu}=\sqrt{\frac{\nu^{\prime}}{\nu}} \cdot \sqrt{\frac{M}{M^{\prime}}}=\sqrt{\frac{5}{\frac{5}{5}} \times \frac{32}{4}}=\sqrt{\frac{25}{21} \times 8}=3.086\)

∴ v’ = 3.086 v

Question 7. Two sound waves having a phase difference of 60° have path difference of

1. \(2 \lambda\)
2. \(\frac{\lambda}{6}\)
3. \(\frac{\lambda}{2}\)
4. \(\frac{\lambda}{3}\)

Answer:

Phase difference, \(\phi=\frac{2 \pi}{\lambda}\) x path difference (Δx)

or, \(60^{\circ}=\frac{\pi}{3}=\frac{2 \pi}{\lambda} \Delta x\)

∴ \(\Delta x=\frac{\lambda}{6}\)

The option B is correct.

Question 8. A stone is dropped from the top of a well of depth 39.2 m and a sound is heard when it reaches the bottom surface after a time of 2.95 s. Find the velocity of sound in the air. (g = 9.8 m/s²)
Answer:

Given

A stone is dropped from the top of a well of depth 39.2 m and a sound is heard when it reaches the bottom surface after a time of 2.95 s.

Depth of the well, h = 39.2 m

The time taken by the stone to travel a 39.2 m distance is,

t = \(\sqrt{\frac{2 h}{g}}=\sqrt{\frac{2 \times 39.2}{9.8}}=2 . \sqrt{2}=2.83 \mathrm{~s}\)

The time taken for sound to travel h distance is, t’ = (2.95 – 2.83) = 0.12 s

∴ The velocity of sound in air, v = \(\frac{h}{t^{\prime}}=\frac{39.2}{0.12}\) = 326.67 m/s

Question 9. The frequency of a particle vibrating In a medium is fHz. How many waves are generated in 5 seconds in the medium?
Answer:

Given

The frequency of a particle vibrating In a medium is fHz.

The number of waves generated in 1 second = f

∴ The number of waves generated in 5 seconds = 5f

Question 10. A tuning fork in air vibrates at 30 Hz with 5 cm amplitude. If the velocity of sound in air is 330 m · s^-1, derive the expression for the generated traveling wave.
Answer:

Given

A tuning fork in air vibrates at 30 Hz with 5 cm amplitude. If the velocity of sound in air is 330 m · s^-1,

Amplitude, A = 5 cm = 0.05 m; frequency n = 30 Hz;

velocity, V = 330 m · s^-1

∴ Wavelength, \(\lambda=\frac{V}{n}=\frac{330}{30}=11 \mathrm{~m}\)

The expression for the generated traveling wave,

y = \(A \sin \frac{2 \pi}{\lambda}(V t-x)=0.05 \sin \frac{2 \pi}{11}(330 t-x) \mathrm{m}\)

Question 11. The path difference of two particles on a wave corresponding to a phase difference of 60° is

1. \(2 \lambda\)
2. \(\frac{\lambda}{2}\)
3. \(\frac{\lambda}{6}\)
4. \(\frac{\lambda}{3}\)

Answer:

The option 3 is correct.

Question 12. When sound enters water from air or vice-versa, which physical quantity of the waves remains unchanged?
Answer:

When sound enters water from air or vice-versa, the frequency of the waves remains unchanged.

Question 13. Two tuning forks vibrating simultaneously produce 5 beats per second. The frequency of one fork is 275 Hz. A small wax is attached to the other fork and 2 beats per second are produced when the two vibrate simultaneously. Find the frequency of the other fork.
Answer:

Given

Two tuning forks vibrating simultaneously produce 5 beats per second. The frequency of one fork is 275 Hz. A small wax is attached to the other fork and 2 beats per second are produced when the two vibrate simultaneously.

Let the frequency of the other fork be x.

Beat frequency = difference in the frequencies of the two tuning forks.

If x< 275 Hz, before waxing we have, 275 – x = 5 H or x = 270 Hz. After waxing, x will decrease, so the beat frequency will increase.

∴ x > 275 Hz and before waxing we have, x – 270 = 5

or x = 280 Hz. After waxing, A will decrease slightly, so the beat frequency will decrease to 2.

Question 14. The velocity of sound in air at 20°C and 1 atm pressure is 344.2 m/s. At 40°C and 2 atm pressure, the velocity of sound in air is approximately

1. 350 m/s
2. 350 m/s
3. 303 m/s
4. 370 m/s

Answer:

If the velocities of sound are c₁ and c₂ at absolute temperatures T₁ and T₂, respectively, then

∴ \(\frac{c_1}{c_2}=\sqrt{\frac{T_1}{T_2}}\)

Here, T₁ = 273 + 20 = 293 K and, T₂ = 273 + 40 = 313 K

∴ \(c_2=c_1 \times \sqrt{\frac{T_2}{T_1}}=344.2 \times \sqrt{\frac{313}{293}}\)

= 355.75 ≈ 356 m/s

The option 2 is correct.

Question 15. If the pressure, temperature, and density of an ideal gas are denoted by p, T, and ρ respectively, the velocity of sound in the gas is

1. Proportional to √p, when T is constant
2. Proportional to √T
3. Proportional to √p, when ρ is constant
4. Proportional to T

Answer:

Velocity of sound in air, \(\nu=\sqrt{\frac{\delta p}{\rho}}\)…(1)

or, \(\nu=\sqrt{\frac{\delta R T}{M}}\)…(2)

⇒ \(\left[because p V=R T \text { or, } p=\frac{R T}{V}=\frac{R T}{\frac{M}{\rho}}\right]\)

According to equation (1), v ∝ √p when ρ is constant.

According to equation (2), v ∝ √T

The options 2 and 3 are correct.

Question 16. A uniform string of length 20 m is suspended from a rigid support. A short wave pulse is introduced at its lowest end. It starts moving up the string. The time taken to reach the support is

1. 2π√2 s
2. 2s
3. 2√2 s
4. √2s

Answer:

Tension at height y from the lowest end of the string, T = \(\frac{Mgy}{L}\)

[M is the mass of the string, L is the length of the string]

∴ The velocity of the wave at the height y of the string, \(\frac{d y}{d t}=\nu=\sqrt{\frac{T}{\mu}}\) [μ = mass per unit length of the string]

= \(\sqrt{\frac{\frac{M g y}{L}}{\frac{M}{L}}}=\sqrt{g y}\)

∴ \(\frac{d y}{\sqrt{y}}=\sqrt{g} d t\)

or, \(\int_0^L \frac{d y}{\sqrt{y}}=\sqrt{g} \int_0^t d t\)

or, \(\left.2 \sqrt{y}\right|_0 ^L=\left.\sqrt{g} t\right|_0 ^t\)

or, \(2 \sqrt{L}=\sqrt{g} t\)

∴ t = \(2 \sqrt{\frac{L}{g}}=2 \sqrt{\frac{20}{10}}=2 \sqrt{2} \mathrm{~s}\)

The option 3 is correct.

Question 17. A uniform rope of length L and mass m₁ hangs vertically from a rigid support. A block of mass m₂ is attached to the free end of the rope. A transverse pulse of wavelength λ₁ is produced at the lower end of the rope. The wavelength of the pulse when it reaches the top of the rope is λ₂. The ratio λ₂/λ₁ is

1. \(\sqrt{\frac{m_1+m_2}{m_2}}\)
2. \(\sqrt{\frac{m_2}{m_1}}\)
3. \(\sqrt{\frac{m_1+m_2}{m_1}}\)
4. \(\sqrt{\frac{m_1}{m_2}}\)

Answer:

Tension at the lower end of the rope, T₁ = m₂g

Tension at the top of the rope, T₂ = (m₁ + m₂)g

The velocity of the transverse wave, \(\nu=n \lambda=\sqrt{\frac{T}{\mu}}\)

[where n = frequency, λ = wavelength and μ = mass per unit length of the rope]

Hence, λ ∝ √T

∴ \(\frac{\lambda_1}{\lambda_2}=\sqrt{\frac{T_1}{T_2}}=\sqrt{\frac{m_2 g}{\left(m_1+m_2\right) g}}\)

= \(\sqrt{\frac{m_2}{m_1+m_2}}\)

or, \(\frac{\lambda_2}{\lambda_1}=\sqrt{\frac{m_1+m_2}{m_2}}\)

The option 1 is correct.

Question 18. A metal rod of 1 m in length, is dropped exactly vertically onto a hard metal floor. With an oscilloscope, it is determined that the impact produces a longitudinal wave of 1.2 kHz frequency. The speed of sound in the metal rod is

1. 600 m/s
2. 2400 m/s
3. 1800 m/s
4. 1200 m/s

Answer:

For the fundamental wave in the rod of length l, l \(=\frac{\lambda}{2}\)

Hence, Aλ =2l [l = wavelength]

∴ Velocity of sound,

V= nλ = (1.2 x 10³) x 2 x 1 = 2400 m/s

The option 2 is correct.

Question 19. The transverse harmonic wave on a string is described by Y(x,t) = 3.0sin(36t + 0.018x + π/4) where x and y are in cm and t is in second, The positive direction of x is from left to right,

1. Is this a traveling wave or a stationary wave? Give its direction of propagation.
2. What are its amplitude and frequency?
3. What is the initial phase at the origin?
4. What is the least distance between two successive crests in the wave?

Answer:

Compared with the general equation of a traveling wave,

y = a sin(ω- kx+ θ), we get, a = 3.0 cm,

ω = 36 Hz, k = -0.018 cm^-1 and \(\theta=\frac{\pi}{4}\)

1. The given equation corresponds to a traveling wave. As k is negative, the wave propagates along the negative x-direction.

2. Amplitude, a = 3.0 cm; frequency, \(\nu=\frac{\omega}{2 \pi}=\frac{36}{2 \times 3.14}=5.73 \mathrm{~Hz} .\)

3. Putting x = 0 and t = 0, we get

y = a sinθ = a sin \(\frac{\pi}{4}\)

So the initial phase at the origin = \(\frac{\pi}{4}\)

4. In magnitude, k = 0.018 cm^-1.

As k =\(\frac{2\pi}{\lambda}\), the wavelength,

∴ \(\frac{2\pi}{k}\) = 349 cm = least distance between two successive crests

Question 20. What change is observed when a wave gets reflected from a hard and rigid support?
Answer:

• The reflected wave would propagate in the reverse direction, and
• Its phase would be reversed, i.e. a 180° phase difference would be introduced between the incident and the reflected waves.

Question 21. Though a violin note and a si tar note may have the same frequency, yet we can distinguish between the two notes. Explain how?
Answer:

• A violin note and a sitar note may have the same frequency or the same combination of frequencies. They may also have the same loudness.
• Yet, we can always distinguish between them because of a third property, which is essentially different for the two notes.
• This is the property called quality or timbre. The difference in timbre arises due to the distinct shapes of the waveforms associated with different musical notes.

Question 22. If y = sin(3.6t + 0.018 x π/4) cm, find the amplitude and velocity of the wave.
Answer:

Given

y = sin(3.6t + 0.018 x π/4) cm

Comparing with the general equation y = asin(ωt -kx+θ) of a progressive wave, we get,

Amplitude, a = 3 cm,

Angular frequency, ω = 2πn = 3.6 Hz

and wave number, k = \(\frac{2 \pi}{\lambda}\) = -0.018 cm^-1

(The negative value means wave propagation in the negative x direction).

The velocity of the wave,

∴ V = \(n \lambda=\frac{2 \pi n}{\frac{2 \pi}{\lambda}}=\frac{\omega}{k}=\frac{3.6}{0.018}=200 \mathrm{~cm} \cdot \mathrm{s}^{-1}\)

Kinetic Theory Of Gases

Question 1. Which one of the following is not true for an ideal gas?

1. The molecules of an ideal gas move randomly
2. The molecules of an ideal gas attract one another
3. The volume of the molecules of an ideal gas is negligible
4. The pressure of an ideal gas increases with an increase in the velocity of its molecules

Answer: 2. The molecules of an ideal gas attract one another is correct

Question 2. A monatomic ideal gas is heated at constant pressure. How much fraction of heat is used to increase internal energy?

1. 2/5
2. 3/5
3. 3/7
4. 3/4

Answer:

If ΔT is the increase in temperature of a monoatomic ideal gas, increase in internal energy, \(\Delta E=\frac{3}{2} R \Delta T\) [for 1 mol gas]

Again, molar specific heat at constant pressure, \(C_p=\frac{5}{2} R\).

So, given heat to increase the temperature by ΔT at constant pressure = \(C_p \Delta T=\frac{5}{2} R \Delta T\)

Hence, the required fraction

= \(\frac{\text { increase in internal energy }}{\text { given heat }}=\frac{\frac{3}{2} R \Delta T}{\frac{5}{2} R \Delta T}=\frac{3}{5}\)

The option 2 is correct.

Question 3. Kinetic theory proves that, the pressure of a gas p = \(\frac{2 E}{3 v}\)(where v = 1 molar volume, E = kinetic = energy).
Answer:

Given

The pressure of a gas p = \(\frac{2 E}{3 v}\)(where v = 1 molar volume, E = kinetic = energy)

Molar mass of the gas = M;

Number of molecules = Avogadro’s number = N;

Molar volume of the gas = v = \(\frac{M}{\rho} \text {. }\)

We know, pressure, \(p=\frac{1}{3} \rho c^2\)

Here, ρ = density of the gas,

c = rms speed of gas molecules

Now, the kinetic energy of the gas molecules,

E = \(\frac{1}{2} m c_1^2+\frac{1}{2} m c_2^2+\cdots+\frac{1}{2} m c_N^2\)

= \(\frac{1}{2} m N \frac{c_1^2+c_2^2+\cdots+c_N^2}{N}\)

= \(\frac{1}{2} M c^2=\frac{1}{2} M \cdot \frac{3 p}{\rho}=\frac{3}{2} p v\)

Hence, \(p=\frac{2 E}{3 v}\)

Question 4. In thermal equilibrium, the rms velocity of a gas molecule is

1. Proportional to T²
2. Proportional to T²
3. Proportional to √T
4. Zero

Answer:

rms speed of gas molecule, = \(\sqrt{\frac{3 R T}{M}}\)

The option 3 is corrections

Question 5. Write the expression for mean free path
Answer:

The expression for mean free path

Mean free path, \(\lambda=\frac{1}{\pi \sigma^2 n}\)

where n = number of molecules in unit volume;

= diameter of each gas molecule.

Question 6. From fonetic theory of gases proves that the pressure of 1 a gas p = 2E/3V, where V = 1 molar volume, E = Kinetic energy.
Answer:

Given

Molar mass of the gas =M, Number of molecules = Avogadro number = N,

Molar volume = V = \(\frac{M}{\rho}\), pressure = p = \(\frac{1}{3} \rho c^2\), here = density of gas, c= rms speed.

Kinetic energy of gas molecules, E

= \(\frac{1}{2} m c_1^2+\frac{1}{2} m c_2^2+\cdots+\frac{1}{2} m c_N^2\)

= \(\frac{1}{2} m N \frac{c_1^2+c_2^2+\cdots+c_N^2}{N}=\frac{1}{2} M c^2=\frac{1}{2} M \cdot \frac{3 p}{c}=\frac{3}{2} p V\)

Therefore, \(p=\frac{2 E}{3 V}\)

Question 7. The volume and pressure of two moles of an ideal gas are V and p respectively Another 1 mol ideal gas haring volume 2V also exerts the same pressure p. Molecular mass of the second gas is 16 times that of the first gas. Compare the rms velocities of the two gases.
Answer:

Given

The volume and pressure of two moles of an ideal gas are V and p respectively Another 1 mol ideal gas haring volume 2V also exerts the same pressure p. Molecular mass of the second gas is 16 times that of the first gas.

If the temperature of the first gas is \(T_1\), then \(p V=2 R T_1\) and the temperature of the second gas \(T_2\), then P.

then \(P \cdot 2 \mathrm{~V}=R T_2 or, p \mathrm{~V}=\frac{1}{2} R T_2\)

According to the question, \(2 R T_1=\frac{1}{2} R T_2 \quad \text { or } \frac{T_1}{T_2}=\frac{1}{4} \)

rms speed, c = \(\sqrt{\frac{3 R T}{M}}\)

∴ \(\frac{c_1}{c_2}=\sqrt{\frac{T_1}{T_2} \cdot \frac{M_2}{M_1}}=\sqrt{\frac{1}{4} \times \frac{16}{1}}=\frac{2}{1}\)

or, \(c_1=2 c_2\)

Question 8. The perfect equation for 4g ofhydroÿst gas is

1. pV = RT
2. pV = 2RT
3. pV = 1/2 RT
4. pV = 4RT

Answer:

The number of moles in 4 g of hydrogen gas.

n = \(\frac{m}{M}=\frac{4}{2}=2 \mathrm{~mol}\)

∴ perfect equation, pV = nRT

on pV=2RT

The option 2 is correct.

Question 9. The relation \(p=\frac{1}{3} m n c_{m s}^2\), shows that the mean kinetic energy of a molecule is the same for all types of gases.
Answer:

Given, \(p=\frac{1}{3} m n c_{\text {rms }}^2=\frac{1}{3} \frac{m N}{V} c_{\text {rms }}^2\)

[N= number of molecules, V= volume]

or, \(\frac{1}{3} m c_{\mathrm{mms}}^2=\frac{p V}{N}=\frac{\left(\frac{N}{N_n}\right) R T}{N}\)

(\(N_A\)= Avogadro’s number)

or \(\frac{1}{2} m c_{\mathrm{rms}}^2=\frac{3 R T}{2} \frac{R}{N_A}\)

∴ mean kinetic energy of a molecule = \(\frac{1}{2} m c_{\text {rms }}^2=\frac{3 R T}{2} \frac{R T}{N_A}\)

Is the same for all types of gases at a particular temperature.

Question 10. The rms speed of oxygen is v at a particular temperature. If the temperature is doubled and oxygen molecules dissociate Into oxygen atoms, the rms speed becomes

1. v
2. √2v
3. 2v
4. 4v

Answer:

The rms speed of oxygen molecule, v = \(\sqrt{\frac{2 R T}{M}}\)

When the molecules dissociate into oxygen atoms, molar mass will change from M to M/2 (when the temperature changes from T to 2 T)

Then, rms speed = \(\sqrt{\frac{3 R \cdot 2 T}{M / 2}}=2 \sqrt{\frac{3 R T}{M}}=2 v\)

The option 3 is correct.

Question 11. If the rms velocity of hydrogen gas at a certain temperature is c, then the rms velocity of oxygen gas at the same temperature is

1. \(\frac{c}{8}\)
2. \(\frac{c}{10}\)
3. \(\frac{c}{4}\)
4. \(\frac{c}{2}\)

Answer:

rms velocity of hydrogen gas, c = \(\sqrt{\frac{3 R T}{2}}\)

rms velocity of oxygen gas, \(c_{\mathrm{O}_2}=\sqrt{\frac{3 R T}{32}}\)

∴ \(c_{\mathrm{O}_2}=\sqrt{\frac{2}{32}} \times c=\frac{c}{4}\)

The option 3 is correct.

Question 12. The temperature of an ideal gas, initially at 27°C, is raised by 6°C. The rms velocity of the gas molecule will,

1. Increase by nearly 2%
2. Decrease by nearly 2%
3. Increase by nearly 1%
4. Decrease by nearly 1%

Answer:

⇒\(\nu_{\mathrm{rms}} \propto \sqrt{T}\)

or, \(\frac{\Delta \nu}{\nu}=\frac{1}{2} \frac{\Delta T}{T}=\frac{1}{2} \times \frac{6}{(273+27)}=\frac{1}{100}=1 \%\)

The option 3 is correct

Question 13. An open pipe made up of glass is immersed in mercury in such a way that a length of 8 cm extends above the mercury level. The open end of the tube is then closed and sealed and the tube is raised vertically up by an additional 46 cm. What will be the length of the air column above the mercury in the tube now? (atm pressure = 76 cm of Hg)

1. 16 cm
2. 22 cm
3. 38 cm
4. 6 cm

Answer:

Let <α = cross-sectional area of the pipe (in cm²)

So, initially, the volume of air above mercury, V₁ = 8α cm³; pressure, p₁ = 76 cm Hg.

When the top of the pipe is (46 + 8) or 54 cm higher than the outer mercury level, let x = rise of mercury level inside the pipe.

Then, the volume of air inside the pipe, v₂ = (54 – x)a cm³

and its pressure, p₂ = (76 – x) cm Hg

From Boyle law, p₁V₁ = p₂V₂

or, 76 x 8a = (76 – x)(54 – x)a

or, 608 = 4104- 130x+ x²

or, x² – 130x + 3496 = 0

or, (x-38)(x-92) = 0;

So, x = 38 cm or, x = 92 cm

The only physically meaningful solution is x = 38 cm Therefore, the length of the air column in the pipe above mercury =54-38 = 16 cm

The option 1 is correct.

Question 14. Consider an ideal gas confined in an isolated closed chamber. As the gas undergoes an adiabatic expansion, the average time of collision between molecules increases as V^q, where V is the volume of the gas. The value of q is  \(\left(\gamma=\frac{C_p}{C_\nu}\right)\)

1. \(\frac{3 \gamma+5}{6}\)
2. \(\frac{3 \gamma-5}{6}\)
3. \(\frac{\gamma+1}{2}\)
4. \(\frac{\gamma-1}{2}\)

Answer:

The rms speed of the molecules, c = \(\sqrt{3 R T / M}\)

Mean free path, \(\lambda=\frac{1}{\sigma^2 n}=\frac{1}{\sigma^2 \frac{N}{\nu}}\)

[N = number of molecules, cr = diameter of the molecules]

Average time of collision between the molecules,

t = \(\frac{\lambda}{c}=\frac{V}{\sigma^2 N} \frac{1}{\sqrt{3 R T / M}}\)

or, \(t \propto \frac{V}{\sqrt{T}}\)

or, \(T \propto \frac{V^2}{t^2}\)

Again, in the adiabatic process, \(T V^{\gamma-1}=\text { constant } \quad \text { or, } T \propto V^{1-\gamma}\)

∴ \(\frac{V^2}{t^2} \propto V^{1-r}\)

or, \(t^2 \propto v^{\gamma+1}\) or, \(t \propto V^{\frac{\gamma+1}{2}}\)

Hence, \(q=\frac{\gamma+1}{2}\)

The option 3 is correct

Question 15. An ideal gas undergoes a quasistatic, reversible process in which its molar heat capacity C remains constant. If during this process the relation of pressure p and volume V is given by pVⁿ = constant, then n is given by (here C_p and C_v are molar specific heat at constant pressure and constant volume, respectively)

1. n = \(\frac{C_p}{C_\nu}\)
2. n = \(\frac{C-C_p}{C-C_\nu}\)
3. n = \(\frac{C_p-C}{C-C_\nu}\)
4. n = \(\frac{C-C_\nu}{C-C_p}\)

Answer:

Here, \(p V^n=k\) (constant)….(1)

For \(1 \mathrm{~mol}\) of ideal gas,

pV = RT…(2)

Dividing (1) by (2) we get, \(V^{n-1} T=\frac{k}{R}\)

∴ \(\left(\frac{d V}{d T}\right)=\frac{V}{(n-1) T}=\frac{V}{(1-n) T}\)

According to first law of thermodynamics,

dQ = \(C_\nu d T+p d V\)

∴ \(\frac{d Q}{d T}=C_\nu+p\left(\frac{d V}{d T}\right)=C_\nu+\frac{p V}{(1-n) T}=C_\nu+\frac{R}{1-n}\)

Hence, thermal capacity, \(C=C_\nu+\frac{R}{1-n}\)

or, \(1-n=\frac{R}{C-C_v}\)

or, \(n=1-\frac{R}{C-C_\nu}=\frac{C-\left(C_\nu+R\right)}{C-C_\nu}=\frac{C-C_p}{C-C_\nu}\)

(because \(C_p-C_v=R\))

The option 2 is correct.

Question 16. The mean free path of a molecule of gas (radius r) is inversely proportional to

1. r³
2. r²
3. r
4. √r

Answer:

Mean free path,

λ = \(\frac{1}{\sqrt{2} \pi d^2 n}=\frac{1 \times 4}{\sqrt{2} \pi r^2 n}\)

∴ \(\lambda \propto \frac{1}{r^2}\)

The option 2 is correct

Question 17. The ratio of the specific heats \(\frac{C_p}{C_v}=\gamma\) in terms of degrees of freedom (n) is given by

1. \(\left(1+\frac{1}{n}\right)\)
2. \(\left(1+\frac{n}{3}\right)\)
3. \(\left(1+\frac{2}{n}\right)\)
4. \(\left(1+\frac{n}{2}\right)\)

Answer:

⇒ \(C_\nu=\frac{n}{2} R, C_P=\frac{n}{2} R+R=\frac{n+2}{2} R\)

∴ \(\gamma=\frac{\frac{n+2}{2} R}{\frac{n}{2} R}=\frac{n+2}{n}=1+\frac{2}{n}\)

The option 3 is correct.

Question 18. The molecules of a given mass of a gas have rms velocity of 200 m · s^-1 at 27 °C and 1.0 x 10⁵ N · m^-2 pressure. When the temperature and pressure of the gas are respectively, 127 °C and 0.05 x 10⁵ N · m^-2, the rms velocity of its molecules in m · s^-1 is

1. \(\frac{400}{\sqrt{3}}\)
2. \(\frac{100 \sqrt{2}}{3}\)
3. \(\frac{100}{3}\)
4. \(100 \sqrt{2}\)

Answer:

rms speed of gas molecule, c ∝ √T

∴ \(\frac{c_1}{c_2}=\sqrt{\frac{T_1}{T_2}}=\sqrt{\frac{27+273}{127+273}}=\sqrt{\frac{300}{400}}=\frac{\sqrt{3}}{2}\)

or, \(c_2=\frac{2}{\sqrt{3}} c_1=\frac{2}{\sqrt{3}} \times 200=\frac{400}{\sqrt{3}} \mathrm{~m} / \mathrm{s}\)

The option 1 is correct.

Question 19. When the temperature of a gas is raised from 30°C to 90°C, the percentage Increase in the rms velocity of the molecules will be

1. 60%
2. 10%
3. 15%
4. 30%

Answer:

30°C = 303K, 90°C = 363K

rms speed, \(c \propto \sqrt{T}\)

Hence, \(\frac{c_2}{c_1}=\sqrt{\frac{T_2}{T_1}}=\sqrt{\frac{363}{303}}=\sqrt{1.2} \approx 1.1\)

∴ The percentage increase

= \(\frac{c_2-c_1}{c_1} \times 100=\left(\frac{c_2}{c_1}-1\right) \times 100\)

= \((1.1-1) \times 100=10 \%\)

The option 2 is correct.

Question 20. At what temperature will the rms speed of oxygen molecules become just sufficient for escaping from the earth’s atmosphere? [Given: Mass of oxygen molecule (m) = 2.76 x 10^-26 kg Boltzmann’s constant k =
1.38 x 10^-23 J • K^-1]

1. 5.016 X10⁴ K
2. 8.360 x 10⁴ K
3. 2.508 x 10⁴K
4. 1.254 x 10⁴ K

Answer:

Escape velocity of a body for the earth, \(v_e=\sqrt{2 g R}=\sqrt{2 \times 9.8 \times 6400 \times 10^3}\)

= 11200 m/s

Let us consider, at temperature TK, the escape velocity of the oxygen molecules will be (v_e).

Then, \(\sqrt{\frac{3 k_B T}{m}}=v_e\)

or, \(\frac{3 \times 1.38 \times 10^{-23}}{2.76 \times 10^{-26}} \cdot T=(11200)^2\)

or, T = \(8.363 \times 10^4 \mathrm{~K}\)

The option 2 is correct.

Question 21. What would be the effect on the rms speed of gas molecules if the temperature of the gas is increased by a factor of 4?
Answer:

We know, rms speed, c = \(\sqrt{\frac{3 R T}{M}}\)

∴ \(c \propto \sqrt{T}\)

Now, \(\frac{c_1}{c_2}=\sqrt{\frac{T_1}{T_2}}=\sqrt{\frac{T_1}{4 T_1}}=\frac{1}{2}\)

(since \(T_2=4 T_1\))

or, \(\quad c_2=2 c_1\)

So, rms speed is doubled

Question 22. A flask contains Argon and Chlorine in the ratio of 2:1 by mass. The temperature of the mixture is 27°C. Obtain the ratio of average kinetic energy per molecule and root mean square speed v_rms of the molecules of the two gases. Given: atomic mass of argon =39,9 U and molecular mass of chlorine = 70.9 u.
Answer:

Given

A flask contains Argon and Chlorine in the ratio of 2:1 by mass. The temperature of the mixture is 27°C.

atomic mass of argon =39,9 U and molecular mass of chlorine = 70.9 u

Average kinetic energy of a molecule of any ideal gas = \(\frac{3}{2}\)kT

1. As both argon and chlorine have the same temperature, the ratio of average kinetic energy per molecule of the two gases =1:1

2. Let m be the mass of a molecule of the gas.

Now, average kinetic energy per molecule = \(\frac{1}{2} m v_{\mathrm{rms}}^2\)

∴ \(\frac{1}{2} m_1 v_1^2=\frac{1}{2} m_2 v_2^2=\frac{3}{2} k T\)

or, \(\frac{v_1^2}{v_2^2}=\frac{m_2}{m_1}=\frac{M_2}{M_1}=\frac{70.9}{39.9}=1.7\)

where M is the atomic or molecular mass of the gas.

So, \(\frac{v_1}{v_2}=\sqrt{1.77}=1.33\)

Here, the composition of a mixture of gases is not relevant since,

⇒ \(\frac{v_1}{v_2}=\sqrt{\frac{\overline{M_2}}{M_1}}=\text { constant }\)

Question 23. Prove that the average kinetic energy of a molecule of an ideal gas is directly proportional to the absolute temperature of the gas.
Solution:

We know that the average kinetic energy of a molecule is directly proportional to the absolute temperature of the gas.

Let one gram mole of an ideal gas occupy a volume V at temperature T.

If m is the mass of each molecule of the gas, then M = m x N_A [where N_A is the Avogadro’s number]

If c is the rms speed of the gas molecules, then the pressure exerted by the ideal gas is p = \(\frac{1}{3} \rho c^2=\frac{1}{3} \frac{M}{V} c^2\)

Now from the equation pV = RT,

⇒ \(\frac{1}{3} M c^2=R T \quad or, M c^2=3 R T \quad or, [latex]\frac{1}{2} M c^2=\frac{3}{2} R T\)

or, \(\frac{1}{2} m N_A c^2=\frac{3}{2} R T \quad\left[because M=m N_A\right]\)

or, \(\frac{1}{2} m c^2=\frac{3}{2}\left(\frac{R}{N_A}\right) T=\frac{3}{2} k_B T\left[because k_B=\frac{R}{N_A}\right]\)

So, the average kinetic energy of translation per molecule of a gas

= \(\frac{1}{2} m c^2=\frac{3}{2} k_B T\)

Hence, the average kinetic energy of a molecule is directly proportional to the absolute temperature of the gas.

Question 24. The absolute temperature of a gas is increased 3 times. What will be the increase in root mean square velocity of the gas molecule?
Answer:

Given

The absolute temperature of a gas is increased 3 times.

rms velocity in terms of Boltzmann constant (k),

c = \(\sqrt{\frac{3 k T}{m}}\)

At temperature, \(T_1, c=\sqrt{\frac{3 k T_1}{m}}\)

Now, \(T_2=3 T_1\)

∴ \(c_2=\sqrt{\frac{3 k T_2}{m}}=\sqrt{\frac{3 k \times 3 T_1}{m}}=\sqrt{3} \cdot \frac{\sqrt{3 k T_1}}{m}\)

or, \(c_2=\sqrt{3} c\)

If the temperature of the gas increases 3 times, then the root mean square velocity of the gas particles increases by a factor of √3.

Work And Energy – Rotational Motion Of A Body In A Vertical Circle

Non-uniform circular motion: When a body of mass m is tied to a string and is rotated in a vertical circle, it will not rotate with uniform speed in that circular path. So, it is an example of a non-uniform circular motion.

The magnitude of its velocity increases when it descends from the highest point B of the circle to the lowest point A. Again, the magnitude of velocity gradually decreases when it rises up to the highest point from the lowest point of the circle.

During its rotation, the weight of the body always acts vertically downwards, but the direction of centripetal force changes continuously. Hence, the tension in the string does not remain constant.

At the highest point B of the circular path, tension in the string and the weight of the body together provide the necessary centripetal force for the rotation of the body. If the velocity of the body is v and the tension in the string is T₂ at point B, then \(T_2+m g=\frac{m v^2}{r}\)

or, \(T_2=\frac{m v^2}{r}-m g\)…(1)

Again, the tension in the string at the lowest point A on the circular path acts in the vertically upward direction, i.e., towards the center of the circle, but the weight of the body acts vertically downwards.

As a result, the difference between the tension in the string and the weight of the body provides the necessary centripetal force. If the velocity of the body is u and the tension in the string is T₁ at point A, then

⇒ \(T_1-m g=\frac{m u^2}{r}\)

or, \(T_1=\frac{m u^2}{r}+m g\)…(2)

If the tension in the string at the point B is zero, i.e., if T₂ = 0, then

⇒ \(\frac{m v^2}{r}-m g=0 \quad \text { or, } \frac{m v^2}{r}=m g\)

or, v = \(\sqrt{r g}\)…(3)

Hence, if the value of v is less than \(\sqrt{r g}\) at the highest point, then the tension in the string becomes negative, i.e., the string gets relaxed. As a result, the body descends in a parabolic path instead of rotating along a circular path.

Hence, the least velocity of the body at the highest point on the circular path to maintain the vertical circular motion should be \(\sqrt{r g}\). This least velocity is called the critical velocity.

Least Velocity At The Lowest Point On The Circular Path Necessary To Maintain The Critical Velocity: If the potential energies of the body at points A and B are V₁ and V₂, respectively, then

total energy at point A = \(\frac{1}{2} m u^2+V_1 \text {. }\)

Similarly, total energy at point B = \(\frac{1}{2} m v^2+V_2 \text {. }\)

According to the principle of conservation of energy, \(\frac{1}{2} m u^2+V_1=\frac{1}{2} m v^2+V_2\)

or, \(V_2-V_1=\frac{1}{2} m\left(u^2-v^2\right)\)…(4)

When a body of mass m attains a height h, the potential energy gained by it = mgh. In the given figure, the height of the point B with respect to the point A is 2r.

So, the change in potential energy = V₂ – V₁ = mg · 2r

∴ From equation (4) we get, mg \(\cdot 2 r=\frac{1}{2} m\left(u^2-v^2\right)\)

or, \(4 g r=u^2-v^2\)

or, \(u^2=4 g r+v^2=4 g r+g r\)

(the least velocity at \(B, v=\sqrt{g r}\))

or, \(u^2=5 g r\)

or, \(u=\sqrt{5 g r}\)…(5)

For rotation in a vertical circle, this is the least value of u. It should be noted that the value of this least velocity is independent of the mass of the body.

For this last value of u, the minimum value of tension at point A can be determined from equation (2).

⇒ \(\left(T_1\right)_{\min }=\frac{m u_{\min }^2}{r}+m g=\frac{m \cdot 5 g r}{r}+m g \)

= \(5 m g+m g=6 \mathrm{mg}\)

or, \(\left(T_1\right)_{\min }=6 \times\) weight of the body…(6)

So, if a body is rotated in a vertical circle, then at the lowest point of the circle, the minimum tension in the string is 6 times the weight of the body.

Rotation Of A Bucket Full Of Water In A Vertical Circle: When a bucket full of water is made to rotate with high speed in a vertical circle, it is seen that at the highest point of the circular path, though the bucket is inverted, water from it does not spill out. It is clear that when the bucket is at the highest position, two forces act simultaneously on the water in the bucket:

1. Downward weight (mg) of water (m = mass of water).
2. The upward reaction of the centripetal force; if the velocity of the bucket at the highest position is v and the radius of the circular path is r, the value of this centripetal force acting on the bucket due to its rotation = \(\frac{m v^2}{r}.\)

Thus, when the upward force becomes greater than the force downward, water in the bucket cannot fall down. So, the condition for water to not fall down is, \(\frac{m v^2}{r}.\) ≥ mg or, v² ≥ rg.

Hence, the minimum value of v should be \(\sqrt{r g}\).

According to equation (6), when this bucket is rotated along a complete circular path, at the lowest position of the bucket, the tension acting on the hand will be 6 times the weight of the bucket filled completely with water. Hence, it is difficult to perform the experiment with a heavy bucket full of water. However, it is easier with a bucket or some other container containing a small amount of water.

Work And Energy – Rotational Motion Of A Body In A Vertical Circle Numerical examples

Example 1. A body of mass 1 kg is tied with a thread and is whirled in a vertical circle of radius 50 cm with a speed of 500 cm · s^-1. What will be the tension in, the thread at the highest and the lowest positions of the body?
Solution:

Given

A body of mass 1 kg is tied with a thread and is whirled in a vertical circle of radius 50 cm with a speed of 500 cm · s^-1

Speed of the body at every point on the circular path is the same and its value, v = 500 cm · s^-1 = 5 m · s^-1.

So, at every point on the circular path, centripetal force, F = \(\frac{m v^2}{r}=\frac{1 \times(5)^2}{0.5}=50 \mathrm{~N}\)

Weight of the body = mg = 1 x 9.8 = 9.8 N

At the highest position, the tension in the thread (T₁) and the weight of the body both act downwards and together they provide the necessary centripetal force.

Hence, F = mg+ T₁ or, T₁ = F- mg = 50 – 9.8 = 40.2 N

At the lowest position, the tension in the thread (T₂) acts in the upward direction but the weight acts downwards, and hence, in this case, (T₂ – mg) provides the necessary centripetal force.

So, T₂ – mg – F or, T₂ = F+ mg = 50 + 9.8 = 59.8 N.

Example 2. A particle of mass 100 g is suspended from the end of a weightless string of length 100 cm and is rotated in a vertical plane. The speed of the particle is 200 cm · s^-1 when the string makes an angle of θ = 60° with the vertical. Determine

1. The tension in the string when θ = 60° and
2. The speed of the particle at the lowest position. Acceleration due to gravity = 980 cm · s^-2.

Solution:

Given

A particle of mass 100 g is suspended from the end of a weightless string of length 100 cm and is rotated in a vertical plane. The speed of the particle is 200 cm · s^-1 when the string makes an angle of θ = 60° with the vertical.

1. The body is rotated in a vertical plane.

At point B, when the string makes an angle of 60° with the vertical, the velocity of the particle is v (say).

If the tension in the string is T, then T – \(mg \cos 60^{\circ}=\frac{m v^2}{r}\)

∴ T = \(m g \cos 60^{\circ}+\frac{m v^2}{r}\)

Here, v = \(200 \mathrm{~cm} \cdot \mathrm{s}^{-1} ; r=100 \mathrm{~cm} ; m=100 \mathrm{~g}\)

∴ T = \(100 \times 980 \times \frac{1}{2}+\frac{100 \times(200)^2}{100}\)

= \(8.9 \times 10^4 \mathrm{dyn} .\)

2. From the principle of conservation of energy we get, the kinetic energy of the body at the lowest point A = sum of the potential and kinetic energies at the point B

or, \(\frac{1}{2} m v_A^2=\frac{1}{2} m \nu^2+m g \cdot A C\)

(\(v_A\) = speed of the particle at the point A)

or, \(v_A^2=\nu^2+2 g \cdot A C\)

Here, \(A C=O A-O C=O A-O B \cos \theta\)

= \(l-l \cos \theta=100-100 \cos 60^{\circ}=50 \mathrm{~cm}\)

∴ \(v_A^2=(200)^2+2 \times 980 \times 50=13.8 \times 10^4\)

or, \(v_A=371.5 \mathrm{~cm} \cdot \mathrm{s}^{-1} \text {. }\)

Example 3. A body slides down an inclined plane after being released from rest from a height h and finally, it describes a circle of radius r instead of travelling along a horizontal floor. For what minimum value of h, can the particle describe that motion? Ignore friction.
Solution:

Given

A body slides down an inclined plane after being released from rest from a height h and finally, it describes a circle of radius r instead of travelling along a horizontal floor.

After descending along the inclined plane, the body can describe the circular path if the velocity of the body at the lowest point B of the circular path is, at least, v = \(\sqrt{5 g r}\).

The kinetic energy of the body at the point B

= \(\frac{1}{2} m v^2=\frac{1}{2} m \cdot 5 g r=\frac{5}{2} m g r\)

The falling body will acquire this kinetic energy in exchange for the potential energy that it possesses at the height h.

So, \(\frac{5}{2} m g r\) = mgh or, h = \(\frac{5}{2}\)r

Example 4. A body starts falling from the top of a smooth sphere of radius r. What angle does the body subtend at the center of the sphere when it just loses contact with the sphere and what will be its velocity then?
Solution:

Given

A body starts falling from the top of a smooth sphere of radius r.

Let the angle subtended by the body at the center of the sphere when it loses contact with the sphere be θ. The body just loses contact with the sphere when the component of gravitational force cannot provide the necessary centripetal force to the body so that it can continue in the circular path.

So, according to mgcosθ = \(\frac{mv^2}{r}\)……(1)

(v = velocity of the body when it just loses contact with the sphere)

According to the principle of conservation of energy, \(\frac{1}{2}\)\({mv^2}\)= mgr(1-cosθ)

[OC = OA – r, OB = rcosθ, BC = OC- OB – r- rcosθ = r(1 – cosθ)]

or, v² = 2gr(1 – cosθ)…(2)

From equations (1) and (2) we get, mg\(\cos \theta=\frac{m}{r} \cdot 2 g r(1-\cos \theta)\)

or, \(\cos \theta=2(1-\cos \theta)\)

or, \(\cos \theta=\frac{2}{3} \quad \text { or, } \theta=\cos ^{-1} \frac{2}{3}=48.2^{\circ}\)

Again, \(v^2=2 g r(1-\cos \theta)=2 g r\left(1-\frac{2}{3}\right)=\frac{2}{3} g r \)

∴ v = \(\sqrt{\frac{2}{3} g r .}\)

Work And Energy – Elastic And Inelastic Collisions

The total momentum of a system of interacting bodies remains constant in the absence of an external force. But generally, the total kinetic energy of the system is not conserved. In most collisions, a fraction of the kinetic energy transforms into heat and sound.

• If the total momentum and the total kinetic energy of a system are conserved, the collision is termed as an elastic collision. The collision of two billiard balls on a smooth board is almost elastic. Collisions in atoms and molecules, or interactions involving protons, electrons, neutrons, etc., are assumed to be elastic.
• On the other hand, if the total momentum is conserved, but the total kinetic energy is not, it is an inelastic collision. Most practical collisions are inelastic.

Derivation Of One-Dimensional Elastic Collision Between Two Particles: Suppose m₁ and m₂ are the masses of two particles that are moving with velocities \(\overrightarrow{u_1}\) and \(\overrightarrow{u_2}\) respectively(ux > 1ÿ) in the same direction along a straight line. They collide elastically, and after collision, move along the same direction with velocities \(\overrightarrow{v_1}\) and \(\overrightarrow{u_2}\) respectively.

Since one-dimensional motion is considered here vector notation can be dropped. Only components can be used with signs to indicate direction.

From the law of conservation of momentum, \(m_1 u_1+m_2 u_2=m_1 v_1+m_2 v_2\)

or, \( m_1\left(u_1-v_1\right)=m_2\left(v_2-u_2\right)\)…(1)

In an elastic collision, the total kinetic energy of the particles will also be conserved.

Hence, kinetic energy before collision = kinetic energy after collision

or, \(\frac{1}{2} m_1 u_1^2+\frac{1}{2} m_2 u_2^2=\frac{1}{2} m_1 v_1^2+\frac{1}{2} m_2 v_2^2\)

or, \(m_1\left(u_1^2-v_1^2\right)=m_2\left(v_2^2-u_2^2\right)\)…(2)

Now, (2) ÷(1) gives, \(u_1+v_1=v_2+u_2 \quad \text { or, } u_1-u_2=v_2-v_1\)…(3)

Hence, \(v_2=u_1+v_1-u_2\)

Substituting the value of \(v_2\) in equation (1) \(m_1\left(u_1-v_1\right)=m_2\left(u_1+v_1-2 u_2\right)\)

or, \(u_1\left(m_1-m_2\right)+2 u_2 m_2=v_1\left(m_1+m_2\right)\)

or, \(v_1=\frac{m_1-m_2}{m_1+m_2} u_1+\frac{2 m_2}{m_1+m_2} u_2\)

Similarly, taking the value of \(v_1\) from equation (3) and substituting it in equation (1), \(v_2=\frac{m_2-m_1}{m_1+m_2} u_2+\frac{2 m_1}{m_1+m_2} u_1\)

It is important to note that, equations (4) and (5) are symmetrical against interchange of the first and the second particles i.e., if the subscripts 1 and 2 are interchanged same equations (4) and (5) will be obtained.

Derivation Of One-Dimensional Elastic Collision Between Two Particles Special Cases:

1. Particles Are Of Equal Mass: In this case m₁ = m₂, hence from equations (4) and (5), v₁ = u₂ and v₂ = u₁.

Particles exchange their velocities after the collision.

2. Particles Are Of Equal Mass, And The Second Particle Is Initially At Rest: Here, m₁ = m₂ and u₂ = 0. Hence v₁ = 0 and v₂= u₁, from equations (4) and (5). The first particle comes to rest and the second particle gains the velocity of the first, after collision. Such events are frequent in games like billiards.

3. Particles Are Of Equal Mass, And The Second Particle Is Initially At Rest: Here, m₁ = m₂ and u₂ = 0. Equations (4) and (5) thus change as,

⇒ \(v_1=\frac{m_1-m_2}{m_1+m_2} u_1 \text {, and } v_2=\frac{2 m_1}{m_1+m_2} u_1\)

If m₁ and m₂ have different values, the first particle will not stop after collision. Hence, when a striker hits a stationary coin in a game of carrom, both the striker and the coin move after collision.

4. The First Particle Is Much Heavier Than The Second Particle And The Second Particle Is Initially At Rest.

In this case, m₁ >> m₂ and u₂ = 0.

Hence, m₁ – m₂ ≅ m₁ + m₂ ≅ m₁

∴ v₁ ≅ u₁ and v_{2 ≅} 2u₁

Therefore, after collision, the velocity of the first (massive) particle practically remains unchanged; but the second particle gains a velocity equal to almost twice the initial velocity of the first particle. In a collision, the velocity acquired by a body cannot be greater than twice the velocity of the collider.

5. The Second Particle Is Much Heavier Than The First And Is Initially At Rest: Here, m₂ >>m₁ and u₂ = 0. Hence m₁ -m₂ ≈ -m₂ and m₁ + m₂ ≈ m₂.

Values of v₁ and v₂ are thus v₁ ≈ -u₁ and v₂ = 0.

Hence, after the collision, the massive body will continue to be at rest; and the collider will recoil with the same magnitude of velocity. A collision between a tennis ball and the earth’s surface is of this type.

Work And Energy – Elastic Collisions Numerical Examples

Example 1. Two particles of equal mass moving towards each other with velocities 20 m · s^-1 and 30 m · s^-1 collide. If the collision is elastic, find their velocities after the collision.
Solution:

Given

Two particles of equal mass moving towards each other with velocities 20 m · s^-1 and 30 m · s^-1 collide. If the collision is elastic

The particles are moving towards each other. If the velocity (u₁) of one is 20 m · s^-1, then the velocity (u₂) of the other particle is -30 m · s^-1.

After the collision, suppose the velocities are v₁ and v₂ respectively

From the law of conservation of linear momentum, m x 20 – m x 30 = mv₁ + mv₂

or, v₁ + v₂ = -10…(1)

For elastic collision, v₂ – v₁ = u₁ – u₂

or, v₂ – v₁ = 20-(-30) = 50…..(2)

Solving equations (1) and (2), v₁ = -30 m · s^-1 and v₂ = 20 m · s^-1.

Example 2. Three balls A, B, and C of masses m₁, m_2, and m₃ respectively are kept at rest along a straight line. Now A moving in that straight line with velocity u₁ strikes B and then B moving with velocity u₂ strikes C. As a result velocity of C becomes u₃. If the collisions are elastic, show that u₃ ≅ 4u₁, when m₁ >> m₂ and m₂ >>m₃. In case A hits C directly, will the velocity of C be higher or lower?
Solution:

Given

Three balls A, B, and C of masses m₁, m_2, and m₃ respectively are kept at rest along a straight line. Now A moving in that straight line with velocity u₁ strikes B and then B moving with velocity u₂ strikes C. As a result velocity of C becomes u₃. If the collisions are elastic,

Suppose A acquires a velocity v₁ after collision with B. From the law of conservation of linear momentum, for collision between A and B, m₁u₁ = m₁v₁ + m₂u₂

or, m₁(u₁ – v₁) = m₂u₂…..(1)

From the law of conservation of kinetic energy for elastic collision \(\frac{1}{2} m_1 u_1^2=\frac{1}{2} m_1 v_1^2+\frac{1}{2} m_2 u_2^2\)

or, \(m_1\left(u_1^2-v_1^2\right)=m_2 u_2^2\)….(2)

Dividing equation (2) by (1), \(u_1+v_1=u_2 \text { or, } v_1=u_2-u_1\)

Inserting the value of \(v_1\) in equation (1), \(m_1\left(2 u_1-u_2\right)=m_2 u_2 \quad \text { or, } 2 m_1 u_1=\left(m_1+m_2\right) u_2\)

Hence, \(u_2=\frac{2 m_1}{m_1+m_2} \cdot u_1=\frac{2}{1+\frac{m_2}{m_1}} \cdot u_1\)

As \(m_1 \gg m_2, \frac{m_2}{m_1} \ll 1\); hence, \(1+\frac{m_2}{m_1} \approx 1\).

Thus, \(u_2 \approx 2 u_1\).

Similarly, for the collision of B with C, \(u_3 \approx 2 u_2\). Thus \(u_3 \approx 4 u_1\).

But, for the direct collision of A with C, the velocity gained by C, u₃‘ ≈ 2u₁

∴ u₃‘ is less than u₃.

Hence, in case of a direct collision between A and C, the speed of C will be less.

One-Dimensional Inelastic Collision Between Two Particles: In an inelastic collision, the total kinetic energy is not conserved, but the total momentum is conserved.

• Two particles are moving in the same direction in a straight line with velocities u₁ and u₂ respectively (u₁ > u₂). Before the collision, the distance between the two particles will decrease, and their velocity of approach will be = (u₁ – u₂).
• After collision, their separation increases with time i.e., (v₂ > v₁), where v₁ = velocity of first particle, v₂ = velocity of second particle after the impact. Therefore, ( v₂ – v₁) = the velocity of separation.

Coefficient Of Restitution: Coefficient of restitution is defined as the ratio of the velocity with which the two bodies separate after collision to the velocity of their approach before collision,

i.e., e = \(\frac{\text { relative velocity of separation after collision }}{\text { relative velocity of approach before collision }}\)

= \(\frac{v_2-v_1}{u_1-u_2}\)…..(2)

The value of e is practically constant for two specific bodies.

Perfectly Inelastic Collision: In this type of collision the two colliding bodies stick together and move as a single body, i.e., they move with a common velocity. A perfectly inelastic collision has been shown.

Suppose, two bodies of masses m₁ and m₂, moving with velocities u₁ and u₂ respectively in the same direction along a straight line, collide perfectly inelastically. After the collision, they stick together and move with velocity v along the same straight line.

From the principle of conservation of linear momentum, we get, \(m_1 u_1+m_2 u_2=\left(m_1+m_2\right) v\)

or, \(v^{\prime}=\frac{m_1 u_1+m_2 u_2}{m_1+m_2}\)…(2)

Perfectly Inelastic Collision Loss Of Kinetic Energy: Total kinetic energy before collision = \(\frac{1}{2} m_1 u_1^2+\frac{1}{2} m_2 u_2^2\)

Total kinetic energy after collision = \(\frac{1}{2}\left(m_1+m_2\right) v^2\)

∴ Loss of kinetic energy

= \(\frac{1}{2} m_1 u_1^2+\frac{1}{2} m_2 u_2^2-\frac{1}{2}\left(m_1+m_2\right) v^2\)

= \(\frac{1}{2} m_1 u_1^2+\frac{1}{2} m_2 u_2^2-\frac{1}{2}\left(m_1+m_2\right)\left(\frac{m_1 u_1+m_2 u_2}{m_1+m_2}\right)^2\)

= \(\frac{1}{2} m_1 u_1^2+\frac{1}{2} m_2 u_2^2-\frac{1}{2} \frac{\left(m_1 u_1+m_2 u_2\right)^2}{m_1+m_2}\)

= \(\frac{1}{2\left(m_1+m_2\right)}\)

= \(\frac{1}{2\left(m_1+m_2\right)}\) (\(m_1^2 u_1^2+m_1 m_2 u_1^2+m_1 m_2 u_2^2\))

= \(\frac{m_1 m_2}{2\left(m_1+m_2\right)}\left[u_1^2+u_2^2-2 u_1^2 u_2^2\right]\)

= \(\frac{m_1 m_2}{2\left(m_1+m_2\right)} \cdot\left(u_1 u_1^2-m_2^2 u_2^2-2 m_1 m_2 u_1 u_2 u_2\right]\)

Here (u₁– u₂) is the relative velocity of the two bodies before collision. So the loss of kinetic energy is proportional to the square of this relative velocity. Again if u₂ = 0 i.e., if the second body is at rest before collision the loss of kinetic energy becomes maximum.

Partially Elastic Collision Definition: In this type of collision the two bodies do not stick, but move separately along the same straight line with different velocities. Here momentum remains conserved while the total kinetic energy decreases.

The relative velocity of the two bodies before collision = u₁ – u₂ and that after collision = v₂ – v₁.

So, the coefficient of restitution, e = \(\frac{v_2-v_1}{u_1-u_2}\)…(4)

According to the law of conservation of momentum, \(m_1 u_1+m_2 u_2=m_1 \nu_1+m_2 v_2\)…(5)

Now from equation (4) we have, \(e\left(u_1-u_2\right)=v_2-v_1\)

Multiplying the above equation by \(m_2\) we have, \(e m_2 u_1-e m_2 u_2=m_2 \nu_2-m_2 v_1\)…(6)

Subtracting equation (6) from equation (5) we get, \(u_1\left(m_1-e m_2\right)+u_2(1+e) m_2=\left(m_1+m_2\right) v_1\)

∴ \(\nu_1=\frac{\left(m_1-e m_2\right) u_1+(1+e) m_2 u_2}{m_1+m_2}\)…(7)

Similarly, \(v_2=\frac{\left(m_2-e m_1\right) u_2+(1+e) m_1 u_1}{m_1+m_2}\)

Partially Elastic Collision Loss Of Kinetic Energy: \(\left(\frac{1}{2} m_1 u_1^2+\frac{1}{2} m_2 u_2^2\right)-\left(\frac{1}{2} m_1 v_1^2+\frac{1}{2} m_2 v_2^2\right)\)

= \(\frac{1}{2\left(m_1+m_2\right)}\left[\left(m_1+m_2\right)\left(m_1 u_1^2+m_2 u_2^2\right)-\right.\) \(\left.\quad\left(m_1+m_2\right)\left(m_1 v_1^2+m_2 v_2^2\right)\right]\)

Now, \(\left(m_1+m_2\right)\left(m_1 u_1^2+m_2 u_2^2\right)\)

= \(m_1^2 u_1^2+m_2^2 u_2^2+m_1 m_2\left(u_1^2+u_2^2\right)\)

= \(\left(m_1 u_1+m_2 u_2\right)^2+m_1 m_2\left(u_1-u_2\right)^2\)

Similarly, \(\left(m_1+m_2\right)\left(m_1 v_1^2+m_2 v_2^2\right)\)

= \(\left(m_1 v_1+m_2 v_2\right)^2+m_1 m_2\left(v_1-v_2\right)^2\)

= \(\left(m_1 u_1+m_2 u_2\right)^2+m_1 m_2 e^2\left(u_1-u_2\right)^2\)

(because \(v_2-v_1=e\) \((u_1-u_2)\))

So, the loss of kinetic energy = \(\frac{1}{2\left(m_1+m_2\right)}[\left(m_1 u_1+m_2 u_2\right)^2+m_1 m_2\left(u_1-u_2\right)^2\)

– \((m_1 u_1+m_2 u_2)^2-m_1 m_2 e^2\left(u_1-u_2\right)^2\)

= \(\frac{m_1 m_2}{2\left(m_1+m_2\right)}\left(1-e^2\right)\left(u_1-u_2\right)^2 \)…(9)

1. If e = 1, i.e., if the collision is perfectly elastic, then the loss of kinetic energy = 0, i.e., in case of a perfectly elastic collision kinetic energy remains conserved.
2. If e = 0 , i.e., if the collision is perfectly inelastic, kinetic energy decreases and the loss of kinetic energy = \(\frac{m_1 m_2}{2\left(m_1+m_2\right)}\left(u_1-u_2\right)^2\)
3. For partially elastic collision, 0 < e < 1.

Inelastic Collisions Numerical Examples

Example 1. Two bodies of masses 5 kg and 10 kg move towards each other with velocities 10 m · s^-1 and 14 m s^-1 respectively. If the coefficient of restitution is 0.8, find their velocities after the collision.
Solution:

Given

Two bodies of masses 5 kg and 10 kg move towards each other with velocities 10 m · s^-1 and 14 m s^-1 respectively. If the coefficient of restitution is 0.8

Let u₁ and u₂ respectively be the velocities of the first and the second body before the collision. After the collision, their velocities will be v₁ and v₂ respectively.

From the law of conservation of linear momentum \(m_1 u_1+m_2 u_2=m_1 v_1+m_2 v_2\)

Here, \(m_1=5 \mathrm{~kg}, \quad m_2=10 \mathrm{~kg}, \quad u_1=10 \mathrm{~m} \cdot \mathrm{s}^{-1}\) and \(u_2=-14 \mathrm{~m} \cdot \mathrm{s}^{-1}\)

[let us assume \(u_1\) to be taken along positive x-axis and \(u_2\) along negative x-axis]

∴ \(5 \times 10+10 \times(-14)=5 v_1+10 v_2\)

or, \(v_1+2 v_2=-18\)….(1)

Also, \(e=\frac{v_2-v_1}{u_1-u_2}\) or, \(0.8=\frac{v_2-v_1}{10-(-14)}\)

or, \(v_2-v_1=0.8 \times 24=19.2\)…..(2)

Hence, from (1) and (2), \(3 v_2=1.2 \text { or, } v_2=\frac{1.2}{3}=0.4 \mathrm{~m} \cdot \mathrm{s}^{-1}\)

and \(v_1=-18.8 \mathrm{~m} \cdot \mathrm{s}^{-1}\) which means after the collision the 5 kg mass advances along the negative x-axis and the 10 kg mass goes along positive x-axis.

Example 2. Two bodies of masses 1 kg and 0.5 kg towards each other with velocities 10 cm · s^-1 and 5 cm · s^-1 respectively. After the collision, the bodies coalesce (Join together). Find the common velocity after collision and the loss in kinetic energy.
Solution:

Given

Two bodies of masses 1 kg and 0.5 kg towards each other with velocities 10 cm · s^-1 and 5 cm · s^-1 respectively. After the collision, the bodies coalesce

Let the velocity of the combined mass after collision be v. From the law of conservation of momentum, \(m_1 u_1+m_2 u_2=\left(m_1+m_2\right)v\)

or, \(1 \times 0.1+0.5 \times(-0.05)=(1+0.5) v\)

or, \(0.1-0.025=1.5 v \quad \text { or, } v=\frac{0.075}{1.5}=0.05 \mathrm{~m} \cdot \mathrm{s}^{-1} .\)

This implies that the velocity of the combined mass is along the direction of the initial velocity of the 1kg mass.

The initial kinetic energy of the bodies = \(\frac{m_1 u_1^2}{2}+\frac{m_2 u_2^2}{2}=\frac{1}{2} \times 1 \times(0.1)^2+\frac{1}{2} \times(0.5) \times(0.05)^2\)

= \(5.625 \times 10^{-3} \mathrm{~J}\)

The final kinetic energy of the combined mass after the collision = \(\left(m_1+m_2\right) \frac{v^2}{2}\)

= \(\frac{1}{2} \times(1+0.5) \times(0.05)^2=1.875 \times 10^{-3} \mathrm{~J}\)

∴ Loss in kinetic energy = \((5.625-1.875) \times 10^{-3}=3.75 \times 10^{-3} \mathrm{~J} \text {. }\)

Example 3. A car of mass 2000 kg collides with a truck of mass 10⁴ kg moving at 48 km • h^-1. After the collision, the car rides up the truck, and the truck-car combination moves at 15 km • h^-1. What was the velocity of the car before the collision?
Solution:

Given

A car of mass 2000 kg collides with a truck of mass 10⁴ kg moving at 48 km • h^-1. After the collision, the car rides up the truck, and the truck-car combination moves at 15 km • h^-1

The collision is inelastic as both bodies move together after the collision. Suppose a body of mass m, moving with velocity u, collides inelastically along a straight line with a body of mass M and velocity v. After collision, the two masses combine and move with velocity V. Applying the law of conservation of momentum,

mu+Mv = (m + M)V or, mu – (m + M)V-Mv

or, u = \(\frac{m+M}{m} V-\frac{M}{m} \nu=\left(1+\frac{M}{m}\right) V-\frac{M}{m} \nu\)

Here \(m=2000 \mathrm{~kg}, \quad \nu=48 \mathrm{~km} \cdot \mathrm{h}^{-1}, \quad M=10000 \mathrm{~kg}\),

V = \(15 \mathrm{~km} \cdot \mathrm{h}^{-1}\)

∴ \(\frac{M}{m}=\frac{10000}{2000}=5\)

u = (1 + 5) x 15- 5 x 48 = 90 – 240 = -150 km · h^-1

The negative sign indicates that, before the collision, the car was moving in the direction opposite to that of the truck.

Example 4. A ball of mass 100 g was thrown vertically upwards with a velocity of 49 m · s^-1. At the same time another identical ball was dropped, to fall along the same path from a height of 98 m. After some time, the two balls collided and got stuck together. This combined mass then fell to the ground. How long was the combined mass in motion?
Solution:

Given

A ball of mass 100 g was thrown vertically upwards with a velocity of 49 m · s^-1. At the same time another identical ball was dropped, to fall along the same path from a height of 98 m. After some time, the two balls collided and got stuck together. This combined mass then fell to the ground.

Suppose the balls collide at a height h above the ground in time t₁. Analysing the upward motion of the 1st ball, we get, h = \(49 t_1-9.8 t_1^2 / 2\)…….(1)

Analysing the downward motion of the 2nd ball, we get, 98-h = \(9.8 t_1^2 / 2\)……(2)

From equations (1) and (2), 98 = 49t₁ or, t₁ = 2 s

At the time of the collision, let the velocity of the 1st ball be v₁ and that of the 2nd ball be v₂.

∴ v₁ = 49 – 9.8 x 2 = 29.4 m · s^-1 (upward) and

v₂ = 9.8 x 2 = 19.6 m · s^-1 (downward)

After the collision, let the velocity of the combined mass be V. Then according to the law of conservation of momentum, 0.1 x 29.4 – 0.1 x 19.6 = 2 x 0.1 x V [indicating downward motion with a negative sign]

or, V = 4.9 m · s^-1 (upward)

From equation (1), we get, h = 49 x 2 – 9.8 x (2)²/2 = 78.4 m

If the combined mass was in motion for a time t, then from the relation h = ut+ 1/2 gt², we get,

-78.4 = 4.9t – 1/2 x 9.8 t², or, t² – t- 16 = 0

∴ t = \(\frac{1}{2}(1 \pm \sqrt{1+4 \times 1 \times 16})\)

As t cannot be negative, we have, t = 4.53 s

Example 5. A bullet of mass 50 g is fired into a wooden block of mass 2 kg resting on a smooth table surface. The bullet enters at 50 m · s^-1 and gets embedded in the block. Find the final velocity of the block. Find the initial and the final kinetic energy of the block-bullet system.
Solution:

Given

A bullet of mass 50 g is fired into a wooden block of mass 2 kg resting on a smooth table surface. The bullet enters at 50 m · s^-1 and gets embedded in the block.

From the law of conservation of momentum, m₁v = (m₁ + m₂)V

or, V = \(\frac{m_1 \nu}{m_1+m_2}\)….(1)

Here, m₁ = 50 g = 0.05 kg, v = 50 m ·  s^-1 and m₂ = 2 kg.

∴ The final velocity of the block, V = \(\frac{0.05 \times 50}{0.05+2}=1.22 \mathrm{~m} \cdot \mathrm{s}^{-1}\)

Initial kinetic energy of the system, \(E_i=\frac{1}{2} \times 0.05 \times 50^2=62.5 \mathrm{~J}.\)

Final kinetic energy, \(E_f=\frac{1}{2} \times(2+0.05) \times(1.22)^2=1.52 \mathrm{~J}\)

Example 6. A bullet of mass 250 g, moving with a horizontal velocity of 400 m · s^-1, gets embedded in a target The target of mass 4.75 kg can move freely. Find the loss of kinetic energy due to this collision. What happens to this energy loss?
Solution:

Given

A bullet of mass 250 g, moving with a horizontal velocity of 400 m · s^-1, gets embedded in a target The target of mass 4.75 kg can move freely.

Mass of the bullet, m = 250 g = 0.25 kg, the velocity of the bullet, v = 400 m · s^-1, and the mass of the target, M = 4.75 kg.

After the impact, the target and the bullet both move with a velocity V (say).

From the law of conservation of momentum, mv = (M+ m)V

∴ V = \(\frac{m v}{M+m}=\frac{0.25 \times 400}{0.25+4.75}=20 \mathrm{~m} \cdot \mathrm{s}^{-1} .\)

Kinetic energy of the target and the bullet after collision = \(\frac{1}{2}(M+m) V^2=\frac{1}{2}(4.75+0.25) \times 20^2=1000 \mathrm{~J}\)

Kinetic energy of the bullet before the collision,

= \(\frac{1}{2} m v^2=\frac{1}{2} \times 0.25 \times(400)^2=20000 \mathrm{~J}\).

Hence, the kinetic energy lost in collision = 20000-1000 = 19000 J.

This kinetic energy transforms into heat and sound energy.

Collisions In Two Dimensions: Let a particle of mass m₁ moving with a velocity u₁ along AO, collide at point O with another particle of mass m₂ moving with velocity u₂ along BO. The point O is chosen as the origin, the direction of AO as the x-axis, and a perpendicular direction on the plane of AO and BO as the y-axis.

Before the collision, it is evident that, the x-component of momentum of the 1st particle = m₁u₁,

y -component of momentum of the 1st particle = 0,

x -component of momentum of the 2nd particle = -m₂u₂ cosθ,

y -component of momentum of the 2nd particle = m₂u₂ sinθ.

So, there is no component of momentum perpendicular to the xy-plane, i.e., along the z-direction. We know that, in both elastic and inelastic collisions, the total momentums of the particles remain conserved.

Thus, even after the collision between the two particles, there will be no z-component of momentum. This means that every two-particle collision, in general, is two-dimensional, i.e., such a collision is always confined in a plane.

In collision experiments, usually, a beam of incident particles hits a stationary target. After the collision, the first particle is observed by placing detector D in a definite orientation, say, at right angles with the direction of the incident beam.

In the figure, m₁, m₂: masses of the particles; u₁= velocity of m₁ before collision; u₂ = 0 = velocity of m₂ before collision;

v₁ = velocity of m₁ after collision deviated by 90°;

v₂ = velocity of m₂ after collision;

θ = angle made by v₂ with the direction of the incident beam.

Now, we may apply the principle of conservation of momentum, with the choice of the x and y-axes as shown

In the figure, x -direction: \(m_1 u_1+0=0+m_2 v_2 \cos \theta\)

or, \(m_2 v_2 \cos \theta=m_1 u_1\)…(1)

y-direction: \(0+0=m_1 v_1-m_2 v_2 \sin \theta\)

or, \(m_2 v_2 \sin \theta=m_1 v_1\)..(2)

Squaring and adding equations (1) and (2), \(m_2 v_2=m_1 \sqrt{u_1^2+v_1^2} \text { or, } v_2=\frac{m_1}{m_2} \sqrt{u_1^2+v_1^2}\)…(3)

Again, dividing equation (2) by equation (1), \(\tan \theta=\frac{v_1}{u_1} \quad \text { or, } \theta=\tan ^{-1} \frac{v_1}{u_1}\)…(4)

Usually, m₁, m_2, and are known prior to the experiment, and v₁ is measured by the detector D. So the magnitude and direction of the velocity (v₂) of the second particle after collision can be calculated using equations (3) and (4).

General Analysis Of Two-Dimensional Collisions: We have already seen that the motion of two particles, both before and after they collide with each other, is confined in a plane. Let that plane be chosen as the xy-plane. We assume that the magnitudes and the directions of the velocities (i.e., of the momenta) of the two particles, before the collision, are known beforehand.

Then, four unknown quantities, associated with the collision, are to be solved. They are the magnitudes and the directions of the velocities of the two particles after the collision.

If the collision is elastic, we can construct at most three equations from the conservation principles:

1. Conservation Of Momentum Along The X-Axis,
2. Conservation Of Momentum Along The Y-Axis,
3. Conservation Of Kinetic Energy

These three equations are not sufficient to solve for four unknown quantities. If the collision is inelastic, the kinetic energy is not conserved. Then we have only two equations at hand. So, for a complete analysis of a two-particle collision, we should have some information on the particles after the collision.

• For example, in the collision experiment described above, we had placed a detector in such a way—at right angles to the initial direction of motion of one particle—that the final direction of the velocity of that particle was already assigned. Moreover, the detector could measure the magnitude of its velocity.
• So, we had to solve for only two unknown quantities—the magnitude and direction of the velocity of the second particle. So, only two equations were sufficient— equations obtained from the momentum conservation along the x- and the y-axes. Even the kinetic energy equation was not necessary; so the treatment was applicable to elastic as well as inelastic collisions.
• One other point is to be noted. In general, both the particles may be in motion before the collision. However, the collision may be observed by assuming one of the particles to be at rest.

The particle 2 is at rest; then the velocity of particle 1 is actually its relative velocity with respect to particle 2. This means that the frame attached to particle 2 has been taken as the frame of reference. So, there is no loss of generality if we describe a two-particle collision as a moving particle colliding with a stationary target.

Expansion Of Gases Relation Between Pressure, Temperature And Density

Relation Between Pressure, Temperature And Density:

Let an ideal gas is taken of molecular weight M (say).

For m₁ mass of the gas, the volume and density be V₁ and ρ₁, respectively at a temperature T₁ K. When its mass is m₂ the volume and density are V₂ and ρ₂ respectively at T₂ K.

∴ \(V_1=\frac{m_1}{\rho_1} \text { and } V_2=\frac{m_2}{\rho_2}\)

If p₁ and p₂ are the pressures at those temperatures, then from the equation pV = nRT = m/M RT we have,

⇒ \(\frac{p_1 V_1}{m_1 T_1}=\frac{p_2 V_2}{m_2 T_2}\) [since for a particular gas M is constant]

or, \(\frac{p_1}{\rho_1 T_1}=\frac{p_2}{\rho_2 T_2}\)

∴ \(\frac{p}{\rho T}=\) constant

1. If the pressure is a constant, \(\rho_1 T_1=\rho_2 T_2 \text { i.e., } \rho T=\text { constant or, } \rho \propto \frac{1}{T}\)

i.e., at constant pressure, the density of a gas is inversely proportional to Its absolute temperature.

2. If the temperature is a constant,

i.e., at a constant temperature, the density of a gas is directly proportional to its pressure.

Expansion Of Gases Relation Between Pressure, Temperature And Density Numerical Examples

Example 1. Temperature and pressure on top of a hill are 7°C and 70 cmHg, and the corresponding values at its base are 27°C and 76 cmHg. Compare the densities of air at the top and the base of the hill.
Solution:

Given

Temperature and pressure on top of a hill are 7°C and 70 cmHg, and the corresponding values at its base are 27°C and 76 cmHg.

We use the relation \(\frac{p_1}{\rho_1 T_1}=\frac{p_2}{\rho_2 T_2}\)

Here, p₁ = 70 cmHg, T₁ = 273 + 7 = 280 K, p₂ = 76 cmHg and T₂ = 273 + 27 = 300 K.

⇒ \(\frac{70}{\rho_1 \times 280}=\frac{76}{\rho_2 \times 300} \text { or, } \frac{\rho_1}{\rho_2}=\frac{300 \times 70}{76 \times 280}=\frac{75}{76} \text {. }\)

Example 2. Density of argon at 27°C and 76 cmHg pressure Is 1.6 g · L^-1. An electric bulb of volume 200 cm3 is filled with argon. The pressure of the gas inside the bulb is 75 cmHg and the average temperature is 127°C. Find the mass of argon gas in the bulb.
Solution:

Given

Density of argon at 27°C and 76 cmHg pressure Is 1.6 g · L^-1. An electric bulb of volume 200 cm3 is filled with argon. The pressure of the gas inside the bulb is 75 cmHg and the average temperature is 127°C.

Let the density of the gas in the bulb at a pressure of 75 cmHg and at 127°C =ρ₂

Here, p₁ = 76 cmHg, ρ₁ = 1.6 g · L^-1,

T₁ = 273 + 27 = 300 K

p₂ = 75 cmHg,

T₂ = 273 + 127 = 400 K

From the relation \(\frac{p_1}{\rho_1 T_1}=\frac{p_2}{\rho_2 T_2}\) we get,

⇒ \(\frac{76}{1.6 \times 300}=\frac{75}{\rho_2 \times 400}\)

∴ \(\rho_2=\frac{75 \times 1.6 \times 300}{76 \times 400}=1.184 \mathrm{~g} \cdot \mathrm{L}^{-1}\)

∴ Mass of argon gas of volume 200cm³

= \(\frac{200}{1000} \times 1.184=0.2368 \mathrm{~g} .\)

Example 3. At a place, air pressure is 75 cmHg and the temperature is 27°C. At another place, the respective values are 70 cmHg and 17°C. Compare the densities of air in the two places.
Solution:

Given

At a place, air pressure is 75 cmHg and the temperature is 27°C. At another place, the respective values are 70 cmHg and 17°C.

We have, \(\frac{p_1}{\rho_1 T_1}=\frac{p_2}{\rho_2 T_2}\)

Given, p₁ = 75 cmHg, T₁ = 273 + 27 = 300 K, p₂ = 70 cmHg

and T₂ = 273 + 17 = 290 K

∴ \(\frac{75}{\rho_1 \times 300}=\frac{70}{\rho_2 \times 290} \text { or, } \frac{\rho_1}{\rho_2}=\frac{75 \times 290}{300 \times 70}=\frac{29}{28}\)

∴ \(\rho_1: \rho_2=29: 28\)

Example 4. When an air bubble rises from the bottom of a lake to the upper surface, its diameter increases from 1 mm to 2 mm. If the atmospheric pressure is 76 cmHg, calculate the depth of the lake. Density of mercury is 13.6 g · cm^-3.
Solution:

Given

When an air bubble rises from the bottom of a lake to the upper surface, its diameter increases from 1 mm to 2 mm. If the atmospheric pressure is 76 cmHg,

Density of mercury is 13.6 g · cm^-3.

Let the depth of the lake be h cm.

Pressure on the air bubble, at the bottom of the lake, p₁ = atmospheric pressure + pressure of water at depth h = (76 x 13.6 x 981 + h x 1 x 981) dyn · cm^-2, and pressure at the surface of the lake, p₂ = atmospheric pressure = 76 x 13.6 x 981 dyn · cm^-2

Volumes of the bubble: at the bottom, \(V_1=\frac{4}{3} \pi\left(\frac{1}{20}\right)^3 \mathrm{~cm}^3\)

at the top, \(V_2=\frac{4}{3} \pi\left(\frac{1}{10}\right)^3 \mathrm{~cm}^3\)

Using Boyle’s law: p₁ V₁ = p₂ V₂, we get,

(h x 1 x 981 + 76 x 13.6 x 981) x \(\times \frac{4}{3} \pi\left(\frac{1}{20}\right)^3\)

= 76 x 13.6 x 981 x \(\times \frac{4}{3} \pi\left(\frac{1}{10}\right)^3\)

or, (h x 76 x 13.6) x 1/8 = 76 x 13.6

or, h = 76 x 13.6 x 8 – 76 x 13.6

∴ The lake is 7235.2 cm or 72.352 m deep

Example 5. An electronic vacuum tube is constructed and sealed at 27°C and 1.2 x 10^-6 cm Hg pressure. The tube has a volume of 100 cm³. Calculate the number of gas molecules left In the tube. Avogadro number is 6.02 x 10²³, and the gas occupies a volume of 22.4 litres at STP.
Solution:

Given

An electronic vacuum tube is constructed and sealed at 27°C and 1.2 x 10^-6 cm Hg pressure. The tube has a volume of 100 cm³.

Avogadro number is 6.02 x 10²³, and the gas occupies a volume of 22.4 litres at STP

Molar density at STP = \(\frac{6.02 \times 10^{23}}{22.4 \times 10^3} \text { molecules } \cdot \mathrm{cm}^{-3}\)

Let the number of molecules left in the tube, after sealing, be n.

From gas equation we have, \(\frac{p_1}{\rho_1 T_1}=\frac{p_2}{\rho_2 T_2}\)

Here, \(p_1=1.2 \times 10^{-6} \mathrm{cmHg}\)

⇒ \(\rho_1=\frac{n}{100} \text { molecules } \cdot \mathrm{cm}^{-3}\)

⇒ \(T_1=273+27=300 \mathrm{~K} \text {, }\)

⇒ \(p_2=76 \mathrm{cmHg} \text {, }\)

⇒ \(\rho_2=\frac{6.02 \times 10^{23}}{22.4 \times 10^3} \text { molecules } \cdot \mathrm{cm}^{-3}\)

and \(T_2=273 \mathrm{~K}\)

∴ \(\frac{1.2 \times 10^{-6}}{\frac{n}{100} \times 300}=\frac{76}{\frac{6.02 \times 10^{23}}{22.4 \times 10^3} \times 273}\)

i.e., n = \(\frac{1.2 \times 10^{-6} \times 100 \times 6.02 \times 10^{23} \times 273}{300 \times 76 \times 22.4 \times 10^3}\)

= \(3.86 \times 10^{13} \text {. }\)

Example 6. While constructing a bulb of volume 250 cm³, it is sealed at 27° C temperature and 10^-3 mmHg pres-sure. Find the number of gas molecules in the bulb. Avogadro number = 6.0 x 10²³.
Solution:

Given

While constructing a bulb of volume 250 cm³, it is sealed at 27° C temperature and 10^-3 mmHg pres-sure.

Volume of air inside the bulb, V₁ = 250 cm³; pressure of air, p₁ = 10^-3 mmHg = 10^-4 cmHg; temperature T₁ = 27 + 273 = 300 K.

Let at STP the volume of air = V₂ cm³; p₂ = 76 cmHg; T₂ = 273 K

∴ Using \(\frac{p_1 V_1}{T_1}=\frac{p_2 V_2}{T_2}\) we get,

⇒ \(V_2=\frac{p_1 V_1 T_2}{T_1 p_2}=\frac{10^{-4} \times 250 \times 273}{300 \times 76}=29.934 \times 10^{-5} \mathrm{~cm}^3\)

At STP, number of molecules in air of volume 22400 cm³ = 6 x 10²³.

So, that in volume 29.934 x 10^-5 cm³

= \(\frac{6 \times 10^{23}}{22400} \times 29.934 \times 10^{-5}=8.018 \times 10^{15}\)

Example 7. Two containers of volume 5 L and 3 L contain air at 3 standard atmospheres and 7 standard atmospheres respectively. The containers are now connected by a short narrow tube. What will be the common pressure in both containers?
Solution:

Given

Two containers of volume 5 L and 3 L contain air at 3 standard atmospheres and 7 standard atmospheres respectively. The containers are now connected by a short narrow tube.

Suppose the containers have n₁ and n₂ number of moles of the gas.

∴ 5×3 = n₁RT….(1)

and 3×7 = n₂RT…(2)

Therefore, 15 + 21 = (n₁ + n₂)RT [adding (1) and (2)]

or, (n₁ + n₂)RT= 36

At constant temperature, let the common pressure attained be p.

At that common pressure p, p(5 + 3) = (n₁ + n₂)RT = 36

∴ p = 36/8 = 4.5 standard atmospheres.

Example 8. Two bulbs of equal volume are connected by a narrow tube of negligible volume and filled with a gas at STP. If one of the bulbs is kept in melting ice and the other in a water bath at 62°C, what will be the new pressure of the gas?
Solution:

Given

Two bulbs of equal volume are connected by a narrow tube of negligible volume and filled with a gas at STP. If one of the bulbs is kept in melting ice and the other in a water bath at 62°C

Let volume of each bulb =Vcm³ and each contains n number of moles of the gas.

∴ Total number of moles contained in the bulb initially

= \(2 n=\frac{76 \times 2 V}{R \times 273} \quad[\text { from } p V=n R T]\)….(1)

Let the final pressure in both the bulbs = p.

Number of moles in one of the bulbs = \(\frac{p V}{R \times 273}\)

and that in the other bulb  = \(\frac{p V}{T(273+62)}=\frac{p V}{R \times 335}\)

∴ Total number of moles contained in the bulbs = \(\frac{p V}{R \times 273}+\frac{p V}{R \times 335}=\frac{p V}{R}\left(\frac{1}{273}+\frac{1}{335}\right)\)…(2)

From (1) and (2) we get, \(\frac{76 \times 2 V}{R \times 273}=\frac{p V}{R}\left[\frac{1}{273}+\frac{1}{335}\right]\)

or, \(\frac{p V}{R}(0.0037+0.0030)=\frac{76 \times V \times 2}{R \times 273}\)

or, \(p \times 0.0067=0.5568 or, p=\frac{0.5568}{0.0066}=83.10 \mathrm{cmHg}\).

Example 9. An air bubble rises from the bottom of a lake to its upper surface. The diameters of the bubble at the bottom and the surface are 3.6 mm and 4 mm respectively. Depth of the lake is 2.5 m and the temperature at the upper surface is 40°C. Find the temperature at the bottom of the lake. Ignore the change in density of water with height. (Atmospheric pressure = 76 cmHg and g = 980cm · s^-2)
Solution:

Given

An air bubble rises from the bottom of a lake to its upper surface. The diameters of the bubble at the bottom and the surface are 3.6 mm and 4 mm respectively. Depth of the lake is 2.5 m and the temperature at the upper surface is 40°C.

Pressure at the bottom of the lake p₁ =76x 13.6×980 + 250x 1×980

= (76 x 13.6 + 250) x 980 dyn · cm^-2

Volume of the air bubble at the bottom \(V_1=\frac{4}{3} \pi(0.18)^3 \mathrm{~cm}^3\)

Let the temperature at the bottom of the lake = T₁

Now, at the surface, pressure p₂ = 76 x 13.6 x 980 dyn · cm^-2

volume of the air bubble \(V_2=\frac{4}{3} \pi(0.2)^3 \mathrm{~cm}^3 \text {; }\); temperature T₂ = 273 + 40 = 313 K

Now using \(\frac{p_1 V_1}{T_1}=\frac{p_2 V_2}{T_2}\), we get

⇒ \(\frac{(76 \times 13.6+250) \times 980 \times \frac{4}{3} \pi(0.18)^3}{T_1 . \quad}=\frac{76 \times 13.6 \times 980 \times \frac{4}{3} \pi(0.2)^3}{313}\)

or, \(T_1=283.37 \mathrm{~K}=283.37-273^{\circ} \mathrm{C}=10.37^{\circ} \mathrm{C}\).

Example 10. A balloon at STP can lift a total mass of 175 kg attached with it When the barometer reads 50 cmHg and the temperature becomes -10°C at an upper point to where the balloon rises, find the maximum mass that can be lifted. Consider the volume of the balloon to be a constant.
Solution:

Given

A balloon at STP can lift a total mass of 175 kg attached with it When the barometer reads 50 cmHg and the temperature becomes -10°C at an upper point to where the balloon rises

The change in lifting capacity is due to the change in the upthrust, as the density of air changes at higher altitude due to the change in temperature and pressure.

Let V = volume of the balloon, ρ₁ = density at STR ρ₂ = density at 50 cmHg and -10°C; M = mass it can carry at the given altitude.

Hence, from Archimedes’ principle, \(V \rho_1=175 \mathrm{~kg} \text { and } V \rho_2=M \mathrm{~kg}\)

Now, \(\frac{p_1}{\rho_1 T_1}=\frac{p_2}{\rho_2 T_2} or, \frac{p_1}{V \rho_1 T_1}=\frac{p_2}{V \rho_2 T_2}\)

Substituting the corresponding values, we get, \(\frac{76}{175 \times 273}=\frac{50}{M \times 263}\)

or, M = \(\frac{175 \times 273 \times 50}{263 \times 76}=119.5 \mathrm{~kg}\).

Example 11. A chamber contains a mass m₁ of a gas at pressure p₁. A second chamber contains a mass m₂ of the same gas at pressure p₂. If the two chambers are now connected, what will be the pressure of the gas mixture?
Solution:

Given

A chamber contains a mass m₁ of a gas at pressure p₁. A second chamber contains a mass m₂ of the same gas at pressure p₂. If the two chambers are now connected,

Let V₁ = volume of the 1st chamber, V₂ = volume of the 2nd chamber,

M = molecular weight of the gas, p = final pressure of the gas mixture,

T = constant temperature before and after mixing of the gases.

We use the relation \(p V=n R T=\frac{m}{M} R T\)

∴ For the 1st chamber, \(p_1 V_1=\frac{m_1}{M} R T\)

and for the 2nd chamber, \(p_2 V_2=\frac{m_2}{M} R T\)

∴ \(p_1 V_1+p_2 V_2=\left(m_1+m_2\right) \frac{R T}{M}\)

Using the relation for the gas mixture, we have \(p\left(V_1+V_2\right)=\left(m_1+m_2\right) \frac{R T}{M}\)

∴ \(p\left(V_1+V_2\right)=p_1 V_1+p_2 V_2 \text { or, } p=\frac{p_1 V_1+p_2 V_2}{V_1+V_2}\)

Now, \(V_1=\frac{m_1}{p_1} \frac{R T}{M} and V_2=\frac{m_2}{p_2} \frac{R T}{M}\).

So, \(V_1+V_2=\left(\frac{m_1}{p_1}+\frac{m_2}{p_2}\right) \frac{R T}{M}=\frac{p_2 m_1+p_1 m_2}{p_1 p_2} \frac{R T}{M}\)

∴ p = \(\left(m_1+m_2\right) \frac{R T}{M} \cdot \frac{p_1 p_2}{p_2 m_1+p_1 m_2} \cdot \frac{M}{R T}\)

= \(\frac{p_1 p_2\left(m_1+m_2\right)}{p_2 m_1+p_1 m_2} .\)

Example 12. A 100 cm long glass capillary tube, closed at both ends, has a mercury thread of length 10 cm. When the tube is horizontal, the mercury thread stays at the middle of the tube with air columns of equal length on either side, at 76 cmHg pressure and 27°C. Now, the temperature of one side is changed to 0°C, and of the other side to 127°C. Find the length and the pressure of the air column kept at 0°C. Neglect expansions of glass and mercury.
Solution:

Given

A 100 cm long glass capillary tube, closed at both ends, has a mercury thread of length 10 cm. When the tube is horizontal, the mercury thread stays at the middle of the tube with air columns of equal length on either side, at 76 cmHg pressure and 27°C. Now, the temperature of one side is changed to 0°C, and of the other side to 127°C.

Let the area of the cross-section of the capillary be α cm².

In the first case, the volume of air on either side of the mercury thread in the tube, V₁ = 45α cm³; temperature,

T₁ = 273 + 27 = 300 K; pressure of air, p₁ = 76 cmHg.

In the second case, let the left side of the Hg thread inside the tube be at 0°C or 273 K and the other side be at 127°C or 400 K.

Let the pressure of the confined air on each side = p cm.

Let l = length of the air column at 0°C; (90 -l) = length of the other air column at 127°C.

Using the equation of state, \(\frac{p_1 V_1}{T_1}=\frac{p_2 V_2}{T_2}, \frac{76 \times 45 \alpha}{300}=\frac{p l \alpha}{273}\)…(1)

and \(\frac{76 \times 45 \alpha}{300}=\frac{p(90-l) \alpha}{400}\)…(2)

From (1) and (2) we get, \(\frac{p \times l \times \alpha}{273}=\frac{p(90-l) \alpha}{400}\)

or, \(400 l=273(90-l) or, 673 l=273 \times 90\)

∴ l = \(\frac{90 \times 273}{673}=36.5 \mathrm{~cm} \text {. }\)

From equation (1), \(\frac{76 \times 45 \alpha}{300}=\frac{p \times 36.5 \alpha}{273} \text { or, } p=85.3 \mathrm{cmHg}\)

∴ The length of the air column at 0°C is 36.5 cm and its pressure is 85.3 cmHg.

Example 13. Two heat proof containers of volumes IL and 3L are connected by a tube. Keeping the valve attached to the tube dosed, the 1st container is filled with nitrogen at 0°C and at 0.5 standard atmosphere pressure, and the 2nd container with argon at 100°C and at 1.5 standard atmosphere pressure. The temperature of the gas mixture becomes 79 °C when the valve is opened. Find the pressure of the gas mixture.
Solution:

Given

Two heat proof containers of volumes IL and 3L are connected by a tube. Keeping the valve attached to the tube dosed, the 1st container is filled with nitrogen at 0°C and at 0.5 standard atmosphere pressure, and the 2nd container with argon at 100°C and at 1.5 standard atmosphere pressure. The temperature of the gas mixture becomes 79 °C when the valve is opened.

In the 1st container, pressure of nitrogen (p₁) = 0.5 standard atmosphere, volume (V₁) = 1 L, temperature (T₁) = 0 + 273 = 273 K;

and in the 2nd container, pressure of argon (p₂) = 1.5 standard atmosphere, volume (V₂) = 3 L, temperature (T₂) = 100 + 273 = 373 K

When the valve is opened, let the pressure of the gas mixture be p volume, V = 4L and temperature, T = 273 + 79 = 352 K

As \(\frac{p V}{T}=n R\),

we have, \(n_1 R+n_2 R=\left(n_1+n_2\right) R\)

or, \(\frac{p_1 V_1}{T_1}+\frac{p_2 V_2}{T_2}=\frac{p V}{T}\)

or, \(\frac{0.5 \times 1}{273}+\frac{1.5 \times 3}{373}=\frac{p \times 4}{352} \)

or, \(\frac{p}{88}=0.0018+0.0120\)

∴ p =0.0138×88 = 1.2144 standard atmosphere.

Example 14. Two glass bulbs of volumes 3 L and 1 L are connected by a narrow tube. The system is filled with air at 30°C temperature and at 76 cmHg pressure. Now the bulb of volume 3 L is immersed in water vapour at temperature 100°C while the other bulb is kept at 30°C. Find the air pressures in the two bulbs. Neglect the volume expansion of the 3 L bulb.
Solution:

Given

Two glass bulbs of volumes 3 L and 1 L are connected by a narrow tube. The system is filled with air at 30°C temperature and at 76 cmHg pressure. Now the bulb of volume 3 L is immersed in water vapour at temperature 100°C while the other bulb is kept at 30°C.

Initially the total volume of the bulbs (V) = 3 + l= 4L,air pressure (p₁) = 76 cmHg, temperature of air (T₁) = 273 + 30 = 303 K

In the 2nd case, air pressure = p₂

temperature of 3 L bulb = 100 + 273 = 373 K

temperature of 1L bulb = 30 + 273 = 303 K

We use the relation, n = \(\frac{p V}{R T}\)

In the 1st case, number of gram-molecules in the gas

= \(\frac{76 \times 3}{R \times 303}+\frac{76 \times 1}{R \times 303}=\frac{76}{R \times 303}(3+1)=\frac{76 \times 4}{R \times 303}\)

Let in the 2nd case, air pressure = p

No. of gram-molecules = \(\frac{p \times 3}{R \times 373}+\frac{p \times 1}{R \times 303}=\frac{p}{R}\left(\frac{3}{373}+\frac{1}{303}\right)\)

∴ No. of gram-molecules is unchanged,

=  \(\frac{76 \times 4}{R \times 303}=\frac{p}{R}\left(\frac{3}{373}+\frac{1}{303}\right) \text { or, } \frac{76 \times 4}{303}=p\left(\frac{909+373}{373 \times 303}\right)\)

or, \(p=\frac{76 \times 4 \times 373}{1282}=88.4 \mathrm{cmHg} .\)

Example 15. A narrow tube of uniform cross-section is closed at one end. Inside this tube a mercury thread of length hem detaches some air from the atmosphere outside. When the tube is held vertical keeping its closed end up, the length of the confined air column becomes l₁ cm. Again the length of the air column becomes l₂ cm when the tube is held vertical keeping its open end up. Find the magnitude of the atmospheric pressure.
Solution:

Given

A narrow tube of uniform cross-section is closed at one end. Inside this tube a mercury thread of length hem detaches some air from the atmosphere outside. When the tube is held vertical keeping its closed end up, the length of the confined air column becomes l₁ cm. Again the length of the air column becomes l₂ cm when the tube is held vertical keeping its open end up.

Let the atmospheric pressure =pcmHg; cross-section of the tube = α cm².

When the closed end of the tube is at the top, the volume of the confined air column V₁ = l₁αcm³;

pressure of confined air p₁ = (p- h) cmHg.

Now when the open end is at the top the volume of the confined air column V₂ = l₂ α cm³

pressure of confined air p₂ = (p + h) cmHg.

According to Boyle’s law p₁V₁ = p₂ V₂

or, \((p-h) l_1 \alpha=(p+h) l_2 \alpha\)

or, \((p-h) l_1=(p+h) l_2 or, p=\frac{l_1+l_2}{l_1-l_2} h\)

Hence, atmospheric pressure = \(\frac{l_1+l_2}{l_1-l_2}\) h cmHg.

Example 16. A glass tube of uniform area of cross-section and open at one end, encloses some air at 27°C by a 4 cm long mercury thread that acts like a piston. When the tube is held vertical with its open end up length of the air column in the tube is 9 cm. When the open end is held downwards by turning the tube, the length of the enclosed air column becomes 10 cm. Find

1. The value of the atmospheric pressure,
2. The temperature at which the length of the air column becomes 9 cm again, while the tube is still held inverted.

Solution:

Given

A glass tube of uniform area of cross-section and open at one end, encloses some air at 27°C by a 4 cm long mercury thread that acts like a piston. When the tube is held vertical with its open end up length of the air column in the tube is 9 cm. When the open end is held downwards by turning the tube, the length of the enclosed air column becomes 10 cm.

Let the atmospheric pressure = p cmHg, and the area of the cross-section of the tube = α cm².

When the open end is kept upwards, the volume of the confined air, V₁ = 9αcm³ and pressure p₁ = (p + 4)cmHg.

On inverting the tube, volume of the confined air V₂ = 10α cm³ and pressure p₂ =(p-4)cmHg.

Assuming temperature to be constant at 27°C, from Boyle’s law p₁V₁ = p₂V₂

or, 9α(p + 4) = 10α(p-4)

or, 9p + 36 = 10p-40, or, p= 76 cmHg

Suppose at T₂ K, the length of the air column becomes 9 cm again when the tube is still held inverted.

Then its pressure, p₂ = 76 – 4 = 72 cmHg and volume V₂ = 9αcm³.

∴ \(\frac{p_1 V_1}{T_1}=\frac{p_2 V_2}{T_2}\)

or, \(\frac{(76+4) \times 9 \alpha}{273+27}=\frac{72 \times 9 \alpha}{T_2} \quad \text { or, } \frac{80}{300}=\frac{72}{T_2}\)

∴ \(T_2=270 \mathrm{~K}=(270-273)^{\circ} \mathrm{C}=-3^{\circ} \mathrm{C}\)

Example 17. A uniform glass tube closed at both ends, encloses air columns of equal lengths on either side of a mercury thread of length 5 cm, when the tube is placed horizontally. The pressure of the enclosed air is p. When the tube is placed at 60° to the vertical, the upper and the lower columns of air are of lengths 46 cm and 44.5 cm respectively. Find the value of p. The Temperature of the system remains constant at 30°C.
Solution:

Given

A uniform glass tube closed at both ends, encloses air columns of equal lengths on either side of a mercury thread of length 5 cm, when the tube is placed horizontally. The pressure of the enclosed air is p. When the tube is placed at 60° to the vertical, the upper and the lower columns of air are of lengths 46 cm and 44.5 cm respectively.

Let in the first case, length of air column on each side of the mercury thread = l.

Again, length of the tube = 44.5 + 46 + 5 = 95.5 cm.

∴ \(l=\frac{95.5-5}{2}=45.25 \mathrm{~cm}\)

In the second case, p₂ = p₁ + 5cos60°

= p₁+ 2 = (p₁+ 2.5) cmHg

If in this case, V₁ and V₂ are the volumes of the upper and the lower air columns in the tube, then according to Boyle’s law,

p₁V₁ = p₂V₂

or, p₁ · 46α’ = (p₁ +2.5) x 44.5α

[where α = cross-sectional area of the tube]

Solving we get, p₁= 74.17 cm

Again, pV = p₁V₁

or, \(p=\frac{p_1 V_1}{V}=\frac{74.17 \times 46 \alpha}{45.25 \alpha}=75.4 \mathrm{~cm}\)

i.e., p = 75.4 cmHg

Example 18. The reading in a barometer changes from 75 cmHg to 25 cmHg when 10 cm³ of air at atmospheric pressure is introduced in the vacuum space of the barometer tube. What is the volume of air In the tube?
Solution:

Given

The reading in a barometer changes from 75 cmHg to 25 cmHg when 10 cm³ of air at atmospheric pressure is introduced in the vacuum space of the barometer tube.

Pressure of confined air =75-25 = 50 cmHg.

Initial volume of air V₁ = 10 cm³ and initial pressure p₁ = 75 cmHg.

Final volume of air = V₂ cm³ and final pressure p₂ = 50 cmHg.

Using Boyle’s law p₁V₁ = p₂V₂ we get,

⇒ \(V_2=\frac{p_1 V_1}{p_2}=\frac{75 \times 10}{50}=15\)

So, the volume occupied by the air inside the barometer is 15 cm³.

Example 19. The reading of a barometer decreases from 75 cmHg to 65 cmHg when some air is introduced in the vacuum space of the tube. The initial length of the space was 6 cm. If the area of the cross-section of the tube is 1 cm², what is the volume of this air at standard pressure?
Solution:

Given

The reading of a barometer decreases from 75 cmHg to 65 cmHg when some air is introduced in the vacuum space of the tube. The initial length of the space was 6 cm. If the area of the cross-section of the tube is 1 cm²

Standard pressure means p₁ = 76 cmHg.

The volume of air at that pressure = V₁ cm³.

Pressure of confined air p₂ = 75 – 65 = 10 cmHg.

Volume of confined air V₂ = {6+ (75-65)} x 1 = 16 cm³.

Using Boyle’s law, p₁V₁ = p₂V₂

or, \(V_1=\frac{p_2 V_2}{p_1}=\frac{10 \times 16}{76}=2.105\)

So the confined air occupies a volume of 2.105 cm³ at standard pressure.

Example 20. An air bubble of volume 20 cm³ forms in a lake at a depth of 40 m below the water surface. What will be its volume when it rises just below the water surface? (Standard atmospheric pressure = 76 cmHg)
Solution:

Given

An air bubble of volume 20 cm³ forms in a lake at a depth of 40 m below the water surface.

Volume of the air bubble at a depth of 40 cm, V₁ = 20 cm³.

Pressure on the bubble there, p₁ = atmospheric pressure + pressure due to 4000 cm of water column

= 76 x 13.6 x 980 + 4000 x 1 x 980

= (76 x 13.6 + 4000) x 980 dyn · cm^-2

Let the volume of the bubble just below die surface be V₂.

Pressure on the bubble just below the surface, p₂ = 76 x 13.6 x 980 dyn • cm²

From Boyle’s law, p₁V₁ = p₂V₂

or, 980 x (76 x 13.6 + 4000) x 20 = V₂ x 76 x 13.6 x 980

∴ \(V_2=\frac{(76 \times 13.6+4000) \times 20}{76 \times 13.6}=97.4 \mathrm{~cm}^3\)

Example 21. In a capillary tube, closed at one end, some air is enclosed by a mercury thread of length 10 cm. When the tube is kept horizontal the length of the air column is 17 cm. When it is held vertical with the open end up, the length changes to 15 cm. What will be the length of the air column when the tube is held vertical with the open end downwards?
Solution:

Given

In a capillary tube, closed at one end, some air is enclosed by a mercury thread of length 10 cm. When the tube is kept horizontal the length of the air column is 17 cm. When it is held vertical with the open end up, the length changes to 15 cm.

Let atmospheric pressure = pcmHg; crosssectional area of the tube = α cm².

When the tube is horizontal, pressure of the confined air, p₁ = p cmHg,

volume of this confined air, V₁ = 17 α cm_³.

When the tube is held vertical with the open end up, pressure of the confined air, p₂ = (p + 10) cmHg,

volume of this confined air, V₂ = 15α cm³

According to Boyle’s law, px 17α = (p + 10) x 15α or, p = 75

∴ Atmospheric pressure = 75 cmHg

Again, when the tube is held vertical with the open end downwards,

pressure of the confined air, p₃ = (p-10) cmHg,

volume of this confined air, V₃ = hα cm³

[where h = length of the air column]

According to Boyle’s law, \(p_1 V_1=p_3 V_3 \text { or, } p \times 17 \alpha=(p-10) \times h \alpha \)

or, h = \(\frac{17 p}{p-10}=\frac{17 \times 75}{65}=19.6 \mathrm{~cm}\).

Example 22. Volume of a room is 15 m x 12 m x 8 m. The room was at 22°C in the morning. What is the percentage of initial volume of air of the room that Is expelled when the room temperature reaches 30° C at noon? The pressure remains constant during die change of temperature.
Solution:

Given

Volume of a room is 15 m x 12 m x 8 m. The room was at 22°C in the morning.

From Charles’ law: \(\frac{V_1}{T_1}=\frac{V_2}{T_2}\) or, \(\frac{V_1}{V_2}=\frac{T_1}{T_2}\)

or, \(\frac{V_1}{V_2-V_1}=\frac{T_1}{T_2-T_1} or, \frac{V_2-V_1}{V_1} \times 100=\frac{T_2-T_1}{T_1} \times 100\)

Given, \(T_1=22^{\circ} \mathrm{C}=295 \mathrm{~K}, T_2=30^{\circ} \mathrm{C}=303 \mathrm{~K}, \quad V_2-V_1\) = volume of expelled air

∴ \(\frac{V_2-V_1}{V_1} \times 100=\frac{T_2-T_1}{T_1} \times 100=\frac{303-295}{295} \times 100=2.7\)

∴ 2.7% of air will be expelled from the room.

Example 23. Air is enclosed in a glass container at 67°C. Find the temperature to which the container is to be raised at constant pressure, so that 1/3 rd of the final volume of the air is expelled from the vessel? (Neglect the expansion of glass)
Solution:

Given

Air is enclosed in a glass container at 67°C.

Let initially the volume of air = V₁

In this case, temperature of air, T₁ = 273 + 67 = 340 K

Suppose the volume of the air becomes V₂ at a temperature T₂.

∴ Volume of expelled air = 1/3 V₂

∴ \(V_2=V_1+\frac{1}{3} V_2 \text { or, } V_2=\frac{3}{2} V_1\)

Since the pressure is constant, we can write \(\frac{V_1}{T_1}=\frac{V_2}{T_2}\)

∴ \(\frac{V_1}{340}=\frac{\frac{3}{2} V_1}{T_2}\)

or, \(T_2=\frac{3}{2} \times 340=510 \mathrm{~K}=(510-273)^{\circ} \mathrm{C}=237^{\circ} \mathrm{C} .\)

Thermodynamics – First And Second Law Of Thermodynamics The First Law Of Thermodynamics

The first law of thermodynamics is nothing but energy conservation applied to a thermodynamic system. From the equation W = JH, we know that an amount of work W is spent to produce an equivalent amount of heat H and from H amount of heat, an equivalent amount of work W is obtained.

Hence, heat and mechanical energy (work) are interconvertible. But in practice, when a system takes some amount of heat from its surroundings, it is used up in two ways:

1. A part of it increases the internal energy of the system and
2. The remaining part is converted into some external work done by the system, i.e., heat absorbed = rise in internal energy + external work done.

This is the first law of thermodynamics.

The first law of thermodynamics is, in essence, the energy conservation law which is applicable to every thermodynamic system together with its surroundings.

Let U_i = initial internal energy of a system,

Q = heat taken by the system from the surroundings,

U_f = final internal energy of the system,

W = external work done by the system.

Here, U_f– U_t = ΔU = change in internal energy. So the first law states that,

Q = \(\left(U_f-U_i\right)+W=\Delta \boldsymbol{U}+\boldsymbol{W}\)….(1)

In this equation, W and Q are expressed in the same unit.

So Joule’s equivalent is J = 1.

When the internal energy of a system does not change, i.e., Uj-= Up we get, Q= W.

If, on the other hand, W is expressed in joule and Q in calorie, then J ≠ 1. We can then write, W= JQ. (the symbol H for heat is replaced by Q here).

So, Joule’s law of mechanical equivalent of heat is a special case of the first law of thermodynamics (where ΔU =0)

If a small amount of heat dQ changes the internal energy of a system by dU and an external work dW is done, then

dQ = dU + dW……(2)

This is the differential form of the first law of thermodynamics, whereas equation (1) is known as the integral form.

In equations (1) and (2), the convention is:

1. Q or dQ is considered positive when heat is absorbed by a system;
2. Q or dQ is considered negative when heat is released by a system.

Significance of the first law of thermodynamics: From the first law of thermodynamics, it is known that mechanical energy can be converted into heat and vice versa. Certain amount of heat is required to do a certain amount of work. On the other hand, a certain amount of work has to be done to generate a certain amount of heat.

• The first law of thermodynamics indicates a new property of a body internal energy. In different thermodynamic processes, if we take only work and heat as the two manifestations of energy, the conservation of energy principle is violated.
• However, the body in every state has some internal energy. In every process, if we consider the change of this internal energy along with heat and work, energy remains conserved in all cases.
• This internal energy (U) is a state function. During the transformation from one state of a body to another, the change of internal energy does not depend on the intermediate path. In thermodynamics, there is no need to identify the source of internal energy.
• But it is seen from the kinetic theory of gases that, the kinetic energy and the potential energy due of to translational, rotational, and vibrational motions of the molecules of a body are the sources of its internal energy.

The perpetual motion of the first kind: it is impossible to get work without dissipation of energy. So, it is impossible to invent a machine that can work indefinitely without any supply of energy. If a machine can work indefinitely without any energy input, its motion is called the perpetual motion of the first kind.

But the first law of thermodynamics, which expresses the law of conservation of energy, states that this is impossible. So, perpetual motion of the first kind does not exist in nature.

Origin of internal energy: For complete conversion between W and Q, we have W = Q. Then from the first law, U_f – U_i = 0 or, U_f = U_i i.e., the internal energy of a system does not change.

• But complete conversion between W and Q is an ideal case, and has some natural restrictions. Thus, directly from the law of conservation of energy, we get the existence of a new thermodynamic property — the internal energy U of a system.
• For example, when a system takes some heat from outside but does no work (W = 0), we have Q = U_f– U_i, or heat absorbed = increase in internal energy.
• So, the effect of heat on a system, doing no work, is a change in its internal energy. These effects are rise in temperature, melting of a solid, vaporization of a liquid, etc. In each of these examples, the change in temperature of the system or the latent heat is directly related to the change in its internal energy.

When the motion of the molecules in a system is considered, we get a clear picture of the internal energy of a system. But thermodynamics does not discuss molecular motions that will be dealt with in kinetic theory.

Internal energy of a gas: Thermal condition of a fixed mass of gas is determined by three quantities temperature, pressure and volume. But internal energy of a gas does not depend on all of these three quantities. It depends only on its temperature in some cases.

• For example, there is no change in internal energy if the pressure or volume of a monatomic ideal gas of a particular mass changes at constant temperature. If we know the rise in temperature of this type of gas, we can determine the increase in internal energy.
• But the change in internal energy does not always depend on the change in temperature. If a gas undergoes a phase change, there is no change in temperature, but its- internal energy changes.

Limitations of the first law of thermodynamics: This law expresses the law of energy conservation. That is, it says that energy cannot be created or destroyed it can only be converted from one form to another. But it cannot predict anything about the direction of natural processes. It fails to explain why

1. Heat can flow only from higher to lower temperature and never from lower to higher temperature
2. Some amount of work can entirely be converted into heat, but complete conversion of heat into work never occurs in nature.

So, the maximum heat or work available from a certain amount of work or heat, respectively, cannot be determined from the first law of thermodynamics. These limitations are overcome by the formulation of another law, the second law of thermodynamics.

Thermodynamics – First And Second Law Of Thermodynamics The First Law Of Thermodynamics Numerical Examples

Example 1. An ideal monatomic gas goes through a cyclic process ABCDA, as shown. Find out the work done and heat supplied in this cyclic process.

Solution:

Work done in the cycle ABCD = area of ABCD

= AB-BC = (2p-p)(2V-V) = pV

Here, the initial state A = the final state A.

So, the change in internal energy, U_f – U_i = U_A – U_A =0

Then, U_f– U_i = Q-W or, Q = (U_f – U_i) + W= 0 + pV = pV

∴ Heat supplied in the cyclic process = pV.

Example 2. The volume of 1 g of water (1 cm³) becomes 1671 cm³ on being converted to steam at standard atmosphere pressure. Find out the work done and rise in internal energy. Given, the latent heat of vaporization of water = 540 cal · g^-1 the standard atmosphere pressure = 1.013 x 10⁵ N · m^-2.
Solution:

W = \(\int_{V_1}^{V_2} p d V=p \int_{V_1}^{V_2} d V=p\left(V_2-V_1\right)\)

Here, \(V_2-V_1=1671-1\)

= \(1670 \mathrm{~cm}^3\)

= \(1670 \times 10^{-6} \mathrm{~m}^3=1.67 \times 10^{-3} \mathrm{~m}^3\)

∴ W = \(\left(1.013 \times 10^5\right) \times\left(1.67 \times 10^{-3}\right) \)

= 169.17 J

∴ \(U_2-U_1=Q-W=540 \times 4.2-169.17\)

(\(540 \mathrm{cal}=540 \times 4.21\))

= 2099 J.

Viscosity And Surface Tension Rate Of Flow Of A Liquid And Continuity

Rate of flow Of liquid definition: For streamline flow of a perfectly incompressible liquid, the amount of liquid flowing through any cross section of a tube in a given time interval remains constant.

• The rate of flow of a liquid through a tube means the volume of liquid flowing through any cross-section of the tube per second.
• Suppose a liquid flows through a tube of cross-sectional area a with a uniform velocity ν. The volume of liquid flowing through any cross-section of the tube per second is equal to the volume of a cylinder of length v and cross-sectional area α.

∴ Volume of liquid flowing per second = the rate of flow of the liquid = velocity of flow x area of cross section of the tube = νa

Therefore, the mass of liquid flowing per second = velocity of flow x area of cross-section of the tube x density of the liquid = ναρ [ρ = density of the liquid]

Continuity of flow definition: For a streamlined flow of a fluid (liquid or gas) through a tube, the mass of the fluid flowing per second through any cross-section of the tube remains constant. This is known as the continuity of flow.

Definition Of Equation Of continuity: Let us consider two sections A and B of a tube having cross-sectional areas a₁ and a₂ respectively. The velocities of the fluid at sections A and B are ν₁ and ν₂, and its densities are ρ₁ and ρ₂ respectively.

The mass of fluid flowing through section A per second = \(v_1 \alpha_1 \rho_1\) and the mass of fluid flowing through section B per second = \(v_2 \alpha_2 \rho_2\)

For streamline flow, the fluid enters through section A and leaves through section B, and does not remain stored in the region between A and B, hence

⇒ \(v_1 \alpha_1 \rho_1\) = \(v_2 \alpha_2 \rho_2\) …..(1)

The product vαρ is the mass flow rate. If the fluid is incompressible (like a liquid), then its density is constant, and in that case ρ₁ = ρ₂.

∴ \(v_1 \alpha_1\) = \(v_2 \alpha_2\)…….(2)

or, να = constant …..(3)

Equations (2) and (3) are known as the equations of continuity of liquid flow.

∴ \(\nu \propto \frac{1}{\alpha},\) which means that the velocity of liquid flow through any cross-section of a tube is inversely proportional to its cross-sectional area.

The equations of continuity essentially express the law of conservation of mass.

Energy of Liquid in Streamline Flow: At any point inside a flowing liquid, there are three forms of energy

1. Kinetic energy,
2. Potential energy and
3. Energy due to pressure.

1. Kinetic energy: If mass m of a liquid flows with a velocity v, then the kinetic energy of that liquid = = \(\frac{1}{2} m v^2\).

Kinetic energy per unit mass = \(\frac{1}{2}v^2\)

Kinetic energy per unit volume = \(\frac{1}{2} \frac{m}{V} v^2\) [volume of the liquid]

= \(\frac{1}{2} \rho v^2\left[\rho=\frac{m}{V}=\text { density of the liquid }\right]\)

2. Potential energy: If mass m of a liquid is at a height h above the surface of the earth, then the potential energy of that liquid = mgh.

Potential energy per unit mass = gh

Potential energy per unit volume = \(\frac{m g h}{V}\) = ρgh.

3. Energy due to pressure: if a liquid is under the action of some applied pressure, then it acquires some energy and this energy is known as energy due to pressure. The liquid can perform work by expending this energy. Let some liquid of density ρ whose free surface is PQ be kept in a container.

A narrow side tube AB of cross sectional area α is attached near the bottom of the container. This tube is fitted with a piston P, which can move freely along the tube. If the pressure of the liquid at rest along the axis of the narrow tube is p, then the force acting on the piston = pα.

If the piston is slowly pushed inside the tube through a distance x, then work done = pαx. As a result, liquid of volume αx or mass αxρ enters the container. Since the piston is moved slowly, the liquid acquires negligible velocity and hence it will possess no kinetic energy.

So the work done pax remains stored as potential energy in mass αxρ of the liquid that has entered the container. This energy is called the energy due to pressure for the liquid.

The energy due to pressure per unit mass of the liquid = = \(\frac{p a x}{a x \rho}=\frac{p}{\rho} .\)

∴ The energy due to pressure per unit volume of the liquid = \(\frac{p a x}{a x}=p\).

Bernoulli’s Theorem: The Swiss mathematician Daniel Bernoulli established a law for the streamline flow of an ideal fluid (which is incompressible and non-viscous). This law is known as Bernoulli’s theorem. It is an important theorem in Hydrodynamics.

Statement Of the Bernoulli’s theorem: For a streamline flow of an ideal liquid, the sum of the potential energy, kinetic energy, and energy due to pressure per unit volume of the liquid always remains constant at every point on the I streamline.

If the kinetic energy per unit volume of the liquid = \(\frac{1}{2} \rho v^2\); potential energy = pgh and energy due to pressure = p, then

⇒ \(\frac{1}{2} \rho v^2+\rho g h+p\) = constant

or, \(\frac{1}{2} v^2+g h+\frac{p}{\rho}\) = constant …..(1)

This is the mathematical form of Bernoulli’s theorem. Dividing equation (1) by g, we get,

⇒ \(\frac{v^2}{2 g}+h+\frac{p}{\rho g}\) = constant……..(2)

This also is a form of Bernoulli’s theorem. Here, \(\frac{v^2}{2 g}\) is called the velocity head, h is the elevation head, and \(\frac{p}{\rho g}\) is the pressure head. Each of these heads has the dimension of length.

So, velocity head + elevation head + pressure head
= constant …..(3)

According to relation (3), Bernoulli’s theorem can also be stated as follows.

For a streamline flow of an ideal liquid, the sum of the velocity head, elevation head and pressure head always remains constant at any point in the liquid.

Bernoulli’s theorem is based on the law of conservation of energy for the streamline motion of an ideal fluid. The theorem states that energy remains conserved along any streamline.

When the flow of liquid is horizontal, the height of each point in the liquid is assumed to be the same, i.e., h = constant. We can rewrite equation (2) as,

⇒ \(\frac{v^2}{2 g}+\frac{p}{\rho g}=\) constant

or, \(p+\frac{1}{\mathrm{a}} \rho v^2\) = constant

Hence, in the horizontal flow of a liquid, the sum of pressure and kinetic energy per unit volume of the liquid at any point is constant. This implies that where the velocity of flow is high, the pressure is low and vice-versa.

Applications of Bernoulli’s theorem

1. Velocity of efflux of a liquid and Torricelli’s theorem: If a small hole is present on the wall of a deep container containing liquid, then the velocity with which the liquid comes out through that small hole is called the velocity of efflux of the liquid.

In Fig, a liquid kept in a large container is emerging with velocity ν through the small hole on the wall of the container. The height of the free surface of the liquid above the hole is h and the depth of the liquid below the hole is h₁.

The total depth of the liquid H = h+ h₁. Let us consider a point B just outside the hole and another point A on the surface of the liquid. Atmospheric pressure p acts on A and B.

If the container is large and the hole is very small, then the free surface of the liquid will come down so slowly that the velocity of the free surface of the liquid would seem to be almost zero.

If we imagine a tube of flow starting from the free surface of the liquid and ending at the point B and apply Bernoulli’s theorem in that tube of flow, then

⇒ \(0+H+\frac{p}{\rho g}=\frac{v^2}{2 g}+h_1+\frac{p}{\rho g}\)

or,\(\frac{v^2}{2 g}=H-h_1=h\)

or, \(v^2=2 g h\)

or, \(v=\sqrt{2 g h}\)

• This is the velocity of efflux of a liquid through a small hole and it is known as Torricelli’s formula. According to this formula, the velocity of efflux of a liquid is the same as that of a body falling freely under gravity through a height h. So, Torricelli’s theorem can be stated as follows:
• The velocity of efflux of a1 liquid through any small hole or orifice is equal to that acquired by a body falling freely from rest under gravity from the free surface of the liquid to the level of the small hole.
• It should be mentioned that this ideal velocity cannot be attained by any liquid in reality because no liquid is non- viscous. It should be remembered that in Bernoulli’s theorem the effect of viscosity of the liquid has been neglected.

Horizontal range: Let, the first drop of liquid emerged from the orifice touches the ground at a distance x after time t. That means, the vertical displacement of the liquid drop is h₁.

Now, considering the motion of the liquid along the vertical direction,

initial velocity = 0, acceleration = g

From the equation h = ut+1/2gt² we get, h = 0 + 1/2gt²

∴ t = \(\sqrt{\frac{2 h_1}{g}}\)

Again, considering the motion of the liquid aong the horizontal direction, the initial velocity, ν = √2gh, acceleration = 0, time = t.

∴ Horizontal range, \(x=v t=\sqrt{2 g h} \times \sqrt{\frac{2 h_1}{g}}=2 \sqrt{h h_1}\)

2. Venturimeter: A venturimeter is used to measure the rate of flow of liquid through a tube. Its working principle is based on Bernoulli’s theorem.

• Shows the action of a venturimeter. The two ends of this tube are equally wide and the middle portion is narrow. Liquid flows through this tube in streamlines. The tube is kept horizontal.
• When a liquid flows through a venturimeter, the velocity of the liquid increases at the narrow part of the tube with consequent decrease in pressure. This decrease in pressure is measured with the help of two vertical tubes attached at the wide and the narrow parts of the venturimeter.

Let the velocity of the liquid at the wider part of the tube be ν₁ and the pressure be p₁ At the narrower part of the tube, the velocity of the liquid is ν₁ and the pressure is p₂.

According to Bernoulli’s theorem, \(\frac{v_1^2}{2 g}+\frac{p_1}{\rho g}=\frac{v_2^2}{2 g}+\frac{p_2}{\rho g}\)

the elevation head, h₁=h₂ since the tube is horizontal.

∴ \(\frac{p_1-p_2}{\rho g}=\frac{1}{2 g}\left(v_2^2-v_1^2\right) \text { or, } p_1-p_2=\frac{\rho}{2}\left(v_2^2-v_1^2\right)\)

or, \(h \rho g=\frac{\rho}{2}\left(v_2^2-v_1^2\right)\)

[h = difference in liquid levels in the vertical tubes attached to the venturimeter]

∴ h = \(\frac{1}{2 g}\left(v_2^2-v_1^2\right)\)

If the cross-sectional areas of the wide and the narrow parts of the venturimeter are α₁ and α₂ respectively, then according to the equation of continuity, we get,

⇒ \(\alpha_1 v_1=\alpha_2 v_2 \text { or, } \frac{v_1}{v_2}=\frac{\alpha_2}{\alpha_1}\)

∴ h = \(\frac{v_2^2}{2 g}\left(1-\frac{v_1^2}{v_2^2}\right)=\frac{v_2^2}{2 g}\left(1-\frac{\alpha_2^2}{\alpha_1^2}\right)\)

or, \(v_2^2=2 g h \cdot \frac{\alpha_1^2}{\alpha_1^2-\alpha_2^2}\)

or, \(v_2=\frac{\alpha_1}{\sqrt{\alpha_1^2-\alpha_2^2}} \cdot \sqrt{2 g h}\) …..(1)

Therefore, the volume of liquid flowing out per second,

V = \(\alpha_2 v_2=\frac{\alpha_1 \alpha_2 \sqrt{2 g h}}{\sqrt{\alpha_1^2-\alpha_2^2}}=\alpha_1 \alpha_2 \sqrt{\frac{2 g h}{\alpha_1^2-\alpha_2^2}}\)…….(2)

So, when α₁ and α₂ are known, by measuring h we can determine the rate of flow of the liquid through the tube with the help of equation (2).

3. Pitot tube: A pitot tube is also used to measure the rate of flow of liquids. Its working principle is similar to that of a venturimeter. Its action also depends on Bernoulli’s theorem.

• In this instrument, two tubes AB and CED, open at both ends are introduced vertically and side by side inside the liquid. The open end B of the tube AB remains parallel to the flow of the liquid. The DE part of the tube CED is so bent that the opening D faces the flowing liquid normally.
• The height of the liquid column in the tube AB expresses the pressure of the liquid at the point B. Since the liquid flow is obstructed at the portion DE of the tube CED, the velocity of flow at point D is zero.

The difference in the liquid levels in the two tubes = h.

Let the velocity of liquid flow be ν.

The points B and D lie on the same horizontal plane; therefore, according to Bernoulli’s theorem,

⇒ \(\frac{1}{2 g} v^2+\frac{p_B}{\rho g}=0+\frac{p_D}{\rho g}\) [ρ= density of the liquid]

or, \(\frac{1}{2 g} v^2=\frac{p_D-p_B}{\rho g}\)

or, \(\frac{1}{2} v^2=\frac{h \rho g}{\rho} v^2=2 g h or, v=\sqrt{2 g h}\)

If the cross-section of the pipe where the two tubes are placed is a, then the volume of liquid flowing per second through that section, V = \(\alpha v=\alpha \sqrt{2 g h} .\)

When an aeroplane is in motion, the velocity of air currents can be determined with the help of a pitot tube.

4. Sprayer or atomizer: A sprayer or atomizer is used for spraying water, insecticides, etc. Its action also depends on Bernoulli’s theorem.

• The liquid to be sprayed is kept in a container A and its mouth is closed with the help of a cork or cap. A narrow tube B passes through the cap of the container. C is another tube through which air is blown. The tube C has a narrow tip.
• When air comes out from this narrow tip O with a high velocity, pressure at O decreases. Since O lies just above the open end of the tube B, the liquid rises through the tube B due to this low pressure, and as it meets the high-velocity air coming out of the tube C, it sprays out in the form of fine droplets.

Explanation of Some Phenomena with Bernoulli’s Theorem

1. It is not safe to stand near a fast-moving train: Due to the very high speed of the train, the air near the train also flows at a very high speed. As a result, pressure in that region decreases compared to the air pressure of the surrounding region. This excess surrounding pressure behind a person pushes him towards the train and may cause a serious accident.

2. The tin roof of a house is sometimes blown off during a storm: Since the velocity of the wind above the roof is very high, pressure becomes very low. The air inside the room is still and so the higher pressure from inside pushes the roof upwards and hence the roof may be lifted and blown off with the wind.

3. Two boats or ships moving side by side have a tendency to come closer: The speed of water in the narrow gap between boats or ships is greater than the speed of water on the other sides of the vessels. So, the pressure in that narrow region decreases. As a result, due to higher water pressure on the other sides of the boats or ships, they experience a lateral force and, hence, come closer.

4. Flying in air of typical-shaped objects: Let us take an object moving through air towards right.

• Its lower surface is flat, but the upper surface is oval-shaped. The relative motions of the streamlines of air moving above and below it are shown by arrows.
• Clearly, the upper streamline traverses a greater distance in any fixed interval of time; so its velocity is higher. Then, according to Bernoulli’s theorem, the air pressure above the object is less than that below it.
• As a result, a net upward pressure acts on the object. This helps the object to fly through air, provided its weight is sufficiently low. This is one of the principles utilised to fly an aeroplane.

5. Magnus effect: When a spinning ball is thrown horizontally with a large velocity, it deviates from its usual parabolic path of spin free motion. This deviation can be explained on the basis of Bernoulli’s principle.

• When a ball moves forward, the air ahead the ball, moving with velocity v (say), rushes to fill up the vacant space behind the ball left evacuated by the motion of the ball.
• As the ball spins, the layer of air surrounding the ball also moves with the ball at a velocity u (say). From the fig, it can be stated that the resultant velocity of air above the ball becomes (v+ u) while that below that ball is (v- u).
• This difference in the velocities of air results in the pressure difference between the lower and upper faces of the ball. This pressure difference exerts a net upward force on the ball due to which it moves along a curved path as shown in Fig.
• If the spin of the ball is opposite to that shown in the Fig. a net downward force will act on it, deviating it from its original path. This effect is known as Magnus effect. If the surface of the ball is rough, more air is dragged and the path of the ball becomes more curved.

6. Blood flow and heart attack: An artery may get constricted due to the accumulation of plaque on its inner walls. In order to drive blood through this constriction the speed of the flow of blood is increased.

• This increased velocity lowers the blood pressure in the constricted region and the artery may collapse due to the external pressure.
• As a result, the heart exerts more pressure to open this artery and forces the blood through. As the blood rushes through the opening, the internal pressure once again drops due to some reasons leading to a repeat collapse which results in heart attack.

Rate Of Flow Of A Liquid Numerical Examples

Example 1. The pressure at an orifice situated at the lower side of a vessel filled with a liquid is greater than the atmospheric pressure by 9.8 x 10³ N · m^-2. What is the velocity of efflux if the density of the liquid is 2500 kg · m^-3? [g = 9.8 m · s^-2]
Solution:

Given

The pressure at an orifice situated at the lower side of a vessel filled with a liquid is greater than the atmospheric pressure by 9.8 x 10³ N · m^-2.

Velocity of efflux of the liquid, v = √2gh

According to the problem, hρg = 9.8 x 10³

or, \(g h=\frac{9.8 \times 10^3}{2500}\)

∴ v = \(\sqrt{\frac{2 \times 9.8 \times 10^3}{2500}}=2.8 \mathrm{~m} \cdot \mathrm{s}^{-1} .\)

Example 2. Water is flowing through a horizontal tube of unequal cross-section. At a point where the velocity of water is 0.4 m · s^-1, the pressure is 0.1 m Hg. What is the pressure at a point where the velocity of water is 0.8 m · s^-1? The density of mercury = 13.6 x 10³ kg · m^-3.
Solution:

Given

Water is flowing through a horizontal tube of unequal cross-section. At a point where the velocity of water is 0.4 m · s^-1, the pressure is 0.1 m Hg.

According to Bernoulli’s theorem,

⇒ \(\frac{1}{2} v_1^2+\frac{p_1}{\rho}=\frac{1}{2} v_2^2+\frac{p_2}{\rho}\) (since the tube is horizontal)

or, \(p_2=p_1+\frac{1}{2} \rho\left(v_1^2-v_2^2\right)= 0.1 \times\left(13.6 \times 10^3\right) \times 9.8\) + \(\frac{1}{2} \times 10^3 \times\left\{(0.4)^2-(0.8)^2\right\}\)

= \(13088 \mathrm{~Pa}=\frac{13088}{\left(13.6 \times 10^3\right) \times 9.8} \mathrm{~m} \mathrm{Hg}\)

= 0.0982 m Hg

Example 3. A pitot tube is connected to a main pipeline of diameter 16 cm. The difference in height of the water columns in the two arms of the tube is 10 cm. Determine the rate of flow of water through the main pipe.
Solution:

Given

A pitot tube is connected to a main pipeline of diameter 16 cm. The difference in height of the water columns in the two arms of the tube is 10 cm.

Volume of water flowing through the pipe per second,

V = αv [v = velocity of water]

V = \(\alpha \sqrt{2 g h} \text { (Here, } \alpha=\pi(8)^2=64 \pi \mathrm{cm}^2, h=10 \mathrm{~cm} \text { ) }\)

∴ V = \(64 \pi \sqrt{2 \times 980 \times 10}=64 \pi \times 140\)

= 2.8 x 10⁴ cm³ = 0.028 m³.

Example 4. A venturimeter has been connected between two points of a pipe. The radii of the pipe at these two points are 5 cm and 3 cm respectively. The difference in pressure between these points is equal to that of a water column of height 5 cm. Determine the rate of flow of water through the pipe.
Solution:

Given

A venturimeter has been connected between two points of a pipe. The radii of the pipe at these two points are 5 cm and 3 cm respectively. The difference in pressure between these points is equal to that of a water column of height 5 cm.

The rate of flow of water,

V = \(\alpha_1 \alpha_2 \sqrt{\frac{2 g h}{\alpha_1^2-\alpha_2^2}}\)

[Here, \(\alpha_1=\pi(5)^2=25 \pi \mathrm{cm}^2, \quad \alpha_2=\pi(3)^2=9 \pi \mathrm{cm}^2\), h=5 \(\mathrm{~cm}, g=980 \mathrm{~cm} \cdot \mathrm{s}^{-2}\)]

= \(25 \pi \cdot 9 \pi \sqrt{\frac{2 \times 980 \times 5}{(25 \pi)^2-(9 \pi)^2}} \approx 3000.6 \mathrm{~cm}^3 \cdot \mathrm{s}^{-1} .\)

Example 5. Water from a tap falls vertically with a velocity of 3 m · s^-1. The area of the cross-section of the mouth of the tap is 2.5 cm². If the flow of water is uniform and steady throughout, then determine the cross-sectional area of the tube of flow of water at a depth of 0.8 m from the mouth of the tap. [g = 10 m · s^-2]
Solution:

Given

Water from a tap falls vertically with a velocity of 3 m · s^-1. The area of the cross-section of the mouth of the tap is 2.5 cm². If the flow of water is uniform and steady throughout

The area of cross-section of the mouth of the tap, A₁ = 2.5 cm², and velocity of water flow there, v₁ = 3 m · s^-1.

Let the area of cross-section of the tube of flow of water at a depth of 0.8 m below the tap be A₂ and the velocity of water flow there be v₂

∴ \(v_2^2=v_1^2+2 g hp\)

= (3)² + 2 x 10 x 0.8

= 9 + 16 = 25

or, v₂ = 5 m · s^-1

We know that, A₁ V₁ = A₂ V₂

or, \(A_2=\frac{A_1 v_1}{v_2}=\frac{2.5 \times 3}{5}=1.5 \mathrm{~cm}^2 .\)

Example 6. Water flows in streamline motion through a vertical tube of non-uniform cross-section. The radius of the tube at point P and at point Q are 1 cm and  0. 5 cm respectively. Point Q is at a distance of 40 cm from point P. The pressure difference between these points is 39.3 cm of water. Determine the rate of flow of water through the tube.

Solution:

Given

Water flows in streamline motion through a vertical tube of non-uniform cross-section. The radius of the tube at point P and at point Q are 1 cm and  0. 5 cm respectively. Point Q is at a distance of 40 cm from point P. The pressure difference between these points is 39.3 cm of water.

Accoring to the equation of continuity,

⇒ \(A_1 v_1=A_2 v_2\)

∴ \(v_2^2=\frac{A_1^2}{A_2^2} v_1^2\)

According to Bernoulli’s theorem,

⇒ \(\frac{v_1^2}{2 g}+h_1+\frac{p_1}{\rho g}=\frac{v_2^2}{2 g}+h_2+\frac{p_2}{\rho g}\)

or, \(\frac{v_2^2-v_1^2}{2 g}=\left(h_1-h_2\right)+\left(\frac{p_1-p_2}{\rho g}\right)\)

or, \(\frac{v_2^2-v_1^2}{2 g}=40+39.3 \frac{\rho g}{\rho g}=79.3 \mathrm{~cm}\)

∴ \(\frac{v_1^2}{2 g}\left[\frac{A_1^2}{A_2^2}-1\right]=79.3\)

or, \(\frac{v_1^2}{2 g}\left[\frac{(\pi)^2}{(0.25 \pi)^2}-1\right]=79.3\)

or, \(v_1=101.79 \mathrm{~cm} / \mathrm{s}\)

The rate of flow of water = \(A_1 v_1=101.79 \times \pi\)

=319.78 cm³

Example 7. A liquid of density 1000 kg/m³ is flowing in streamline motion through a tube of the non-uniform cross-section. The tube is inclined with the ground, The area of cross sections at points P and Q of the tube are 5 x 10^-3 m² and 10 x 10^-3 m² respectively. The height of the points P and Q from the ground are 3 m and 6 m respectively. The velocity of the liquid at point P is 1 m/s. Calculate the work done per unit volume due to

1. the pressure and
2. gravitational force for the flow of liquid from point P to point Q.

Solution: From equation  of continuity, \(A_1 v_1=A_2 v_2\)

or, \(v_2=\left(\frac{A_1}{A_2}\right) v_1=\left(\frac{5 \times 10^{-3}}{10 \times 10^{-3}}\right) \cdot(1)=\frac{1}{2} \mathrm{~m} / \mathrm{s}\)

Applying Bernoulli’s theorem we get,

⇒ \(p_1+\frac{1}{2} \rho v_1^2+\rho g h_1=\rho_2+\frac{1}{2} \rho v_2^2+\rho g h_2\)

or, \(\rho_1-\rho_2=\rho g\left(h_2-h_1\right)+\frac{1}{2} \rho\left(v_2^2-v_1^2\right)\)……….(1)

1. Work done per unit volume of the liquid due to the pressure of the streamline flow from P to Q is,

⇒ \(W_p =p_1-p_2\)

⇒ \(W_p =\rho g\left(h_2-h_1\right)+\frac{1}{2} \rho\left(\nu_2^2-v_1^2\right)[\text { from equation (1)] }\)

= \(\left[(1000)(9.8)(6-3)+\frac{1}{2}(1000)\left(\frac{1}{4}-1\right)\right]\)

= \(\left[3 \times 9.8-\frac{3}{8}\right] \times 10^3=29025 \mathrm{~J} / \mathrm{m}^3\)

2. Work done due to gravitational force for the streamline motion from P to Q is,

W_g = ρg(h₁ – h₂) = 1000 x 9.8 x (3 – 6)

= -29400 J/m³

Example 8. A big container is kept on a horizontal surface of uniform area of cross-section, A. Two non-viscous liquids of densities d and 2d which are not mixed with each other and not compressed, are kept in the container. The height of both liquid column is H/2 and the atmospheric pressure on the open surface of the liquid is p₀. A small orifice is created at a height h from the bottom of the container,

1. Calculate the initial velocity of efflux of the liquid through the orifice,
2. Calculate the horizontal distance x at which the first liquid drop emerged from the orifice will reach,
3. What will be the value of h if the value of the horizontal distance x to be maximum x_m? Also, calculate the value of x_m neglecting the air resistance.

Solution:

1. Let, the initial velocity of efflux of the liquid = v.

According to Bernoulli’s theorem,

⇒ \(p_0+d g\left(\frac{H}{2}\right)+2 d g\left(\frac{H}{2}-h\right)=p_0+\frac{1}{2}(2 d) v^2\)

or, \(v^2=\left(\frac{H}{2}+\frac{2 H}{2}-2 h\right) g or, v=\sqrt{(3 H-4 h)_2^g}\)

2. The time required to reach the ground of the first liquid drop emerged from the orifice is,

t = \(\sqrt{\frac{2 h}{g}}\)

∴ The horizontal distance traversed by the liquid is,

x = vt = \(\sqrt{(3 H-4 h)_2^g} \sqrt{\frac{2 h}{g}}\)

= \(\sqrt{h(3 H-4 h)}\)

3. The condition for x to be maximum (x_m):

x = \(\sqrt{h(3 H-4 h)}=\sqrt{-\left(4 h^2-3 h H\right)}\)

= \(\sqrt{\left\{(2 h)^2-2 \cdot 2 h \cdot \frac{3}{4} H+\left(\frac{3}{4} H\right)^2-\left(\frac{3}{4} H\right)^2\right\}}\)

= \(\sqrt{\frac{9}{8} H^2-\left(2 h-\frac{3}{4} H\right)^2}\)

∴ x = \(x_m \text { when } 2 h-\frac{3}{4} H=0\)

Hence, h= \(\frac{3}{8}H\)

∴ \(x_m=\sqrt{\frac{9}{8} H^2}=\frac{3}{2 \sqrt{2}} H\)

Capillary Action

The word ‘capillary’ has originated from Camillus, a Latin word that means ‘hair’. So a tube with a uniform fine small bore like a hair is known as a capillary tube.

When a capillary tube is inserted vertically upright into the water, the water rises within the tube and its level inside the tube stands higher than the water level outside.

• Moreover, the surface of water inside the tube takes a concave shape instead of a horizontal one. Liquids like water, alcohol, copper sulphate solution, etc. can wet glass. If a capillary tube made of glass is dipped into these liquids, then the same phenomenon occurs.
• On the other hand, if a capillary glass tube is dipped into mercury (or into some other liquid that does not wet glass), then the level of mercury in the tube falls slightly, i.e., the mercury stands at a lower level inside the capillary tube than that outside it. Moreover, the upper meniscus of mercury takes a convex shape.
• The narrower the bore of the capillary tube, the greater the ascent or descent of the liquid level.
• This type of rise or fall of a liquid in a capillary tube is called capillary action. The curvature of a liquid surface in contact with a solid surface is also due to this action. The surface tension of a liquid is the origin of these phenomena.

Examples of Copillaritty Action: Due to capillary action, a blotting paper absorbs ink, a sponge absorbs water, oil rises through a wick and a towel absorbs water droplets from our body. These objects contain small pores and when water or any liquid comes in contact with them, the liquid enters these pores.

• Underground water rises through the small pores of soil and travels upwards so that the soil remains moist. The stems of a tree sends water from the moist soil to the leaves through capillary action.
• The pores of sandy soil are quite large and hence water cannot rise up to a great height through it. Therefore the upper layer of sandy soil remains dry.

Angle of contort Definition: When a liquid is in contact with a solid, the angle between the solid surface and the tangent to the free surface of the liquid at the point of contact, measured from inside the liquid is called the angle of contact for that specific pair of solid and liquid.

In Fig, the upper surface of the liquid touches the solid at the point A. The tangent drawn on the liquid surface is AB and this tangent makes the angle ∠BAC with the solid inside the liquid.

This angle ∠BAC is the angle of contact (θ) between the solid and the liquid. The angle of contact for various liquid-solid pairs ranges from 0° to 180°.

The angle of contact does not depend upon the inclination of the solid with the liquid surface. This angle between a glass plate and water remains the same irrespective of whether the plate is immersed straight or in an oblique manner.

• The rise or fall of a liquid inside a capillary tube depends on the angle of contact (θ). If the value of θ is less than 90°, then the liquid rises through a capillary tube (for example, water rises in a capillary glass tube) and the upper meniscus is concave.
• On the other hand, if the value of θ is more than 90°, then a liquid is depressed inside a capillary tube (for example, mercury level is lower inside a capillary tube) and the upper meniscus is convex.

Factors on which the angle of contact depends:

1. Nature of solid and liquid: if the adhesive force, between a liquid and the material of its solid container, is stronger than the cohesive force among the liquid molecules, then the liquid wets the solid.

• For example, water wets glass because the water-glass adhesive force is stronger than the cohesive force of water molecules. On the other hand, the mercury-glass adhesive force is weaker than the cohesive force among mercury molecules so mercury does not wet its glass container.
• If a drop of pure water is poured on a clean piece of glass, the water spreads quickly on the glass surface. In this case, the angle of contact is very small— almost zero, i.e., θ ≈ 0.
• On the other hand, if a drop of water falls on a lotus petal, the drop does not spread. Instead, it rests on the petal like a dewdrop. Here, the angle of contact, θ > 90°. So, if θ < 90°, the liquid wets the solid in contact with it, but, if θ > 90°, then it does not wet the solid.
• For example, a drop of mercury on a piece of glass does not wet it as θ = 140°. Again, the angle of contact (θ) of water and silver is 90°, so, when water is kept in a vertical silver container, the surface of water remains horizontal.

2. Medium in contact with the exposed liquid surface: If the medium in contact with the surface of mercury is air, then the value of the angle of contact between mercury and glass is not the same as when air is replaced by water as the medium in contact with mercury.

3. Impurities present in the liquid: For instance, the angle of contact between clear glass and very pure water is nearly zero, but that between ordinary water and glass is nearly 8°.

• Rise of liquid in a capillary tube: Suppose, a capillary tube of glass having a uniform bore of radius r is dipped vertically in water or any liquid which wets it. The liquid immediately rises up into it up to the height ‘h’ (say) and the shape of the liquid meniscus is spherical and it concaves upwards.
• The liquid meniscus in the tube is along a circle of circumference 2πr which is in contact with the glass tube. Let T be the surface tension of the liquid and θ be the angle of contact for the liquid and the glass of the tube.
• The surface tension T of the liquid acts inwards along the tangent to the liquid meniscus at every point of its contact with the inner surface of the tube, making an angle θ with the wall of the tube.
• Thus an inward pull is exerted on the glass in this direction at all these points. Since in accordance with Newton’s third law of motion, action, and reaction are equal and opposite, the wall. of the tube also exerts an equal (reaction) force T on the liquid. This reaction force T may be resolved into two perpendicular components

1. T cos θ acting vertically upwards and
2. T sin θ acting horizontally outwards.

Taking the whole meniscus into consideration, for each horizontal component T sinθ, there is an equal and opposite component so that the two neutralise each other. The vertical components being in the same direction get added up. It is this force which supports the weight of the liquid column so raised.

Hence, total upward force on the liquid in the tube = 2πr- T cosθ

It is this force which supports the weight of the column h of the liquid in the tube plus the weight of a volume V of the liquid in the meniscus itself, h is the length of the liquid column from the horizontal surface of the liquid in the container to the bottom of the meniscus at E.

So, the weight of the total volume of the liquid  = \(\left(\pi r^2 h+V\right) \rho g\)

∴ \(2 \pi r \cdot T \cos \theta=\left(\pi r^2 h+V\right) \rho g\)

or, \(T=\left(\frac{\pi r^2 h+V}{2 \pi r \cos \theta}\right) \rho g\) ……..(1)

If the volume of the liquid in the meniscus is negligible in comparison with that in the column h i.e., if the tube is of a very fine bore we have,

T = \(\frac{\pi r^2 h \rho g}{2 \pi r \cos \theta}=\frac{r h \rho g}{2 \cos \theta}\)

or, h = \(\frac{2 T \cos \theta}{r \rho g}\)

However if V is not negligible and θ = 0, the meniscus of the liquid in the tube is hemispherical in shape of radius equal to r.

So the volume of the liquid in the meniscus is equal to the difference between the volumes of a cylinder of radius r and length r and a hemisphere of radius r.

∴ V = volume of the cylinder ABCD – volume of the hemisphere AEB

= \(\pi r^2 \cdot r-\frac{1}{2} \cdot \frac{4}{3} \pi r^3\)

= \(\frac{1}{3} \pi r^3\)

So from equation (1) we have,

T = \(\left(\frac{\pi r^2 h+\frac{1}{3} \pi r^3}{2 \pi r \cos \theta}\right) \rho g=\frac{\pi r^2\left(h+\frac{1}{3} r\right) \rho g}{2 \pi r \cos \theta}\)

or, \(T=\frac{r\left(h+\frac{1}{3} r\right) \rho g}{2}\)….(3)

[taking cosθ = 1 ]

If the temperature of the liquid remains unchanged, T, ρ, and θ are constants. In that case it can be said from equation (2),

hr = constant ……….. (4)

This relation is known as Jurin’s law. The graph between the radius (r) and the rise in height (h) of the liquid is a rectangular hyperbola. The same relation holds good for depression of a liquid in a capillary tube but with a negative sign.

Jurin’s law Definition: The ascent or descent of a liquid inside a capillary tube is inversely proportional to the radius of the tube.

• From this law, it is concluded that smaller the radius (r) of the tube, the higher is the rise (h) of the liquid level inside the tube. But, if the radius of the tube is large, i.e., if the capillarity of the tube is not maintained, then this law becomes erroneous.
• If the angle of contact is less than 90°, then cos θ is positive, and hence the value of h as indicated by equation (1) is also positive; in this case, the liquid rises upwards in the capillary tube.
• On the other hand, if the angle of contact is more than 90°, then cos θ becomes negative and hence h will also be negative. In this case, the liquid level falls downwards in the capillary tube.

In the case of clean glass and pure water, the angle of contact, θ ≈ 0°; so cos θ ≈ 1 and hence from equation (2),

⇒ \(h=\frac{2 T}{r \rho g} \quad \text { or, } T=\frac{1}{2} r h \rho g\)

Measuring the rise of water in a capillary tube, its surface tension can be determined from this equation.

Rise of a liquid in a capillary tube of insufficient height: The expression for capillary rise h of a liquid in a tube is

h = \(\frac{2 T \cos \theta}{r \rho g}\)

where, r = radius of the tube

ρ, T = density and surface tension of the liquid, respectively

θ = angle of contact;

g = acceleration due to gravity.

When the height of the capillary tube is more than h, the liquid rises to the full height h to attain equilibrium. If R be the radius of the concave upper surface, then

r = R cosθ

So, h \(=\frac{2 T \cos \theta}{(R \cos \theta) \rho g}=\frac{2 T}{R \rho g}$\)

or, hR = \(\frac{2 T}{\rho g}\) = constant

Now, if a capillary tube of the same radius has a shorter height h’ above the external liquid surface, i.e., if h’ < h, then the liquid can rise only upto that height hf. In this case, the liquid adjusts the curvature of its upper concave surface in such a way that the radius of curvature R’ satisfies the relation,

h’R’ = hR, because hR is a constant.

Naturally, \(R^{\prime}=R \frac{h}{h^{\prime}}>R\), as h’ < h. So the liquid in the capillary attains equilibrium by increasing the radius of curvature of its upper concave surface. This means that the curvature of the surface becomes less. It is to be noted that, the liquid will not spill over the top due to the insufficient height of the capillary tube.

Viscosity And Surface Tension Capillary Action Numerical Examples

Example 1. A liquid of density 830 kg · m^-3 rises through 0.0893 m in a capillary tube of diameter 1.68 x 10^-4 m. Determine the surface tension of the liquid. Take the angle of contact as 0°.
Solution:

Given

A liquid of density 830 kg · m^-3 rises through 0.0893 m in a capillary tube of diameter 1.68 x 10^-4 m.

We know that, \(T=\frac{r h \rho g}{2 \cos \theta}\)

Here, \(r=\frac{1.68 \times 10^{-4}}{2}=8.4 \times 10^{-5} \mathrm{~m} \text {, }\)

h = 0.0893 m, ρ = 830 kg · m^-3 and θ = 0°.

∴ T = \(\frac{8.4 \times 10^{-5} \times 0.0893 \times 830 \times 9.8}{2 \cos 0^{\circ}}=0.0305 \mathrm{~N} \cdot \mathrm{m}^{-1}\)

Example 2. A capillary tube with a bore diameter of 0.5 mm is dipped upright in a liquid having surface tension 0.03 N · m^-1. The specific gravity of the liquid is 0.8 and the liquid wets the surface of the tube completely. Up to what height will the liquid rise in the tube?
Solution:

Given

A capillary tube with a bore diameter of 0.5 mm is dipped upright in a liquid having surface tension 0.03 N · m^-1. The specific gravity of the liquid is 0.8 and the liquid wets the surface of the tube completely.

If a liquid wets the wall of the tube completely, then the angle of contact becomes approximately zero.

∴ T = \(\frac{r h \rho g}{2}\) [As θ is very small, cosθ ≅1 ]

h = \(\frac{2 T}{r \rho g}\left[\text { Here, } T=0.03 \mathrm{~N} \cdot \mathrm{m}^{-1}, r=0.25 \times 10^{-3} \mathrm{~m}\right. \text {, }\left.\rho=0.8 \times 10^3 \mathrm{~kg} \cdot \mathrm{m}^{-3}\right]\)

= \(\frac{2 \times 0.03}{0.25 \times 10^{-3} \times 0.8 \times 10^3 \times 9.8}=0.0306 \mathrm{~m} \text {. }\)

Example 3. The diameter of a barometer tube is 3 mm. What will be the error in the barometer reading due to surface tension? For mercury in the glass tube, the surface tension = 0.647 N · m^-1; the angle of contact = 128°, and the density of mercury = 13500 kg · m^-3.
Solution:

Given

The surface tension = 0.647 N · m^-1; the angle of contact = 128°, and the density of mercury = 13500 kg · m^-3

The diameter of a barometer tube is 3 mm.

We know that h = \(\frac{2 T \cos \theta}{r \rho g} .\)

Here, T = 0.647 N · m^-1, θ = 128°,

r = \(\frac{3 \times 10^{-3}}{2}=1.5 \times 10^{-3} \mathrm{~m}\)

ρ = 13500 kg · m^-3, g = 9.8 m · s^-2

cosθ = cos 128° = cos(90° + 38°)

= -sin38° = -0.6157

∴ h = \(-\frac{2 \times 0.647 \times 0.6157}{1.5 \times 10^{-3} \times 13500 \times 9.8}\)

=-0.004 m

So, to obtain the actual atmospheric pressure, 0.004 m has to be added to the barometric reading.

Example 4. The diameters of the two arms of a U-tube are 1 mm and 2 mm. The tube is filled partly with water and is kept vertical. If the surface tension of water is 70 dyn · cm^-1, then find the difference in the level of water in the two arms of the U-tube. [Angle of contact =8°]
Solution:

Given

The diameters of the two arms of a U-tube are 1 mm and 2 mm. The tube is filled partly with water and is kept vertical. If the surface tension of water is 70 dyn · cm^-1,

Suppose due to capillary action, the water level rises in one arm of the U-tube to h₁ and in the other arm to h₂.

∴ \(h_1=\frac{2 T \cos \theta}{\rho g r_1}=\frac{2 \times 70 \times \cos 8^{\circ}}{1 \times 980 \times 0.05}\)

and \(h_2=\frac{2 T \cos \theta}{\rho g r_2}=\frac{2 \times 70 \times \cos 8^{\circ}}{1 \times 980 \times 0.1}\)

∴ The difference in the level of water in two arms of U tube is

⇒ \(h_1-h_2=\frac{2 \times 70 \times \cos 8^{\circ}}{1 \times 980} \times\left(\frac{1}{0.05}-\frac{1}{0.1}\right)\) = 1.415 cm.

Example 5. There are two parallel glass plates having some water between them. The surface tension of water is 0.07 N/m. When the two plates are moved apart from each other by 0.2 mm, a water layer is formed between the plates. In this case what is the minimum force to be applied to separate the plates. Consider, the angle of contact between water and glass plate is zero, and area of contact is 4 cm².

Solution:

In two two-dimensional picture, the shape of the free surface of the water layer formed between two parallel plates separated by the distance d is concave.

If the radius of curvature of the concave surface is r, then

d = 2rcosθ [where, θ = the angle of contact between water and glass]

The pressure difference between point A (outside the water layer) and point B (inside the water layer) is

⇒ \(p_A-p_B=\frac{T}{r}\) (where T – surface tension of water]

But, p_A = atmospheric pressure (p₀)

i.e., \(p_0-p_B=\frac{T}{r}\)

Atmospheric pressure is also exerted on the two glass plates.

Therefore, the minimum force required to separate the two plates is

F = (p₀ – p_B)α [where α = area of contact surface]

= \(\frac{T \alpha}{r}\)

= \(\frac{2 T \alpha \cos \theta}{d}\)

Here, T = 0.07N/m, α = 4 cm² = 4 x 10^-4 m² , θ= 0, d = 0.2 mm = 0.2 x 10^-3 m

∴ F = \(\frac{2 \times 0.07 \times 4 \times 10^{-4} \times \cos \theta}{0.2 \times 10^{-3}}\)

= 0.28 N.

Example 6. Water reaches upto a height 2 cm along a capillary tube. Calculate the height upto which water can reach along another capillary tube of radius 1/3rd of the previous one.
Solution:

Given

Water reaches upto a height 2 cm along a capillary tube.

Let the rise of liquid in the capillary tube of radius 3 be r/3 be h₁.

According to Jurin’s law, hr = constant.

Hence, \(h_1\left(\frac{r}{3}\right)=h r\)

or, h₁ = 3h = 3 x 2 = 6 cm

∴ The rise of the liquid surface in the second tube = 6 cm.

Question 21. The oscillation of a body on a smooth horizontal surface is represented by the equation, x = Acosωt, where x = displacement at time t, ω = frequency of oscillation. Which one of the following graphs shows correctly the variation a with t?

Here, a = acceleration at time t and T = time period.

Answer:

x = Acosωt.

So, velocity, \(v=\frac{d x}{d t}=-A \omega \sin \omega t\)

and acceleration, a = \(\frac{d v}{d t}=-A \omega^2 \cos \omega t\)

Now at t = 0, a = -Aω²

and at t = T, a = -Aω²

The option 3 is correct.

Question 22. Two similar springs P and Q have spring constants K_P and K_Q, such that K_P > K_Q. They are stretched, first by the same amount (case a), then by the same force (case b), The work done by the springs W_P and W_Q are related as, in case  1 and case 2, respectively

1. W_P = W_Q; W_P> W_Q
2. W_P = W_Q; W_P= w_Q
3. W_P>W_Q; WQ>W_P
4. W_P< W_Q; W_Q< W_P

Answer:

Work done by the spring (W) = change in potential energy

= \(\frac{1}{2}\) Kx2 (x = Increase in length)

Case 1: The increase in length is the same for P and Q.

Then, \(W_P=\frac{1}{2} K_P x^2 and W_Q=\frac{1}{2} K_Q x^2\)

∴ \(W_P>W_Q\)

Again, if F is the applied force, then the spring constant

K = \(\frac{F}{x} \text { or, } x=\frac{F}{K}\)

∴ W = \(\frac{1}{2} K\left(\frac{F}{K}\right)^2=\frac{1}{2} \frac{F^2}{K}\)

Case 2: \(W_P=\frac{1}{2} \frac{F^2}{K_P} \text { and } W_Q=\frac{1}{2} \frac{F^2}{K_Q}\)

∴ \(W_Q>W_P\)

The option 3 is correct.

Question 23. When two displacements represented by y₁ = a sinωt and y₂ = bcosωt are superimposed, the motion is

1. Not a simple harmonic
2. Simple harmonic with amplitude a/b
3. Simple harmonic with amplitude \(\sqrt{a^2+b^2}\)
4. Simple harmonic with amplitude \(\frac{(a+b)}{2}\)

Answer:

y = y₁ + y₂ = asinωt + bcosωt

= \(\sqrt{a^2+b^2}\left(\sin \omega t \cdot \frac{a}{\sqrt{a^2+b^2}}+\cos \omega t \cdot \frac{b}{\sqrt{a^2+b^2}}\right)\)

= \(\sqrt{a^2+b^2}(\sin \omega t \cos \theta+\cos \omega t \sin \theta)\)

= \(\sqrt{a^2+b^2} \sin (\omega t+\theta)\)

Hence, this is a simple harmonic motion, whose amplitude \(\sqrt{a^2+b^2}\)

The option 3 is correct.

Question 24. A particle is executing SHM along a straight line. Its velocities at distances x₁ and x₂ from the mean position are V₁ and V₂, respectively. Its time period is

1. \(2 \pi \sqrt{\frac{x_1^2+x_2^2}{V_1^2+V_2^2}}\)
2. \(2 \pi \sqrt{\frac{x_2^2-x_1^2}{V_1^2-V_2^2}}\)
3. \(2 \pi \sqrt{\frac{V_1^2+V_2^2}{x_1^2+x_2^2}}\)
4. \(2 \pi \sqrt{\frac{V_1^2-V_2^2}{x_1^2-x_2^2}}\)

Answer:

⇒ \(V_1=\omega \sqrt{a^2-x_1^2}$ or, $\frac{V_1^2}{\omega^2}=a^2-x_1^2\)

Similarly, \(\frac{V_2^2}{\omega^2}=a^2-x_2^2\)

∴ \(\frac{V_1^2}{\omega^2}-\frac{V_2^2}{\omega^2}=x_2^2-x_1^2 \quad or, \omega=\sqrt{\frac{V_1^2-V_2^2}{x_2^2-x_1^2}}\)

i.e., T = \(\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{x_2^2-x_1^2}{V_1^2-V_2^2}}\)

Question 25. A particle moves so that its position vector is given by \(\vec{r}=\cos \omega t \hat{x}+\sin \omega t \hat{y}\), where co is a constant. Which of the following is true?

1. Velocity and acceleration both are parallel to \(\vec{r}\)
2. Velocity is perpendicular to r and acceleration is directed towards the origin \(\vec{r}\)
3. Velocity is perpendicular to r and acceleration is directed away from the origin \(\vec{r}\)
4. Velocity and acceleration both are perpendicular to \(\vec{r}\)

Answer:

⇒ \(\vec{r}=\cos \omega t \hat{x}+\sin \omega t \hat{y}\)

velocity of the particle, \(\vec{v}=\frac{d \vec{r}}{d t}=-\omega(\sin \omega t) \hat{x}+\omega(\cos \omega t) \hat{y}\)

Acceleration of the particle, \(\vec{a}=\frac{d \vec{v}}{d t}=-\omega^2(\cos \omega t) \hat{x}-\omega^2(\sin \omega t) \hat{y}\)

= \(-\omega^2(\cos \omega t \hat{x}+\sin \omega t \hat{y})=-\omega^2 \vec{r}\)

∴ Acceleration is directed toward the origin.

Again, \(\vec{r}\) . \(\vec{v}\) = -ω sinωt cosωt + ω sinωt cosωt = 0

Thus, \(\vec{v}\) is perpendicular to \(\vec{r}\).

The option 2 is correct.

Question 26. A pendulum is hung from the roof of a sufficiently high building and is moving freely to and fro like a simple harmonic oscillator. The acceleration of the bob of the pendulum is 20 m/s² at a distance of 5 m from the mean position. The time period of oscillation is

1. 2s
2. s
3. 2s
4. 1s

Answer:

⇒ \(\vec{a}=-\omega^2 \vec{x}\)

∴ \(|a|=\omega^2 x \text { or, } 20=\omega^2 \times 5 \text { or, } \omega^2=4 \text { or, } \omega=2\)

T = \(\frac{2 \pi}{\omega}=\frac{2 \pi}{2}=\pi \mathrm{s}\)

Option 2 is correct.

Question 27. Explain the relation in phase between displacement, velocity, and acceleration in SHM, graphically as well as theoretically.
Answer:

The relation in phase between displacement, velocity, and acceleration in SHM can be shown graphically.

Displacement x = Asin(ωt + ø₀),

Velocity, v = ωAsin(ωt + ø₀),

Acceleration a = -ω²Asin(ωt + ø₀),

Here,x and v differ in phase by \(\frac{\pi}{2}\), v and a differ in
phase by \(\frac{\pi}{2}\).

Question 28. A particle is in linear simple harmonic motion between two points A and B, 10 cm apart. Take the direction from A to B, as a positive direction and give the signs of velocity and acceleration on the particle when it is

1. At the end of B,
2. At 3 cm away from A going towards B.

Answer:

1. At the end of B, the velocity of the particle is zero. Here acceleration and force both are negative as they are directed along BR i.e., along negative direction.
2. At 3 cm away from A going towards B, the particle is at position R, with a tendency to move along RP which is a positive direction. Here, velocity and acceleration both are positive.

Question 29. The time period of a particle in SHM depends on the force constant k and mass m of the particle as follows: T = \(2 \pi \sqrt{\frac{m}{k}}\). A simple pendulum executes SHM approximately, why then is the time period of a simple pendulum independent of the mass of the pendulum?
Answer:

Given

The time period of a particle in SHM depends on the force constant k and mass m of the particle as follows: T = \(2 \pi \sqrt{\frac{m}{k}}\). A simple pendulum executes SHM approximately,

In case of simple pendulum, the force constant, k = \(\frac{m g}{l}\)

So, time period, T = \(2 \pi \sqrt{\frac{m}{k}}=2 \pi \sqrt{\frac{m}{\frac{m g}{l}}}=2 \pi \sqrt{\frac{l}{g}}\)

Hence, T is independent of m.

Question 30. One end of the U-tube containing mercury is connected to a suction pump and the other end to the atmosphere. A small pressure difference is maintained between the two columns. Show that, when the suction pump is removed, the column of mercury in the U-tube executes simple harmonic motion.
Answer:

Given

One end of the U-tube containing mercury is connected to a suction pump and the other end to the atmosphere. A small pressure difference is maintained between the two columns

The restoring force,

F =- weight of mercury column at height 2y

=-(volume) x density x g

=-(A x 2yxρxg) =-2Aρgy

where A = area of cross-section of the tube, ρ = density of mercury.

Force constant k = \(-\frac{F}{y}=2 A \rho \cdot g\)

So, time period T = \(2 \pi \sqrt{\frac{m}{k}}=2 \pi \sqrt{\frac{m}{2 A \rho g}}\)

Let L = length of the whole mercury column.

Therefore, mass of mercury, m = volume x density = ALρ

∴ T = \(2 \pi \sqrt{\frac{A L \rho}{2 A \rho g}}=2 \pi \sqrt{\frac{L}{2 g}}\)

where L is the total length of the mercury column =2h, and h is the height of the mercury column in the U-tube. It shows that the mercury column executes SHM.

Question 31. A simple harmonic motion is described by a = -16x where a is acceleration and x is the displacement in meters. What id the time period?
Answer:

Given

A simple harmonic motion is described by a = -16x where a is acceleration and x is the displacement in meters.

Comparing given equation with a = -ω²x, we get

Now, \(\omega=\frac{2 \pi}{T}=\sqrt{16}=4\)

Hence the time period,

T = \(\frac{2 \pi}{\omega}=\frac{2 \pi}{4}=\frac{\pi}{2}=\frac{3.14}{2}=1.57 \mathrm{~s}\)

Question 32. For a particle executing simple harmonic motion, find the distance from the mean position at which its potential and kinetic energies are equal.
Answer:

Let x₀ = required distance from the mean position

Then, \(\mathrm{KE}=E_x=\frac{1}{2} m \omega^2\left(a^2-x_0^2\right) and \quad \mathrm{PE}=E_p=\frac{1}{2} m \omega^2 x_0^2\)

From the given condition, \(E_k=E_p\)

∴ \(\frac{1}{2} m \omega^2\left(a^2-x_0^2\right)=\frac{1}{2} m \omega^2 x_0^2 or, a^2-x_0^2=x_0^2\)

or, \(2 x_0^2=a^2 \quad or, x_0^2=\frac{a^2}{2} \quad or, x_0= \pm \frac{a}{\sqrt{2}}\)

Question 33. Write the phase difference between the velocity and acceleration of a particle undergoing a simple harmonic motion.
Answer:

In SHM, the phase difference between velocity and acceleration = 90°

Question 33. A simple pendulum of length l and having a bob of mass M is suspended in a car. The car is moving on a circular track of radius R with a uniform speed V. If the pendulum makes a small oscillation in a radial direction about its equilibrium position, what will be its time period?
Answer:

Given

A simple pendulum of length l and having a bob of mass M is suspended in a car. The car is moving on a circular track of radius R with a uniform speed V. If the pendulum makes a small oscillation in a radial direction about its equilibrium position

The pendulum bob Is subject to two distinct accelerations:

1. The acceleration due to gravity, \(\overrightarrow{\mathrm{g}}\), acting downwards;
2. The centripetal acceleration, \(\vec{a}=\frac{v^2}{2}\), acting in a horizontal direction. So, the effective, or relative, acceleration due to gravity on the pendulum bob is \(\overrightarrow{g^{\prime}}=\vec{g}-\vec{a}=\vec{g}+(\overrightarrow{-a})\)

Its magnitude, \(g^{\prime}=\sqrt{g^2+\frac{V^{\prime}}{R^2}} \text {. }\)

Then, the time period of the pendulum becomes \(T^{\prime}=2 \pi \sqrt{\frac{l}{g^{\prime}}}=\frac{2 \pi \sqrt{l}}{\left(g^2+\frac{V^4}{R^2}\right)^{1 / 4}}\)

Question 34. What will be the change in the time period of a loaded spring, when taken to the moon?
Answer:

Time period, \(T \propto \frac{1}{\sqrt{g}} .\). As g, the acceleration due to gravity is less on the surface of the moon, the time period would increase.

Question 35. Show the variation of kinetic energy, potential energy, and total energy with

1. Time and
2. Displacement graphically.

Answer:

The displacement is x = acosωt; this means that the motion starts at one end of the path, comes to the mean position at \(\frac{T}{4}\) (T = time period), goes to the other end at \(\frac{T}{2}\), then again comes to the mean position at \(\frac{3T}{4}\), and finally comes back to the starting point at time T, thus completing a complete oscillation.

The variation of the KE, the PE, and the total energy

1. With time t is shown in the graph
2. With displacement, x is shown in the graph

Question 36. How will the time period of a simple pendulum change when its length is doubled?
Answer:

Time period of a simple pendulum, T = \(2 \pi \sqrt{\frac{l}{g}}\)

So for two different lengths, \(\frac{T_1}{T_2}=\sqrt{\frac{l_1}{L_2}}\)

or, \(T_2=T_1 \sqrt{\frac{l_2}{l_1}}=T_1 \sqrt{\frac{2}{1}}=\sqrt{2} T_1\) (given, \(l_2=2 l_1\))

Hence, the time period becomes √2 times when the length is doubled.

Question 36. Two identical springs of spring constant k are attached to a block of mass m and to fixed supports as shown. Show that the mass executes simple harmonic motion when displaced from its rest position on either side.

Answer:

Let the block of mass be displaced by a small distance x to the right side of the equilibrium position as shown. Under this situation the spring on the left side of the mass gets elongated by a length equal to x and that on the right side gets compressed by the same length

The forces acting on the mass are F₁ = -kx

(Exerted by the spring on the left side, trying to pull the mass towards the mean position)

F₂ = -kx

(exerted by the spring on the right side, trying to push the mass towards the mean position)

Hence, the net force acting on the block of mass is F = -2kx which is proportional to the displacement x and is directed towards the mean position. Therefore, the motion executed by the block of mass is simple harmonic.

Question 37. y(t) = (sinωt- cosωt) represents simple harmonic motion, determining the period of SHM.
Answer:

Given

y(t) = (sinωt- cosωt)

= \(\sqrt{2}\left(\frac{1}{\sqrt{2}} \sin \omega t-\frac{1}{\sqrt{2}} \cos \omega t\right)\)

= \(\sqrt{2}\left(\sin \omega t \cos \frac{\pi}{4}-\cos \omega t \sin \frac{\pi}{4}\right)=\sqrt{2} \sin \left(\omega t-\frac{\pi}{4}\right)\)

It is an equation of SHM. Where angular frequency is and time period T = \(\frac{2 \pi}{\omega}\)

Question 38. At what displacement

1. The PE and
2. KE of a simple harmonic oscillator is maximum?

Answer:

1. PE of a simple harmonic oscillator is maximum at x = ±A

PE at that instant, U = \(\frac{1}{2} m \omega^2 A^2\)

KE of a simple harmonic oscillator is maximum at the position of equilibrium, x = 0.

KE at that instant, K = \(\frac{1}{2} m \omega^2 A^2\)

Oscillation And Waves Simple Harmonic Motion Long Answer Type Questions

Question 21. Show that the equation x = Acos(ωt- α) is a mathematical representation of simple harmonic motion. Find the time period and the maximum speed of the motion.
Asnwer:

Given equation

x = Acos(ωt-α)

∴ Velocity, v = \(\frac{d x}{d t}=-A \omega \sin (\omega t-\alpha)\)

The magnitude of sin (ωt-α) is maximum when it is ±1. This corresponds to the maximum velocity, or simply, the maximum speed.

∴ Maximum speed = ±Aω.

Again, cos (ωt-α) = cos(ωt- α + 2π) .

So, a particle executing SHM returns to its initial phase when the phase angle changes by 2π. The time elapsed between them is the time period (T) of the SHM.

∴ ω(t + T)-α = ωt-α + 2π

or, ωT = 2π or, T = \(\frac{2 \pi}{\omega}\)

Now, acceleration, a = \(\frac{d v}{d t}=-A \omega^2 \cos (\omega t-\alpha)\),

i.e., a = -ω²x

This relation signifies that the given equation is a mathematical representation of SHM.

Question 22. If the frequency of a simple harmonic motion is n then find the frequency of its kinetic energy.
Answer:

Given that the frequency of a simple harmonic motion is n

If the particle executes the simple harmonic motion, then displacement, x = Asin(ωt+θ)

x = \(A \sin (\omega t+\theta)\)

= \(A \sin (2 \pi n t+\theta)\) (because \(n=\frac{\omega}{2 \pi}\))

∴ Velocity, v = \(\frac{dx}{dt}\) = Aω cos(ωt+θ)

So, the kinetic energy, if the particle has a mass m, is

K = \(\frac{1}{2} m v^2=\frac{1}{2} m A^2 \omega^2 \cos ^2(\omega t+\theta)\)

= \(\frac{1}{4} m A^2 \omega^2[1+\cos 2(\omega t+\theta)]\)

= \(\frac{1}{4} m A^2 \omega^2+\frac{1}{4} m A^2 \omega^2 \cos (2 \omega t+2 \theta)\)

On the right-hand side, the first term is a constant it ensures that the kinetic energy K is never negative. The cosine factor in the second term actually signifies that K is an oscillating quantity. The frequency of the oscillation is \(n^{\prime}=\frac{2 \omega}{2 \pi}=2 \cdot \frac{\omega}{2 \pi}=2 n .\)

The potential energy of an oscillating particle also oscillates with a frequency 2n.

Question 23. The kinetic energy of a particle is 1/2 mω²(A² -x²), where m, ω and A are constants. Prove that the motion of the particle is simple harmonic.
Answer:

Given

The kinetic energy of a particle is 1/2 mω²(A² -x²), where m, ω and A are constants.

Kinetic energy, K = \(\frac{1}{2} m v^2=\frac{1}{2} m \omega^2\left(A^2-x^2\right)\)

So, the velocity of the particle is,

v = \(\pm \omega \sqrt{A^2-x^2}= \pm \omega\left(A^2-x^2\right)^{1 / 2}\)

∴ \(\frac{d v}{d x}= \pm \omega \cdot \frac{1}{2}\left(A^2-x^2\right)^{-\frac{1}{2}} \cdot(-2 x)=\mp \frac{\omega x}{\sqrt{A^2-x^2}} .\)

Now, acceleration,

a = \(\frac{d v}{d t}=\frac{d x}{d t} \frac{d v}{d x}=v \frac{d v}{d x}\)

= \(\pm \omega \sqrt{A^2-x^2} \cdot \text { (Ғ) } \frac{\omega x}{\sqrt{A^2-x^2}}=-\omega^2 x .\)

So. a = -ω²xt which signifies that the motion is simple harmonic.

Question 24. The speed of a body of mass m, when it is at a distance x from the origin, is v. Its total energy is \(\frac{1}{2} m v^2+\frac{1}{2} k x^2\), where k is constant. Prove that the body executes a simple harmonic motion.
Answer:

Given

The speed of a body of mass m, when it is at a distance x from the origin, is v. Its total energy is \(\frac{1}{2} m v^2+\frac{1}{2} k x^2\), where k is constant.

Total energy, E = \(\frac{1}{2} m v^2+\frac{1}{2} k x^2\)

∴ \(\frac{d E}{d t}=\frac{1}{2} m \frac{d}{d t}\left(v^2\right)+\frac{1}{2} k \frac{d}{d t}\left(x^2\right)\)

= \(\frac{1}{2} m \frac{d}{d v}\left(v^2\right) \frac{d v}{d t}+\frac{1}{2} k \frac{d}{d x}\left(x^2\right) \frac{d x}{d t}\)

= \(\frac{1}{2} m \cdot 2 v \cdot a+\frac{1}{2} k \cdot 2 x \cdot v\)

= \(m v a+k x v \quad[a=\text { acceleration }]\)

If E = constant, \(\frac{d E}{d t}=0\).

∴ \(m v a+k x v=0 \text { or, } a=-\frac{k}{m} x=-\omega^2 x\left[\omega= \pm \sqrt{\frac{k}{m}}\right]\)

This is the equation of a simple harmonic motion.

Question 25. The mass and radius of a satellite is twice that of the Earth. If a second pendulum is taken to that satellite, what will be Its time period?
Answer:

Given

The mass and radius of a satellite is twice that of the Earth. If a second pendulum is taken to that satellite

Acceleration due to gravity on the earth’s surface,  g = \(\frac{GM}{R^2}\)

Acceleration due to gravity on the surface of the satellite in question,  \(g^{\prime}=\frac{G \cdot 2 M}{(2 R)^2}=\frac{1}{2} \frac{G M}{R^2}\)

∴ \(\frac{g}{g^{\prime}}=2\)

Time period on the surface of the earth T = \(2 \pi \sqrt{\frac{L}{g}}\)

The time period on the surface of that satellite \(T^{\prime}=2 \pi \sqrt{\frac{L}{g^{\prime}}}\)

Hence, \(\frac{T}{T^{\prime}}=\sqrt{\frac{g}{g}}\)

or, \(T^{\prime}=T \times \sqrt{\frac{g}{g^{\prime}}}=2 \sqrt{2}=2.828 \mathrm{~s} \quad(because T=2 \mathrm{~s}) .\)

Question 26. A lift is moving vertically upwards with an acceleration a. What will be the changed time period of a simple pendulum suspended from the roof of the lift? If the lift becomes free and starts falling down with the acceleration due to gravity, what will be the change in the time period?
Answer:

Given

A lift is moving vertically upwards with an acceleration a.

Acceleration due to gravity (downwards) = g

Acceleration of the lift (upwards) = -a

Hence, the effective acceleration due to gravity with respect to the lift, g’ = g- (-a) = g + a

Therefore, the time period of the pendulum in the lift, \(T^{\prime}=2 \pi \sqrt{\frac{L}{g+a}}\)

We compare it with the time period inside the lift with no acceleration: \(T=2 \pi \sqrt{\frac{L}{g}}\)

∴ \(\frac{T}{T^{\prime}}=\sqrt{\frac{g+a}{g}} \text { or, } T^{\prime}=T \sqrt{\frac{g}{g+a}}\)

As g< (g+ a), T’ <T; hence, in this case, the time period will decrease. When the lift is falling freely downwards, its acceleration a = g.

Hence, the acceleration due to gravity with respect to the lift, g’ = g- a = g- g = 0.

Hence, the time period, in this case, T’ = \(2 \pi \sqrt{\frac{L}{g^{\prime}}}\) = infinity.

An infinite time period signifies that the pendulum will not oscillate at all.

Question 27. The bob of a simple pendulum is hollow. If half of it is filled with water, what will be the effect on Its time period?
Answer:

Given

The bob of a simple pendulum is hollow. If half of it is filled with water

The center of gravity of the hollow bob is at its center. Thus, the distance from the point of suspension to the center of the bob is the effective length (L) of the pendulum. When the bob is half filled with water, the center of gravity of the bob is lowered. Thus, the effective length of the pendulum increases. Hence, the time period (T) of oscillator of the pendulum also increases as T ∝ √L.

Question 28. A hollow sphere is filled up with water and is suspended using a long string. As water trickles out slowly through an orifice at the lower part of the sphere, it is observed that the time period of the pendulum first increases and. later continues to decrease. Explain these observations.
Answer:

Given

A hollow sphere is filled up with water and is suspended using a long string. As water trickles out slowly through an orifice at the lower part of the sphere, it is observed that the time period of the pendulum first increases and. later continues to decrease.

• Time period of a simple pendulum at a fixed place is directly proportional to the square root of the effective length of the pendulum. The distance between the point of suspension and the center of gravity of the bob is the effective length. The time period of oscillation increases or decreases with the increase or decrease of its effective length.
• The Centre of gravity of the spherical bob, completely filled with water, lies at the center of the sphere. As water trickles out through the orifice, the upper part of the sphere starts getting emptied making the lower half heavier than the upper half.
• As a result, the center of gravity of the bob starts going down from the center of the sphere and the effective length begins to increase. The time period also starts increasing.
• When more water trickles out, the center of gravity of the bob again starts shifting towards the center of the sphere resulting in a decrease in the effective length, and consequently, a lowering of the time period.

When the hollow sphere becomes completely empty, the center of gravity shifts back to the center of the sphere and the effective length regains its initial value. As a result, the final value of the time period equals its initial value.

Question 29. The bob of a simple pendulum is made of iron. Exactly below the equilibrium position of the pendulum, the pole of a strong magnet is placed. What will be the change in the time period of the pendulum?
Answer:

Given

The bob of a simple pendulum is made of iron. Exactly below the equilibrium position of the pendulum, the pole of a strong magnet is placed.

In addition to the gravitational force, the magnetic force of attraction of the magnet acts on the pendulum bob. If acceleration produced by the magnetic field on the bob is a, the total acceleration of the pendulum bob g’ = g+a

∴ Time period of the simple pendulum T = \(x2 \pi \sqrt{\frac{L}{g+a}}\)…(1)

It is understood from equation (1) that the time period decreases due to the increase in acceleration of the pendulum bob.

Question 30. If the point of suspension of a simple pendulum is in a horizontal motion, with constant acceleration, what will be the effect on the time period?
Answer:

Given

The point of suspension of a simple pendulum is in a horizontal motion, with constant acceleration

When the point of suspension moves horizontally with a constant acceleration a, the effective acceleration due to gravity with respect to the point of suspension, \(g^{\prime}=\sqrt{g^2+a^2}\)

∴ Time period, T = \(2 \pi \sqrt{\frac{L}{\left(g^2+a^2\right)^{1 / 2}}}\)…(1)

From equation (1), it is understood that the time period decreases due to the increase in the value of the acceleration of the bob.

Question 31. A simple pendulum, suspended from the roof of a car, is oscillating. What will be the time period of the pendulum when the car moves in a circular path at a constant speed?
Answer:

Given

A simple pendulum, suspended from the roof of a car, is oscillating.

Two forces act on the pendulum bob when the car moves in a circular path at constant speed:

1. Weight of the bob mg vertically downwards,
2. Centrifugal force in the bob mω²r horizontally. (M=mass of bob, ω angular velocity of bob, r=radius of the circular path)

∴ Resultant effective force on the bob, F = \(\sqrt{m^2 g^2+m^2 \omega^4 r^2}=m \sqrt{g^2+\omega^4 r^2}\)

Hence, acceleration of the bob a = \(\frac{F}{m}=\sqrt{g^2+\omega^4 r^2}\)

∴ Time period T = \(2 \pi \sqrt{\frac{L}{a}}=2 \pi \sqrt{\frac{L}{\left(g^2+\omega^4 r^2\right)^{1 / 2}}}\)

Alternative Method: When the car along with the point of suspension of the pen¬dulum moves on a circular path horizontally, a freely falling body (along the vertical line through the point of suspension) is subjected to centripetal acceleration horizontally and acceleration clue to gravity vertically. The centripetal acceleration is ω²r (ω= angular velocity of point of suspension, r = radius of the circular path).

∴ Acceleration of the body falling freely with respect point of suspension = \(\sqrt{g^2+\omega^4 r^2}\)

Hence, time period of oscillation of the pendulum T = \(2 \pi \sqrt{\frac{L}{\left(g^2+\omega^4 r^2\right)^{1 / 2}}}\)

Question 32. State the changes in the time period of a pendulum, when

1. The pendulum is taken to a mountaintop from the earth’s surface
2. The pendulum is set on the floor of a mine
3. The diameter of the bob of the pendulum is increased
4. Keeping the radius of the bob unchanged, its mass is increased. Give reasons for your answer

Answer:

Time period of a simple pendulum T = \(2 \pi \sqrt{\frac{L}{g}}\)

1. Taking the pendulum from the earth’s surface or a mountain top, the value of g decreases. Hence, the time period increases.
2. On the floor of a mine, the value of g decreases. Hence, the time period of a pendulum increases.
3. On increasing the diameter of the bob, its effective length increases. Hence, the time period increases
4. Time period is independent of the mass of the pendulum bob. Hence, even if mass is increased, keeping the diameter the same, the time period remains unchanged.