JEE Physics Waves MCQs Set 21

Answer: (a) \( \sqrt{5/7}\,v_0 \)

Explanation:
1. Hydrogen: \( \gamma=7/5,\ M=2 \). Helium: \( \gamma=5/3,\ M=4 \).
2. Equimolar mixture: \( M=3 \); \( C_v=\tfrac12(\tfrac52R+\tfrac32R)=2R,\ C_p=3R,\ \gamma=1.5 \).
3. \( v\propto\sqrt{\gamma/M} \): \( \dfrac{v}{v_0}=\sqrt{\dfrac{1.5/3}{1.4/2}}=\sqrt{\dfrac57} \).

Teacher's Note:
1. For a mixture find the average molar mass and the average gamma from Cp and Cv. Then compare the two values of gamma over M.

Answer: (b) 24 cm/s

Explanation:
1. Phase lag of 1 cm = \( \pi/8 \), so \( k=\pi/8\ \text{cm}^{-1} \).
2. \( \omega=3\pi\Rightarrow v=\omega/k=3\pi\times\dfrac8\pi=24\ \text{cm/s} \).

Teacher's Note:
1. Phase difference between two points is k times their separation. Speed is omega over k.

Answer: (b) 5.0 m/s

Explanation:
1. Highest and lowest at the same instant means a phase difference of \( \pi \), i.e. separation \( \lambda/2=10 \) m, so \( \lambda=20 \) m.
2. \( v=\lambda/T=20/4=5\ \text{m/s} \).

Teacher's Note:
1. Opposite phase means half a wavelength apart. 'Within a single cycle' rules out three half wavelengths and so on.

Answer: (a) 100 cm/s and 2 cm

Explanation:
1. \( k=6.28\times0.5=3.14\ \text{cm}^{-1}\Rightarrow\lambda=2\pi/k=2\ \text{cm} \).
2. \( \omega=6.28\times50=314\ \text{rad/s}\Rightarrow v=\omega/k=100\ \text{cm/s} \).

Teacher's Note:
1. Factor out the common number so that the bracket reads k x minus omega t. Then read k and omega directly.

Answer: (d) 80 N

Explanation:
1. \( v=\omega/k=420/21=20\ \text{m/s} \).
2. \( T=\mu v^2=0.200\times400=80\ \text{N} \).

Teacher's Note:
1. The wave speed on a string is root of tension over linear density. Square it to get the tension.

Answer: (a) \( \text{H}_2 \)

Explanation:
1. \( v=\sqrt{\gamma RT/M} \): at the same T the speed is largest for the smallest molar mass (with similar gamma).
2. H2 has molar mass 2 against 28, 4 and 44, so sound is fastest in hydrogen.

Teacher's Note:
1. Lighter gases carry sound faster. Helium has a higher gamma but the root of the ratio still keeps hydrogen on top.

Answer: (c) 927 °C

Explanation:
1. \( v\propto\sqrt T \) (T in kelvin). Doubling v needs 4 times the temperature.
2. New T = 4 × 300 = 1200 K = 927 °C.

Teacher's Note:
1. Always convert to kelvin first. Speed doubles when absolute temperature becomes four times.

Answer: (a) 33 m

Explanation:
1. \( \lambda=v/f=352/384=0.9167\ \text{m} \).
2. Distance in 36 vibrations = \( 36\lambda=33\ \text{m} \).

Teacher's Note:
1. In one vibration the wave advances one wavelength. Multiply by the number of vibrations.

Answer: (a) \( V_u,V_i \) and \( V_a \) are equal

Explanation:
1. The speed of sound depends only on the medium and its temperature, not on the frequency.
2. So all three types of wave have the same speed.

Teacher's Note:
1. Speed in a medium is fixed by its elastic and inertial properties. Frequency changes only the wavelength.

Answer: (b) (b) and (d) only

Explanation:
1. The argument has the form \( \omega t-kx \), so it is a travelling wave and (a) is false.
2. \( v=\omega/k=500/70=50/7\ \text{m/s} \): (b) true.
3. \( f=\omega/2\pi=500/2\pi \) Hz, so (c) is false.
4. \( \lambda=2\pi/70\ \text{m}=\dfrac{20\pi}{7}\ \text{cm} \): (d) true.

Teacher's Note:
1. Frequency is omega divided by two pi, not multiplied. Nearest points in the same phase are one wavelength apart.

Answer: (a) 200 Hz

Explanation:
1. Speed in the medium \( v=f\lambda=300\times1=300\ \text{m/s} \).
2. Child: \( f=v/\lambda=300/1.5=200\ \text{Hz} \).

Teacher's Note:
1. The speed is the same for both voices, so frequency and wavelength are inversely proportional.

Answer: (c) \( A\sin(\omega t-kx) \)

Explanation:
1. A progressive wave depends on t and x only through the combination \( \omega t\mp kx \).
2. Option (b) is a standing wave; (a) and (d) depend on only one variable.

Teacher's Note:
1. Look for the argument ωt minus kx. A product of separate functions of x and t is a standing wave.

Answer: (c) 332.3 m/s

Explanation:
1. \( v=\sqrt{\dfrac{\gamma P}{\rho}}=\sqrt{\dfrac{1.41\times1.013\times10^5}{1.293}}=\sqrt{1.105\times10^5}\approx332\ \text{m/s} \).

Teacher's Note:
1. Newton's formula corrected by Laplace uses gamma. About 332 m/s at 0 °C is a number worth remembering.

Answer: (b) light waves are electromagnetic in nature

Explanation:
1. Electromagnetic waves are oscillating electric and magnetic fields and need no medium.
2. Sound is a mechanical wave and needs a medium.

Teacher's Note:
1. Mechanical waves need a medium to carry the disturbance. Electromagnetic waves do not.

Answer: (c) 5 μm

Explanation:
1. Treat the amplitudes as phasors at 0°, 90° and 180°.
2. The 10 and 7 phasors are opposite: net 10 - 7 = 3 μm along the first direction; 4 μm is perpendicular.
3. Resultant = \( \sqrt{3^2+4^2}=5\ \mu\text{m} \).

Teacher's Note:
1. Add the first and third phasors, which are opposite, and then combine with the second at right angles.

Answer: (c) 9 : 4

Explanation:
1. \( \dfrac{I_{max}}{I_{min}}=\dfrac{(a_1+a_2)^2}{(a_1-a_2)^2}=25\Rightarrow\dfrac{a_1+a_2}{a_1-a_2}=5 \).
2. So \( a_1+a_2=5a_1-5a_2 \), giving \( a_1/a_2=3/2 \).
3. \( I_1:I_2=a_1^2:a_2^2=9:4 \).

Teacher's Note:
1. Intensity goes as amplitude squared. Take the square root of 25 first.

Answer: (b) 252 Hz

Explanation:
1. With 4 beats, \( n_B=256\pm4=252\ \text{or}\ 260 \) Hz.
2. Loading with wax lowers the frequency of A. Beats decrease only if B is already below A, so \( n_B=252 \) Hz.

Teacher's Note:
1. Wax lowers the frequency of a fork. If the beats then fall, the loaded fork was above the other one.

Answer: (a) \( (\sqrt{I_1}+\sqrt{I_2})^2 \) and \( (\sqrt{I_1}-\sqrt{I_2})^2 \)

Explanation:
1. Intensity is proportional to amplitude squared. The amplitude fluctuates between \( a_1+a_2 \) and \( |a_1-a_2| \).
2. So the intensity swings between \( (\sqrt{I_1}+\sqrt{I_2})^2 \) and \( (\sqrt{I_1}-\sqrt{I_2})^2 \).

Teacher's Note:
1. Combine amplitudes first and then square. This is the beat pattern in terms of intensities.

Answer: (b) 1000 Hz

Explanation:
1. Moving the piston by d changes the path by 2d: \( 2\times8.75=17.5\ \text{cm} \).
2. Maximum to minimum needs a path change of \( \lambda/2 \): \( \lambda=35\ \text{cm} \).
3. \( n=v/\lambda=350/0.35=1000\ \text{Hz} \).

Teacher's Note:
1. The path change is twice the displacement of the piston because the sound goes to the piston and back. Maximum to minimum is half a wavelength.

Answer: (b) \( \dfrac{\phi+\pi/2}{2\pi}\lambda \)

Explanation:
1. \( y_2=a_2\sin(\omega t-kx+\phi+\pi/2) \), so the phase difference is \( \Delta\phi=\phi+\pi/2 \).
2. \( \Delta x=\dfrac{\lambda}{2\pi}\Delta\phi=\dfrac{\phi+\pi/2}{2\pi}\lambda \).

Teacher's Note:
1. Convert cosine into sine first. Path difference equals lambda over two pi times the phase difference.

Answer: (c) \( n_2\pm n \)

Explanation:
1. Beat frequency = difference of the frequencies: \( n=|n_1-n_2| \).
2. So \( n_1=n_2\pm n \).

Teacher's Note:
1. The beat frequency is the absolute difference. Without more information both signs are possible.

Answer: (c) \( 2I_0 \)

Explanation:
1. S2D = \( \sqrt{3^2+4^2}=5 \) m, so path difference = 5 - 4 = 1 m.
2. Phase difference \( =\dfrac{2\pi}{4}\times1=\dfrac\pi2 \).
3. \( I=I_0+I_0+2I_0\cos\dfrac\pi2=2I_0 \).

Teacher's Note:
1. Find the path difference from the right triangle. Then use the general interference formula.