Welcome to Lecture 120 (Master Index Entry: L114) of our comprehensive JEE Physics Master Series! Advancing through Unit 10: Oscillations and Waves, we present the definitive masterclass on propagation dynamics: Speed of a Travelling Wave — Transverse Waves on Strings, Longitudinal Waves in Fluids, Newton’s Hypothesis & Laplace’s Correction. Wave speed represents the intrinsic rate of phase and energy advancement through a continuous physical medium. In this masterclass, we derive the string tension formula $v = \sqrt{T/\mu}$, solve wave kinematics in heavy vertically hanging ropes with variable tension, examine longitudinal wave speeds in solids and fluids, dissect Newton’s isothermal failure versus Laplace’s adiabatic success for the speed of sound, rigorously analyze the effects of temperature, pressure, humidity, and wind, and master high-yield JEE Advanced problem patterns.
Detailed Concept Card: Wave Speed Mechanics & Governing Formulations
1. The Universal Wave Speed Principle
For any mechanical wave in an ideal, non-dispersive medium, the propagation speed is governed by the dynamic competition between the restoring elastic property and the inertial mass property of the medium:
$$\mathbf{v = \sqrt{\frac{\text{Elastic Property (Restoring Factor)}}{\text{Inertial Property (Mass Factor)}}}}$$
- Independence of Source Parameters: The propagation speed $v$ is an intrinsic property of the medium. In a non-dispersive medium, $v$ is completely independent of the wave’s frequency $f$, wavelength $\lambda$, and amplitude $A$.
2. Transverse Waves on a Stretched String
Consider a taut, flexible string under tension $T$ with linear mass density $\mu = \frac{m}{L}$ (mass per unit length):
$$\mathbf{v = \sqrt{\frac{T}{\mu}}}$$
- In Terms of Volume Density ($\rho$) and Cross-Sectional Area ($A$): $$\mu = \frac{m}{L} = \frac{\rho (A L)}{L} = \rho A = \rho (\pi r^2)$$ $$\mathbf{v = \sqrt{\frac{T}{\rho A}} = \sqrt{\frac{\text{Tensile Stress}}{\rho}} = \frac{1}{r} \sqrt{\frac{T}{\pi \rho}}}$$
- Dimensional Verification: $$[T] = [M L T^{-2}], \quad [\mu] = [M L^{-1}] \implies \left[\frac{T}{\mu}\right] = \left[\frac{M L T^{-2}}{M L^{-1}}\right] = [L^2 T^{-2}] \implies \left[\sqrt{\frac{T}{\mu}}\right] = [L T^{-1}]$$
3. Vertically Suspended Heavy Rope (Variable Tension Dynamics)
When a heavy uniform rope of mass $M$ and length $L$ hangs vertically under gravity from a ceiling, tension varies continuously with height:
- At distance $x$ measured from the free bottom end ($x = 0$), the tension supports only the rope segment of length $x$ below it: $$T(x) = (\mu x) g = \mu g x$$
- The local wave speed at coordinate $x$ is: $$\mathbf{v(x) = \sqrt{\frac{T(x)}{\mu}} = \sqrt{\frac{\mu g x}{\mu}} = \sqrt{gx}}$$ Notice that $v = 0$ at the free bottom end ($x = 0$) and increases to $v_{\text{top}} = \sqrt{gL}$ at the ceiling ($x = L$).
- Transit Time ($t_{\text{transit}}$) from Bottom to Top: $$v(x) = \frac{dx}{dt} = \sqrt{gx} \implies dt = \frac{dx}{\sqrt{gx}}$$ $$\mathbf{t = \int_0^L \frac{dx}{\sqrt{gx}} = \frac{1}{\sqrt{g}} \left[2\sqrt{x}\right]_0^L = 2\sqrt{\frac{L}{g}}}$$
- Constant Pulse Acceleration: $$\mathbf{a = v \frac{dv}{dx} = \sqrt{gx} \cdot \left(\frac{g}{2\sqrt{gx}}\right) = \frac{g}{2} = \text{constant}}$$ The pulse ascends the hanging rope with a constant upward acceleration of $a = g/2$.
4. Longitudinal Waves: Newton’s Formula & Laplace’s Correction
In any continuous fluid (liquid or gas), the general speed of longitudinal sound waves is $v = \sqrt{B/\rho}$, where $B$ is the volumetric bulk modulus.
| Model / Formulation | Thermodynamic Premise | Effective Bulk Modulus ($B$) | Resulting Speed Formula ($v$) | Predicted Speed in Air at STP |
|---|---|---|---|---|
| Newton’s Formula (1687) | Isothermal Process: Compressions and rarefactions occur slowly with rapid heat conduction ($T = \text{const}$). | $$B_{\text{iso}} = P$$ | $$v_{\text{Newton}} = \sqrt{\frac{P}{\rho}}$$ | $$v \approx 280\text{ m/s}$$ ($\approx 16\%$ error below experiment) |
| Laplace’s Correction (1816) | Adiabatic Process: Pressure variations occur so rapidly that negligible heat conducts across wavelengths ($P V^\gamma = \text{const}$). | $$B_{\text{adia}} = \gamma P$$ | $$\mathbf{v = \sqrt{\frac{\gamma P}{\rho}} = \sqrt{\frac{\gamma R T}{M}}}$$ | $$v \approx 331.3\text{ m/s}$$ (Exact experimental agreement!) |
5. Factors Influencing the Speed of Sound in a Gas
- 1. Effect of Pressure at Constant Temperature: From the ideal gas equation, $P = \frac{\rho R T}{M} \implies \frac{P}{\rho} = \frac{RT}{M}$. Therefore, at constant temperature, $\frac{P}{\rho}$ is invariant. A change in gas pressure has NO effect on the speed of sound!
- 2. Effect of Temperature: $$\mathbf{v \propto \sqrt{T}} \quad (T \text{ in Kelvin})$$ $$\frac{v_1}{v_2} = \sqrt{\frac{T_1}{T_2}}$$ Near room temperature ($0^\circ\text{C}$ to $30^\circ\text{C}$), expanding $v(t) = v_0 \sqrt{1 + t/273} \approx v_0 (1 + \frac{t}{546})$ yields: $$\mathbf{v_t \approx v_0 + 0.61 t}$$ Speed of sound in air increases by approximately $\mathbf{0.61\text{ m/s}}$ for each $1^\circ\text{C}$ rise in temperature.
- 3. Effect of Humidity (Moist Air vs Dry Air): The molar mass of water vapor ($M_{\text{H}_2\text{O}} = 18\text{ g/mol}$) is significantly less than that of dry air ($M_{\text{air}} \approx 29\text{ g/mol}$). Consequently, moist air has a lower density than dry air at the same temperature and pressure ($\rho_{\text{humid}} < \rho_{\text{dry}}$). $$\mathbf{v_{\text{humid}} > v_{\text{dry}}}$$ Sound travels faster through humid air than through dry air.
- 4. Effect of Wind Velocity: If wind blows with speed $w$ at an angle $\theta$ relative to the sound propagation vector: $$\mathbf{v_{\text{observed}} = v \pm w \cos\theta}$$
Four Solved Examples (Graded Basic to Advanced)
Example 1: Basic Application — Transverse Wave Speed on a Stretched Wire
Problem Statement: A uniform steel wire of length $L = 2.0\text{ m}$ and total mass $m = 0.040\text{ kg} = 40\text{ g}$ is stretched under a constant mechanical tension $T = 200\text{ N}$.
(a) Determine the linear mass density $\mu$ of the wire.
(b) Calculate the propagation speed $v$ of a transverse wave on the wire.
(c) Find the time required for a wave pulse to travel from one end of the wire to the other.
(d) If the tension in the wire is increased by $44\%$, calculate the percentage increase in the wave speed.
Solution:
(a) Linear Mass Density:
$$\mathbf{\mu = \frac{m}{L} = \frac{0.040\text{ kg}}{2.0\text{ m}} = 0.020\text{ kg/m}}$$
(b) Wave Speed:
$$\mathbf{v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{200\text{ N}}{0.020\text{ kg/m}}} = \sqrt{10000} = 100\text{ m/s}}$$
(c) Transit Time:
$$\mathbf{t = \frac{L}{v} = \frac{2.0\text{ m}}{100\text{ m/s}} = 0.020\text{ s} = 20\text{ ms}}$$
(d) Tension Scaling:
New tension $T’ = T + 0.44 T = 1.44 T$.
$$v’ = \sqrt{\frac{T’}{\mu}} = \sqrt{\frac{1.44 T}{\mu}} = \sqrt{1.44} \sqrt{\frac{T}{\mu}} = 1.20 v$$
Percentage increase:
$$\mathbf{\frac{v’ – v}{v} \times 100\% = (1.20 – 1.0) \times 100\% = 20\%}$$
Example 2: Intermediate Application — Wave Pulse on a Heavy Rope Supporting an End Load
Problem Statement: A heavy climbing rope of mass $M = 6.0\text{ kg}$ and length $L = 12.0\text{ m}$ is suspended vertically from a rigid ceiling. A block of mass $m = 54.0\text{ kg}$ is attached to the lower free end of the rope ($g = 10\text{ m/s}^2$). A transverse wave pulse is initiated at the bottom end.
(a) Derive the expression for the tension $T(x)$ at distance $x$ from the bottom end.
(b) Calculate the wave speed at the bottom end ($x = 0$) and at the ceiling ($x = L$).
(c) Determine the total time required for the pulse to ascend from the bottom to the ceiling.
Solution:
(a) Tension Function:
Linear mass density $\mu = \frac{M}{L} = \frac{6.0\text{ kg}}{12.0\text{ m}} = 0.50\text{ kg/m}$.
At distance $x$ from the bottom, the tension supports the suspended load $m$ plus the rope segment of length $x$ (mass $\mu x$):
$$T(x) = (m + \mu x) g = [54.0 + 0.50 x](10) = \mathbf{(540 + 5.0 x)\text{ N}}$$
(b) Wave Speeds at Extremities:
At the bottom end ($x = 0$):
$$\mathbf{v_{\text{bottom}} = \sqrt{\frac{T(0)}{\mu}} = \sqrt{\frac{540\text{ N}}{0.50\text{ kg/m}}} = \sqrt{1080} \approx 32.86\text{ m/s}}$$
At the ceiling ($x = L = 12.0\text{ m}$):
$$\mathbf{v_{\text{top}} = \sqrt{\frac{T(L)}{\mu}} = \sqrt{\frac{(m+M)g}{\mu}} = \sqrt{\frac{(54.0+6.0)(10)}{0.50}} = \sqrt{\frac{600}{0.50}} = \sqrt{1200} \approx 34.64\text{ m/s}}$$
(c) Ascent Time:
$$v(x) = \frac{dx}{dt} = \sqrt{\frac{(m + \mu x)g}{\mu}} = \sqrt{g\left(\frac{m}{\mu} + x\right)}$$
Let $x_0 = \frac{m}{\mu} = \frac{54.0\text{ kg}}{0.50\text{ kg/m}} = 108.0\text{ m}$.
$$dt = \frac{dx}{\sqrt{g(x + x_0)}} \implies t = \frac{1}{\sqrt{g}} \int_0^L (x + x_0)^{-1/2} \, dx = \frac{2}{\sqrt{g}} \left[ \sqrt{L + x_0} – \sqrt{x_0} \right]$$
Substituting $L = 12.0\text{ m}, x_0 = 108.0\text{ m},$ and $g = 10\text{ m/s}^2$:
$$t = \frac{2}{\sqrt{10}} \left[ \sqrt{12.0 + 108.0} – \sqrt{108.0} \right] = \frac{2}{\sqrt{10}} \left[ \sqrt{120.0} – \sqrt{108.0} \right]$$
$$t = 2 \left[ \sqrt{\frac{120}{10}} – \sqrt{\frac{108}{10}} \right] = 2 \left[ \sqrt{12} – \sqrt{10.8} \right] = 2 [3.4641 – 3.2863] = 2(0.1778) \approx \mathbf{0.356\text{ s}}$$
Example 3: Advanced Application — Speed of Sound in a Mixed Gas Atmosphere
Problem Statement: A gaseous mixture is prepared containing $n_1 = 2.0\text{ moles}$ of helium (monoatomic gas, $\gamma_1 = 5/3, M_1 = 4.0\text{ g/mol}$) and $n_2 = 1.0\text{ mole}$ of oxygen (diatomic gas, $\gamma_2 = 7/5, M_2 = 32.0\text{ g/mol}$) maintained at temperature $T_K = 300\text{ K}$ ($R = 8.314\text{ J/(mol}\cdot\text{K)}$).
(a) Determine the effective molar mass $M_{\text{mix}}$ of the gas mixture.
(b) Determine the effective adiabatic exponent $\gamma_{\text{mix}}$ of the mixture.
(c) Calculate the speed of sound $v_{\text{mix}}$ in this mixture at $300\text{ K}$.
(d) At what temperature $T_{\text{N}_2}$ would the speed of sound in pure nitrogen ($M_{\text{N}_2} = 28.0\text{ g/mol}, \gamma = 7/5$) equal $v_{\text{mix}}$?
Solution:
(a) Equivalent Molar Mass:
$$\mathbf{M_{\text{mix}} = \frac{n_1 M_1 + n_2 M_2}{n_1 + n_2} = \frac{2(4.0) + 1(32.0)}{2 + 1} = \frac{40.0}{3}\text{ g/mol} = \frac{0.040}{3}\text{ kg/mol} \approx 0.01333\text{ kg/mol}}$$
(b) Effective Adiabatic Exponent:
$$C_{v,\text{mix}} = \frac{n_1 C_{v1} + n_2 C_{v2}}{n_1 + n_2} = \frac{2\left(\frac{3}{2}R\right) + 1\left(\frac{5}{2}R\right)}{3} = \frac{3R + 2.5R}{3} = \frac{5.5}{3}R = \frac{11}{6}R$$
$$C_{p,\text{mix}} = C_{v,\text{mix}} + R = \frac{11}{6}R + R = \frac{17}{6}R$$
$$\mathbf{\gamma_{\text{mix}} = \frac{C_{p,\text{mix}}}{C_{v,\text{mix}}} = \frac{17/6}{11/6} = \frac{17}{11} \approx 1.5455}$$
(c) Speed of Sound in Mixture:
$$v_{\text{mix}} = \sqrt{\frac{\gamma_{\text{mix}} R T_K}{M_{\text{mix}}}} = \sqrt{\frac{(17/11)(8.314\text{ J/mol}\cdot\text{K})(300\text{ K})}{(0.040/3\text{ kg/mol})}}$$
$$v_{\text{mix}} = \sqrt{\frac{(1.54545)(2494.2)}{0.013333}} = \sqrt{289102} \approx \mathbf{537.7\text{ m/s}}$$
(d) Equivalent Temperature for Nitrogen:
Setting $v_{\text{N}_2}^2 = v_{\text{mix}}^2$:
$$\frac{\gamma_{\text{N}_2} R T_{\text{N}_2}}{M_{\text{N}_2}} = 289102 \implies T_{\text{N}_2} = \frac{(289102) M_{\text{N}_2}}{\gamma_{\text{N}_2} R}$$
$$\mathbf{T_{\text{N}_2} = \frac{(289102)(0.028\text{ kg/mol})}{(1.40)(8.314\text{ J/mol}\cdot\text{K})} = \frac{8094.86}{11.6396} \approx \mathbf{695.4\text{ K} \approx 422.3^\circ\text{C}}}$$
Example 4: JEE Advanced Scenario — Composite String Boundary Reflection & Transmission
Problem Statement: Two uniform strings $1$ and $2$ of identical length $L = 1.0\text{ m}$ each, but with differing linear mass densities $\mu_1 = 0.010\text{ kg/m}$ and $\mu_2 = 0.040\text{ kg/m}$, are joined seamlessly end-to-end. The composite system is maintained under a uniform mechanical tension $T = 100\text{ N}$. A transverse wave pulse of amplitude $A_i = 3.0\text{ mm}$ is launched from the free end of string $1$ toward string $2$.
(a) Determine the wave speeds $v_1$ and $v_2$ in strings $1$ and $2$, respectively.
(b) Calculate the total transit time for the pulse to travel through the entire composite string.
(c) Determine the amplitude of the reflected pulse $A_r$ and the transmitted pulse $A_t$ at the interface.
(d) State the phase shift experienced by the reflected and transmitted pulses.
Solution:
(a) Wave Speeds:
$$\mathbf{v_1 = \sqrt{\frac{T}{\mu_1}} = \sqrt{\frac{100\text{ N}}{0.010\text{ kg/m}}} = 100\text{ m/s}}$$
$$\mathbf{v_2 = \sqrt{\frac{T}{\mu_2}} = \sqrt{\frac{100\text{ N}}{0.040\text{ kg/m}}} = 50\text{ m/s}}$$
(b) Transit Time:
$$t_1 = \frac{L}{v_1} = \frac{1.0\text{ m}}{100\text{ m/s}} = 0.010\text{ s}, \quad t_2 = \frac{L}{v_2} = \frac{1.0\text{ m}}{50\text{ m/s}} = 0.020\text{ s}$$
$$\mathbf{t_{\text{total}} = t_1 + t_2 = 0.010\text{ s} + 0.020\text{ s} = 0.030\text{ s} = 30\text{ ms}}$$
(c) Reflected & Transmitted Amplitudes:
From boundary continuity of wave displacement and slope:
$$A_r = \left(\frac{v_2 – v_1}{v_2 + v_1}\right) A_i = \left(\frac{50 – 100}{50 + 100}\right) (3.0\text{ mm}) = \left(-\frac{50}{150}\right)(3.0\text{ mm}) = \mathbf{-1.0\text{ mm}}$$
$$A_t = \left(\frac{2v_2}{v_2 + v_1}\right) A_i = \left(\frac{2 \times 50}{50 + 100}\right) (3.0\text{ mm}) = \left(\frac{100}{150}\right)(3.0\text{ mm}) = \mathbf{+2.0\text{ mm}}$$
(d) Boundary Phase Shifts:
– The negative sign in $A_r$ indicates that the reflected wave experiences a phase reversal of $\pi\text{ rad} \ (180^\circ)$ upon reflection from the denser medium ($\mu_2 > \mu_1$).
– The positive sign in $A_t$ indicates that the transmitted wave undergoes zero phase shift ($0^\circ$).
Practice Worksheet: 10 Target JEE Problems
Test your mastery of transverse string speeds, Laplace’s sound equations, and gas mixture parameters. Solve all 10 problems before consulting the detailed solutions at the end.
Section I: Single Correct Choice Questions (Q1–Q5)
Q1. The speed of a transverse mechanical wave propagating along a stretched string is determined exclusively by:
(A) The frequency and amplitude of the oscillating source
(B) The mechanical tension and the linear mass density of the string
(C) The total length of the string alone
(D) The geometric shape of the initial wave pulse
Q2. At a constant temperature, if the pressure of an ideal gas is doubled, the speed of sound in the gas:
(A) Doubles
(B) Is halved
(C) Becomes $\sqrt{2}$ times its initial value
(D) Remains strictly unchanged
Q3. A uniform heavy rope of mass $M$ and length $L$ hangs vertically under its own weight from a ceiling. The time taken by a transverse wave pulse to travel from the bottom end to the ceiling is:
(A) $\sqrt{\frac{L}{g}}$
(B) $2\sqrt{\frac{L}{g}}$
(C) $\pi\sqrt{\frac{L}{g}}$
(D) $\sqrt{\frac{2L}{g}}$
Q4. If the absolute temperature of an ideal gas is increased by $21\%$, the percentage increase in the speed of sound in that gas is:
(A) $10\%$
(B) $21\%$
(C) $10.5\%$
(D) $4.5\%$
Q5. The ratio of the speed of sound in hydrogen gas ($\gamma_1 = 7/5, M_1 = 2.0\text{ g/mol}$) to that in helium gas ($\gamma_2 = 5/3, M_2 = 4.0\text{ g/mol}$) at the same absolute temperature is:
(A) $\sqrt{21} / 5$
(B) $\sqrt{35} / 4$
(C) $\sqrt{42} / 5$
(D) $\sqrt{14} / 3$
Section II: One or More Than One Correct Questions (Q6–Q8)
Q6. Which of the following statements is/are correct regarding the speed of sound in air?
(A) Laplace corrected Newton’s formula by treating the compressions and rarefactions as adiabatic processes rather than isothermal processes.
(B) Sound travels faster on a humid day than on a dry day at the same ambient temperature and pressure.
(C) The speed of sound in air increases by approximately $0.61\text{ m/s}$ for each $1^\circ\text{C}$ rise in temperature near $0^\circ\text{C}$.
(D) High-frequency treble sound waves propagate faster through air than low-frequency bass sound waves.
Q7. A metal wire under tension $T$ and linear mass density $\mu$ carries a transverse wave. If the tension is increased by a factor of 4 and the radius of the wire is halved while keeping material density $\rho$ unchanged:
(A) The wave speed quadruples ($v’ = 4v$).
(B) The linear mass density of the wire decreases to $1/4$ of its initial value.
(C) The transit time of a pulse along a fixed length $L$ drops to $1/4$ of its initial value.
(D) The fundamental resonance frequency of the fixed-ended wire becomes 4 times its initial value.
Q8. When a transverse wave pulse travelling on a light string under tension $T$ strikes a junction connecting it to a heavier string under the same tension:
(A) The speed of the wave in the heavier string is smaller than in the light string.
(B) The reflected wave pulse undergoes a phase shift of $\pi$ radians ($180^\circ$).
(C) The transmitted wave pulse undergoes zero phase shift ($0^\circ$).
(D) The frequency of the transmitted wave equals the frequency of the incident wave.
Section III: Numerical / Integer Type Questions (Q9–Q10)
Q9. At what absolute temperature (in Kelvin) will the speed of sound in oxygen gas ($M = 32.0\text{ g/mol}$) be equal to the speed of sound in nitrogen gas ($M = 28.0\text{ g/mol}$) at temperature $T = 273\text{ K}$? Both gases are diatomic ($\gamma = 1.40$).
Q10. A uniform heavy rope of mass $M = 2.0\text{ kg}$ and length $L = 40.0\text{ m}$ hangs vertically from a ceiling under gravity. Taking $g = 10\text{ m/s}^2$, calculate the time (in seconds) required for a transverse wave pulse to travel from the bottom end of the rope to the ceiling.
Comprehensive Solutions & Answer Key
Answer Key
| Question No. | Correct Answer | Key Concept / Formula Tested |
|---|---|---|
| Q1 | (B) | $v = \sqrt{T/\mu}$; independent of source frequency and amplitude |
| Q2 | (D) | $P/\rho = RT/M = \text{const} \implies$ pressure has zero effect at constant temperature |
| Q3 | (B) | $t = \int_0^L \frac{dx}{\sqrt{gx}} = 2\sqrt{L/g}$ |
| Q4 | (A) | $T’ = 1.21 T \implies v’ = \sqrt{1.21}v = 1.10 v \implies +10\%$ |
| Q5 | (C) | $v \propto \sqrt{\gamma/M} \implies v_1/v_2 = \sqrt{\frac{7/10}{5/12}} = \sqrt{42}/5$ |
| Q6 | (A, B, C) | Laplace adiabatic law, lower density of moist air, $0.61\text{ m/s}/^\circ\text{C}$ |
| Q7 | (A, B, C, D) | $\mu = \rho \pi r^2 \implies \mu’ = \mu/4; v’ = \sqrt{4T/(\mu/4)} = 4v$ |
| Q8 | (A, B, C, D) | Boundary reflection from denser medium: inversion by $\pi$, transmission $0^\circ$ |
| Q9 | 312 | $T_1/M_1 = T_2/M_2 \implies T_{\text{O}_2} = 32(273/28) = 312\text{ K}$ |
| Q10 | 4 | $t = 2\sqrt{L/g} = 2\sqrt{40/10} = 2(2) = 4\text{ s}$ |
Detailed Step-by-Step Solutions
Solution to Q1:
The speed of a transverse mechanical wave on a stretched string is given by $v = \sqrt{T/\mu}$. It is determined entirely by the elastic restoring force (tension $T$) and the inertial resistance (linear mass density $\mu$). In an ideal non-dispersive string, the wave speed is completely independent of the frequency, amplitude, or waveform generated by the source.
Hence, the correct choice is (B).
Solution to Q2:
From Laplace’s formula:
$$v = \sqrt{\frac{\gamma P}{\rho}}$$
From the ideal gas equation of state:
$$P = \frac{\rho R T}{M} \implies \frac{P}{\rho} = \frac{RT}{M}$$
Substituting into the velocity formula gives $v = \sqrt{\frac{\gamma R T}{M}}$.
When the pressure is doubled at constant temperature, the density $\rho$ also doubles proportionally by Boyle’s law, leaving the ratio $P/\rho$ exactly constant. Hence, pressure variations at constant temperature produce zero change in the speed of sound.
Hence, the correct choice is (D).
Solution to Q3:
At distance $x$ from the free bottom end, the tension is $T(x) = \mu g x$. The local wave speed is:
$$v(x) = \sqrt{\frac{\mu g x}{\mu}} = \sqrt{gx}$$
Since $v = \frac{dx}{dt}$:
$$dt = \frac{dx}{\sqrt{gx}} \implies t = \frac{1}{\sqrt{g}} \int_0^L x^{-1/2} \, dx = \frac{1}{\sqrt{g}} \left[ 2\sqrt{x} \right]_0^L = \mathbf{2\sqrt{\frac{L}{g}}}$$
Hence, the correct choice is (B).
Solution to Q4:
The speed of sound in an ideal gas is proportional to the square root of the absolute temperature: $v \propto \sqrt{T}$.
If temperature increases by $21\%$:
$$T’ = T + 0.21 T = 1.21 T$$
$$v’ = \sqrt{\frac{\gamma R T’}{M}} = \sqrt{1.21} \sqrt{\frac{\gamma R T}{M}} = 1.10 v$$
The percentage increase in wave speed is:
$$\frac{v’ – v}{v} \times 100\% = (1.10 – 1.0) \times 100\% = \mathbf{10\%}$$
Hence, the correct choice is (A).
Solution to Q5:
Using $v = \sqrt{\frac{\gamma R T}{M}}$:
$$\frac{v_{\text{H}_2}}{v_{\text{He}}} = \sqrt{\frac{\gamma_1 / M_1}{\gamma_2 / M_2}} = \sqrt{\frac{(7/5) / 2.0}{(5/3) / 4.0}} = \sqrt{\frac{7/10}{5/12}} = \sqrt{\frac{7}{10} \times \frac{12}{5}} = \sqrt{\frac{84}{50}} = \sqrt{\frac{42}{25}} = \mathbf{\frac{\sqrt{42}}{5}}$$
Hence, the correct choice is (C).
Solution to Q6:
Let us evaluate each statement:
(A) Newton assumed isothermal conditions ($v = \sqrt{P/\rho}$), which gave $280\text{ m/s}$. Laplace corrected this by establishing that rapid pressure cycles are adiabatic ($v = \sqrt{\gamma P/\rho}$), matching the measured $332\text{ m/s}$. (A) is correct.
(B) Moist air contains lighter $\text{H}_2\text{O}$ molecules ($M = 18$) replacing heavier $\text{N}_2$ ($M = 28$) and $\text{O}_2$ ($M = 32$) molecules, reducing the air density $\rho$. Since $v \propto 1/\sqrt{\rho}$, sound travels faster in humid air. (B) is correct.
(C) $v_t \approx v_0 + 0.61 t$, meaning speed increases by $0.61\text{ m/s}$ per $^\circ\text{C}$. (C) is correct.
(D) Air is non-dispersive for audible sound; all frequencies travel at the exact same speed. Thus (D) is incorrect.
Hence, correct choices are (A, B, C).
Solution to Q7:
Let the initial radius be $r$ and tension be $T$.
Initial linear density: $\mu = \rho (\pi r^2)$.
When the radius is halved ($r’ = r/2$):
$$\mu’ = \rho [\pi (r/2)^2] = \frac{\mu}{4}$$
Statement (B) is correct.
New tension $T’ = 4T$.
New wave speed:
$$v’ = \sqrt{\frac{T’}{\mu’}} = \sqrt{\frac{4T}{\mu/4}} = \sqrt{16 \frac{T}{\mu}} = 4 \sqrt{\frac{T}{\mu}} = 4v$$
Statement (A) is correct.
The transit time is $t’ = L/v’ = L/(4v) = t/4$. Statement (C) is correct.
The fundamental frequency is $f’ = \frac{v’}{2L} = \frac{4v}{2L} = 4f$. Statement (D) is correct.
Hence, all choices (A, B, C, D) are correct.
Solution to Q8:
When travelling from a light string ($\mu_1$) to a heavier string ($\mu_2 > \mu_1$) under the same tension $T$:
(A) $v_2 = \sqrt{T/\mu_2} < v_1 = \sqrt{T/\mu_1}$. (A) is correct.
(B) The reflection coefficient is $r = \frac{v_2 – v_1}{v_2 + v_1} < 0$. The reflected pulse suffers a phase reversal of $\pi$ radians ($180^\circ$). (B) is correct.
(C) The transmission coefficient is $t = \frac{2v_2}{v_2 + v_1} > 0$. The transmitted wave maintains identical phase ($0^\circ$). (C) is correct.
(D) Boundary continuity requires identical oscillation frequency across the junction. (D) is correct.
Hence, all choices (A, B, C, D) are correct.
Solution to Q9:
The condition for equal sound speeds in two diatomic gases ($\gamma = 1.40$) is:
$$\sqrt{\frac{\gamma R T_{\text{O}_2}}{M_{\text{O}_2}}} = \sqrt{\frac{\gamma R T_{\text{N}_2}}{M_{\text{N}_2}}} \implies \frac{T_{\text{O}_2}}{M_{\text{O}_2}} = \frac{T_{\text{N}_2}}{M_{\text{N}_2}}$$
Substituting $M_{\text{O}_2} = 32.0\text{ g/mol}, M_{\text{N}_2} = 28.0\text{ g/mol},$ and $T_{\text{N}_2} = 273\text{ K}$:
$$T_{\text{O}_2} = M_{\text{O}_2} \left(\frac{T_{\text{N}_2}}{M_{\text{N}_2}}\right) = 32.0 \left(\frac{273\text{ K}}{28.0}\right) = 32.0 (9.75) = \mathbf{312\text{ K}}$$
Answer: 312
Solution to Q10:
For a vertically suspended heavy rope of length $L$ under gravity $g$, the time for a pulse to ascend from bottom to ceiling is:
$$t = 2\sqrt{\frac{L}{g}}$$
Given $L = 40.0\text{ m}$ and $g = 10\text{ m/s}^2$:
$$t = 2\sqrt{\frac{40.0\text{ m}}{10\text{ m/s}^2}} = 2\sqrt{4.0} = 2(2.0) = \mathbf{4\text{ s}}$$
Answer: 4