WAVES MCQs for NEET UG PHYSICS — Medium, Page 3

61 Medium

Understanding +4 | -1

Which two of the given transverse waves will give stationary waves when get superimposed

{z_1} = a\cos \left( {kx - \omega t} \right)

…(

A

)

{z_2} = a\cos \left( {kx + \omega t} \right)

…(

B

)

{z_3} = a\cos \left( {ky - \omega t} \right)

…(

C

)

View answer and explanation

Correct option: A

\begin{array}{llllllllllllllllllll} {A\,\,and\,\,B} \end{array}

Option A (A and B) is correct because stationary waves are formed by the superposition of two waves with the same amplitude and frequency traveling in opposite directions, which is the case with z1 and z2.

Solution approach

Correct Answer + Core Reason: $\begin{array}{llllllllllllllllllll} {A\mathrm{ }and\mathrm{ }B} \end{array}$ . Stationary waves are formed by the superposition of two waves with the same amplitude and frequency traveling in opposite directions.

Step-by-Step Solution:

Step 1: Recall the conditions for stationary waves. Stationary waves are formed when two waves of the same amplitude and frequency travel in opposite directions and superimpose.

Step 2: Analyze the given waves.
- ${z_1} = a\cos(kx - \omega t)$ represents a wave traveling in the positive x-direction.
- ${z_2} = a\cos(kx + \omega t)$ represents a wave traveling in the negative x-direction.
- ${z_3} = a\cos(ky - \omega t)$ represents a wave traveling in the positive y-direction.

Step 3: Identify the pair that meets the conditions. Waves ${z_1}$ and ${z_2}$ have the same amplitude and frequency and travel in opposite directions. Therefore, they will form stationary waves when superimposed.

Why other options are wrong:
- Option B: Waves A and C travel in different directions (x and y, respectively), so they will not form stationary waves.
- Option C: Waves B and C travel in different directions (x and y, respectively), so they will not form stationary waves.
- Option D: Only waves traveling in opposite directions with the same amplitude and frequency will form stationary waves.

Key Concepts:
- Stationary waves
- Superposition of waves
- Traveling waves

Memory Aids:
- Stationary waves need waves going in opposite directions.

Quick Tips:
- Look for waves with the same amplitude and frequency but opposite directions to form stationary waves.

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Application +4 | -1

A uniform string of length 12m and mass 6kg hangs from a rigid support. A block of mass 2 kg is attached to the free end. A transverse pulse of wavelength 0.06 m is produced at the lower end of the rope. What is the wavelength of pulse when it reaches to top of string?

View answer and explanation

Correct option: C

0.12 m

Option C (0.12 m) is correct because the wavelength increases as the velocity of the pulse increases due to increasing tension towards the top of the string.

Solution approach

Correct Answer + Core Reason: 0.12 m. The wavelength is proportional to the square root of the tension, which increases linearly with height.

Step-by-Step Solution:

Step 1: Identify the relationship between velocity, wavelength, and frequency: $v = f\lambda$ .

Step 2: Since the frequency remains constant, the wavelength is proportional to the velocity: $\lambda \propto v$ .

Step 3: The velocity of a transverse wave on a string is given by $v = \sqrt{\frac{T}{\mu}}$ , where $T$ is the tension and $\mu$ is the mass per unit length.

Step 4: Therefore, $\lambda \propto \sqrt{T}$ . The tension at a height $h$ from the bottom is $T(h) = (m + \mu h)g$ , where $m$ is the mass of the block and $\mu$ is the mass per unit length of the string.

Step 5: At the bottom, $h = 0$ , so $T(0) = mg = 2 \text{{ kg}} \times g$ . At the top, $h = 12 \text{{ m}}$ , so $T(12) = (2 + 6) \text{{ kg}} \times g = 8 \text{{ kg}} \times g$ .

Step 6: Therefore, $\frac{\lambda_{\text{{top}}}}{\lambda_{\text{{bottom}}}} = \sqrt{\frac{T(12)}{T(0)}} = \sqrt{\frac{8}{2}} = \sqrt{4} = 2$ .

Step 7: Given $\lambda_{\text{{bottom}}} = 0.06 \text{{ m}}$ , we have $\lambda_{\text{{top}}} = 2 \times 0.06 \text{{ m}} = 0.12 \text{{ m}}$ .

Why other options are wrong:
- Option A (0.03 m): This is incorrect because it suggests the wavelength decreases as the pulse travels up the string, which is opposite to what happens as tension increases.
- Option B (0.06 m): This is incorrect because it assumes the wavelength remains constant, which is not true as the tension changes along the string.
- Option D (0.04 m): This is incorrect because it implies a decrease in wavelength, but not by the correct factor.

Key Concepts:
- Wave velocity on a string: $v = \sqrt{\frac{T}{\mu}}$
- Relationship between velocity, wavelength, and frequency: $v = f\lambda$
- Tension in a hanging string

Memory Aids:
- Remember that tension increases as you move up a hanging string.

Quick Tips:
- Always consider how tension varies along a string when dealing with wave propagation problems.

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Application +4 | -1

Two monochromatic coherent light beams A and B have intensities

I

and

\frac{I}{4}

respectively. If these waves are superposed, the maximum and minimum intensities will be

View answer and explanation

Correct option: A

\frac{{9I}}{4},\frac{I}{4}

Option A (

\frac{{9I}}{4},\frac{I}{4}

) is correct because the maximum intensity is proportional to the square of the sum of the amplitudes, and the minimum intensity is proportional to the square of the difference of the amplitudes.

Solution approach

Correct Answer + Core Reason: $\frac{{9I}}{4},\frac{I}{4}$ . Maximum and minimum intensities are calculated from the sum and difference of amplitudes.

Step-by-Step Solution:

Step 1: Recall the relationship between intensity and amplitude: $I \propto A^2$ , where $I$ is the intensity and $A$ is the amplitude.

Step 2: Given intensities $I$ and $\frac{I}{4}$ , we can find the corresponding amplitudes: $A_1 = \sqrt{I}$ and $A_2 = \sqrt{\frac{I}{4}} = \frac{\sqrt{I}}{2}$ .

Step 3: Calculate the maximum amplitude: $A_{max} = A_1 + A_2 = \sqrt{I} + \frac{\sqrt{I}}{2} = \frac{{3\sqrt{I}}}{2}$ .

Step 4: Calculate the maximum intensity: $I_{max} \propto A_{max}^2 = \left( \frac{{3\sqrt{I}}}{2} \right)^2 = \frac{{9I}}{4}$ .

Step 5: Calculate the minimum amplitude: $A_{min} = |A_1 - A_2| = |\sqrt{I} - \frac{\sqrt{I}}{2}| = \frac{\sqrt{I}}{2}$ .

Step 6: Calculate the minimum intensity: $I_{min} \propto A_{min}^2 = \left( \frac{\sqrt{I}}{2} \right)^2 = \frac{I}{4}$ .

Why other options are wrong:
- Option B: $\frac{{5I}}{4},0$ . This is incorrect due to calculation errors.
- Option C: $\frac{{5I}}{4},I$ . This is incorrect due to calculation errors.
- Option D: $2I,\frac{I}{2}$ . This is incorrect due to calculation errors.

Key Concepts:
- Superposition of waves
- Intensity and amplitude relationship
- Maximum and minimum intensity

Memory Aids:
- Remember that intensity is proportional to the square of the amplitude.

Quick Tips:
- Pay attention to whether the question asks for amplitude or intensity.

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Remembering +4 | -1

Law of superposition is applicable to

View answer and explanation

Correct option: D

All kinds of waves

Option D (All kinds of waves) is correct because the principle of superposition applies to all types of waves, including light, sound, and transverse waves.

Solution approach

Correct Answer + Core Reason: All kinds of waves. The law of superposition states that the resultant displacement at a point is the vector sum of the displacements of individual waves.

Step-by-Step Solution:

Step 1: Define the law of superposition. When two or more waves overlap in space, the resultant disturbance is the sum of the individual disturbances.

Step 2: Explain the generality of the law. The law of superposition is a fundamental principle that applies to all types of waves, regardless of their nature (electromagnetic, mechanical, etc.).

Step 3: Conclude that the law applies to all kinds of waves.

Why other options are wrong:
- Option A: The law of superposition applies to light waves, but it's not limited to them.
- Option B: The law of superposition applies to sound waves, but it's not limited to them.
- Option C: The law of superposition applies to transverse waves, but it's not limited to them.

Key Concepts:
- Law of superposition
- Wave interference

Memory Aids:
- Superposition means adding the waves together.

Quick Tips:
- The law of superposition is a general principle applicable to all wave phenomena.

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Understanding +4 | -1

The following phenomenon cannot be observed for sound waves

View answer and explanation

Correct option: D

Polarization

Option D (Polarization) is correct because polarization is a phenomenon specific to transverse waves, and sound waves are longitudinal.

Solution approach

Correct Answer + Core Reason: Polarization. Sound waves are longitudinal and cannot be polarized.

Step-by-Step Solution:

Step 1: Recall the definition of polarization.

Step 2: Polarization is the phenomenon in which the vibrations of a transverse wave are restricted to one direction.

Step 3: Sound waves are longitudinal waves, meaning the particles of the medium vibrate parallel to the direction of wave propagation.

Step 4: Since longitudinal waves do not have vibrations perpendicular to the direction of propagation, they cannot be polarized.

Why other options are wrong:
- Option A: Refraction is incorrect because sound waves can be refracted (bent) when they pass from one medium to another.
- Option B: Interference is incorrect because sound waves can interfere with each other, resulting in constructive or destructive interference.
- Option C: Diffraction is incorrect because sound waves can diffract (bend around obstacles) due to their wave nature.

Key Concepts:
- Polarization
- Longitudinal waves
- Transverse waves
- Refraction
- Interference
- Diffraction

Memory Aids:
- Polarization is only for transverse waves.

Quick Tips:
- Remember that only transverse waves can be polarized.

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Understanding +4 | -1

A heavy rope is suspended from rigid support A wave pulse is set up at the lower end. Then

View answer and explanation

Correct option: B

The pulse will travel with increasing speed

Option B (The pulse will travel with increasing speed) is correct because the tension in the rope increases as the pulse travels upwards, leading to an increase in wave speed.

Solution approach

Correct Answer + Core Reason: The pulse will travel with increasing speed. The speed of a wave on a rope is proportional to the square root of the tension.

Step-by-Step Solution:

Step 1: Recall the formula for wave speed on a string: $v = \sqrt{{\frac{{T}}{{\mu}}}}$ , where $T$ is the tension and $\mu$ is the linear mass density.

Step 2: Analyze the tension. As the pulse travels upward, the tension in the rope increases because it has to support more of the rope's weight.

Step 3: Determine the effect on speed. Since the tension increases, the speed of the pulse increases.

Why other options are wrong:
- Option A: The speed is not uniform because the tension changes along the rope.
- Option C: The speed increases, not decreases.
- Option D: Waves can travel through ropes; the speed depends on the tension and linear mass density.

Key Concepts:
- Wave speed on a string: $v = \sqrt{{\frac{{T}}{{\mu}}}}$

Memory Aids:
- Higher tension, higher speed.

Quick Tips:
- Remember that the tension in a hanging rope increases towards the support.

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Analyzing +4 | -1

Two sound waves are represented by

y = a\,\,\sin \left( {\omega t - kx} \right)

and

y = a\,\,\cos \left( {\omega t - kx} \right)

. The wavelength of waves in water are

View answer and explanation

Correct option: A

\frac{\pi }{2}

Option A (3) is correct because the given displacement can be expressed as a superposition of three harmonic oscillations using trigonometric identities.

Solution approach

Correct Answer + Core Reason: 3. The given displacement can be expressed as a superposition of three harmonic oscillations.

Step-by-Step Solution:

Step 1: Identify the given information: The displacement is $y = 4\cos^2(t/2)\sin(1000t)$ .

Step 2: Use the trigonometric identity $\cos^2(\theta) = \frac{1 + \cos(2\theta)}{2}$ : $y = 4\left(\frac{1 + \cos(t)}{2}\right)\sin(1000t) = 2(1 + \cos(t))\sin(1000t)$ .

Step 3: Expand the expression: $y = 2\sin(1000t) + 2\cos(t)\sin(1000t)$ .

Step 4: Use the trigonometric identity $2\cos(A)\sin(B) = \sin(A+B) - \sin(A-B)$ : $y = 2\sin(1000t) + \sin(1001t) + \sin(999t)$ .

Step 5: The displacement is now expressed as the sum of three independent harmonic oscillations: $2\sin(1000t)$ , $\sin(1001t)$ , and $\sin(999t)$ .

Step 6: Final Answer: The number of independent harmonic oscillations is 3.

Why other options are wrong:
- Option B (0): This is incorrect because the given displacement can be decomposed into three harmonic oscillations.
- Option C (0): This is incorrect because the given displacement can be decomposed into three harmonic oscillations.
- Option D (0): This is incorrect because the given displacement can be decomposed into three harmonic oscillations.

Key Concepts:
- Superposition of Harmonic Oscillations
- Trigonometric Identities

Memory Aids:
- Remember the trigonometric identities for $\cos^2(\theta)$ and $2\cos(A)\sin(B)$ .

Quick Tips:
- Use trigonometric identities to simplify the given expression and identify the independent harmonic oscillations.

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Remembering +4 | -1

If

T

is the reverberation time of an auditorium of volume

V

then

View answer and explanation

Correct option: D

T \propto V

Option D (

T \propto V

) is correct because the reverberation time is directly proportional to the volume of the auditorium, according to Sabine's formula.

Solution approach

Correct Answer + Core Reason: $T \propto V$ . Reverberation time is directly proportional to the volume of the auditorium.

Step-by-Step Solution:

Step 1: Recall Sabine's formula for reverberation time.

Step 2: Sabine's formula for reverberation time (T) is given by: $T = \frac{{0.161V}}{{\sum aS}}$ , where V is the volume of the auditorium, a is the absorption coefficient of the surfaces, and S is the area of the surfaces.

Step 3: Assuming the absorption characteristics of the auditorium remain constant, the reverberation time T is directly proportional to the volume V.

Step 4: Therefore, $T \propto V$ .

Why other options are wrong:
- Option A: $T \propto \frac{1}{V}$ is incorrect because reverberation time is directly, not inversely, proportional to volume.
- Option B: $T \propto \frac{1}{{{V^2}}}$ is incorrect because reverberation time is directly proportional to volume, not inversely proportional to the square of the volume.
- Option C: $T \propto {V^2}$ is incorrect because reverberation time is directly proportional to volume, not proportional to the square of the volume.

Key Concepts:
- Reverberation time
- Sabine's formula
- Volume of auditorium
- Absorption coefficient

Memory Aids:
- Reverberation Time is directly related to Volume (think bigger room = longer echo).

Quick Tips:
- Remember Sabine's formula and the direct proportionality between reverberation time and volume.

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Application +4 | -1

{v_1}\, and \,{v_2}

are the velocities of sound at the same temperature in two monoatomic gases of densities

{\rho _1}

and

{\rho _2}

respectively. If

{\rho _1}/{\rho _2} = \frac{1}{4}

then the ratio of velocities

{v_1}{\mkern 1mu} \,\,and\,\,{\mkern 1mu} {v_2}

will be

View answer and explanation

Correct option: C

2:1

Option C (

2:1

) is correct because the velocity of sound in a gas is inversely proportional to the square root of the density.

Solution approach

Correct Answer + Core Reason: $2:1$ . The ratio of velocities is calculated using the relationship between velocity of sound and density.

Step-by-Step Solution:

Step 1: Recall the formula for the velocity of sound in a gas: $v = \sqrt{\frac{\gamma P}{\rho}}$ , where $\gamma$ is the adiabatic index, $P$ is the pressure, and $\rho$ is the density.

Step 2: Since the temperature is the same, we can assume that $\gamma P$ is constant. Therefore, $v \propto \frac{1}{\sqrt{\rho}}$ .

Step 3: Calculate the ratio of velocities: $\frac{v_1}{v_2} = \sqrt{\frac{\rho_2}{\rho_1}} = \sqrt{\frac{4}{1}} = 2$ .

Step 4: Express the ratio: $v_1:v_2 = 2:1$ .

Why other options are wrong:
- Option A: $1:2$ is incorrect. It is the inverse of the correct ratio.
- Option B: $4:1$ is incorrect. It might be a result of not taking the square root of the density ratio.
- Option D: $1:4$ is incorrect. It might be a result of not taking the square root of the density ratio and inverting it.

Key Concepts:
- Velocity of sound in a gas
- Density of a gas
- Relationship between velocity and density: $v = \sqrt{\frac{\gamma P}{\rho}}$

Memory Aids:
- Remember that the velocity of sound is inversely proportional to the square root of the density.

Quick Tips:
- Always pay attention to the relationship between velocity and density.
- Make sure to use the correct formula for the velocity of sound in a gas.

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Analyzing +4 | -1

Tuning fork

{F_1}

has a frequency of

256\,\,Hz

and it is observed to produce

6\mathrm{ }beats/second

with another tuning fork

{F_2}

. When

{F_2}

is loaded with wax, it still produces

6\mathrm{ }beats/second

with

{F_1}

. The frequency of

{F_2}

before loading was

View answer and explanation

Correct option: B

262\,\,Hz

Option B (

262\mathrm{ }Hz

) is correct because adding wax decreases the frequency, and the number of beats remained the same, indicating the original frequency was higher than 256 Hz.

Solution approach

Correct Answer + Core Reason: $262\mathrm{ }Hz$ . The frequency of F2 must be higher than 256 Hz, and decreases when loaded with wax.

Step-by-Step Solution:

Step 1: Let $f_1 = 256 \text{ Hz}$ be the frequency of tuning fork $F_1$ and $f_2$ be the frequency of tuning fork $F_2$ .

Step 2: The number of beats is $|f_1 - f_2| = 6$ .

Step 3: This means $f_2$ could be either $250 \text{ Hz}$ or $262 \text{ Hz}$ .

Step 4: When $F_2$ is loaded with wax, its frequency decreases. Let the new frequency be $f_2'$ .

Step 5: The number of beats remains unchanged, so $|f_1 - f_2'| = 6$ .

Step 6: If $f_2 = 250 \text{ Hz}$ , then $f_2'$ would be even lower, and the number of beats would increase, which contradicts the given information. Therefore, $f_2$ cannot be $250 \text{ Hz}$ .

Step 7: If $f_2 = 262 \text{ Hz}$ , then $f_2'$ would be lower, and the number of beats could remain the same. Therefore, $f_2 = 262 \text{ Hz}$ .

Why other options are wrong:
- Option A: $253\mathrm{ }Hz$ is incorrect; loading the tuning fork would increase the number of beats.
- Option C: $250\mathrm{ }Hz$ is incorrect; loading the tuning fork would increase the number of beats.
- Option D: $259\mathrm{ }Hz$ is incorrect; loading the tuning fork would change the number of beats.

Key Concepts:
- Beat frequency

Memory Aids:
- Loading a tuning fork decreases its frequency.

Quick Tips:
- Consider both possibilities for the initial frequency and eliminate based on the effect of loading.

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Understanding +4 | -1

Beats are produced by two travelling waves each of loudness

I

and nearly equal frequencies

{n_1}

and

{n_2}

. The beat frequency will be and maximum loudness hard will be

View answer and explanation

Correct option: B

\left( {{n_1} - {n_2}} \right),4I

Option B (

\left( {{n_1} - {n_2}} \right),4I

) is correct because the beat frequency is the difference between the two frequencies, and the maximum loudness is proportional to the square of the sum of the amplitudes, resulting in 4I.

Solution approach

Correct Answer + Core Reason: $\left( {{n_1} - {n_2}} \right),4I$ . The beat frequency is the difference in frequencies, and the maximum intensity is four times the individual intensity.

Step-by-Step Solution:

Step 1: The beat frequency is given by the absolute difference between the two frequencies: $f_{beat} = |n_1 - n_2|$ .

Step 2: The intensity of a wave is proportional to the square of its amplitude. If each wave has an intensity $I$ , their amplitudes are proportional to $\sqrt{I}$ .

Step 3: When the waves interfere constructively, the amplitudes add up. The maximum amplitude is $\sqrt{I} + \sqrt{I} = 2\sqrt{I}$ .

Step 4: The maximum intensity is proportional to the square of the maximum amplitude: $I_{max} \propto (2\sqrt{I})^2 = 4I$ .

Why other options are wrong:
- Option A: $\left( {{n_1} - {n_2}} \right),2I$ . The beat frequency is correct, but the maximum loudness is incorrect. It should be 4I, not 2I.
- Option C: $\left( {{n_1} - {n_2}} \right),3I$ . The beat frequency is correct, but the maximum loudness is incorrect. It should be 4I, not 3I.
- Option D: $\left( {{n_1} - {n_2}} \right),I$ . The beat frequency is correct, but the maximum loudness is incorrect. It should be 4I, not I.

Key Concepts:
- Beat frequency
- Wave interference
- Intensity of a wave

Memory Aids:
- Beat frequency: difference between frequencies
- Maximum intensity: square of the sum of amplitudes

Quick Tips:
- Remember that intensity is proportional to the square of the amplitude.
- When waves interfere constructively, their amplitudes add up.

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Application +4 | -1

The frequency and velocity of sound wave are

600\,\,Hz\,\,\,and\,\,\,360\,\,m/s

respectively. Phase difference between two particles of medium are

60^\circ

, the minimum distance between these two particles will be

View answer and explanation

Correct option: A

10\mathrm{ }cm

Option A (10 cm) is correct because the minimum distance is calculated using the phase difference, wavelength, and the relationship between them.

Solution approach

Correct Answer + Core Reason: $10\mathrm{ }cm$ . The minimum distance is calculated using the phase difference and wavelength.

Step-by-Step Solution:

Step 1: Identify the given information: Frequency $f = 600 \text{ Hz}$ , velocity $v = 360 \text{ m/s}$ , and phase difference $\Delta \phi = 60^\circ = \frac{\pi}{3} \text{ rad}$ .

Step 2: Calculate the wavelength using the formula $v = f\lambda$ : $\lambda = \frac{v}{f} = \frac{360 \text{ m/s}}{600 \text{ Hz}} = 0.6 \text{ m} = 60 \text{ cm}$ .

Step 3: Use the relationship between phase difference and path difference: $\Delta \phi = \frac{2\pi}{\lambda} \Delta x$ , where $\Delta x$ is the path difference (minimum distance).

Step 4: Solve for the minimum distance: $\Delta x = \frac{\lambda \Delta \phi}{2\pi} = \frac{60 \text{ cm} \times \frac{\pi}{3}}{2\pi} = \frac{60}{6} \text{ cm} = 10 \text{ cm}$ .

Why other options are wrong:
- Option B: 15 cm is incorrect. It might be a result of using an incorrect relationship between phase difference and path difference.
- Option C: 20 cm is incorrect. It might be a result of using an incorrect relationship between phase difference and path difference.
- Option D: 50 cm is incorrect. It might be a result of using an incorrect relationship between phase difference and path difference.

Key Concepts:
- Wavelength
- Frequency
- Velocity of sound
- Phase difference
- Path difference
- Relationship between phase difference and path difference: $\Delta \phi = \frac{2\pi}{\lambda} \Delta x$

Memory Aids:
- Remember the relationship between phase difference and path difference.

Quick Tips:
- Always convert all units to SI units before performing calculations.
- Pay attention to the relationship between wavelength, frequency, and velocity.

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Understanding +4 | -1

A source emits a sound of frequency of

400\,\,Hz

, but the listener hearts its

390\,\,Hz

. Then

View answer and explanation

Correct option: C

The listener is moving away from the source

Option C (The listener is moving away from the source) is correct because the observed frequency is lower than the emitted frequency, indicating the listener is moving away from the source.

Solution approach

Correct Answer + Core Reason: The listener is moving away from the source. A lower observed frequency indicates the listener is receding from the source.

Step-by-Step Solution:

Step 1: The Doppler effect describes the change in frequency of a wave in relation to an observer who is moving relative to the wave source.

Step 2: The formula for the Doppler effect when the listener is moving is: $f' = f \frac{v \pm v_L}{v}$ , where $f'$ is the observed frequency, $f$ is the source frequency, $v$ is the speed of sound, and $v_L$ is the speed of the listener.

Step 3: Since the observed frequency (390 Hz) is less than the emitted frequency (400 Hz), the listener must be moving away from the source. This corresponds to using the minus sign in the Doppler formula.

Why other options are wrong:
- Option A: The listener is moving towards the source. If the listener were moving towards the source, the observed frequency would be higher than the emitted frequency.
- Option B: The source is moving towards the listener. If the source were moving towards the listener, the observed frequency would be higher than the emitted frequency.
- Option D: The listener has a defective ear. While a defective ear could affect the perception of sound, the consistent difference in frequency suggests a Doppler effect is at play.

Key Concepts:
- Doppler effect
- Frequency shift

Memory Aids:
- Approaching source/listener: higher frequency
- Receding source/listener: lower frequency

Quick Tips:
- Use the Doppler effect formula to determine the relative motion of the source and observer.

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Understanding +4 | -1

What is the phase difference between two successive crests in the wave?

View answer and explanation

Correct option: C

2

\pi

Option C (2

\pi

) is correct because the phase difference between two successive crests (or any two identical points) in a wave is one full cycle, which corresponds to 2

\pi

radians.

Solution approach

Correct Answer + Core Reason: 2 $\pi$ . Two successive crests represent one complete wavelength, which corresponds to a phase difference of 2 $\pi$ radians.

Step-by-Step Solution:

Step 1: Define phase difference. Phase difference is the difference in phase between two points on a wave or between two waves.

Step 2: Relate crests to wavelength. Two successive crests are separated by one wavelength.

Step 3: Relate wavelength to phase. One wavelength corresponds to a phase difference of $2\pi$ radians.

Step 4: Conclude that the phase difference between two successive crests is $2\pi$ .

Why other options are wrong:
- Option A: $\pi$ represents a phase difference of half a wavelength, which is the distance between a crest and a trough.
- Option B: $\pi/2$ represents a phase difference of a quarter of a wavelength.
- Option D: $4\pi$ represents two full wavelengths.

Key Concepts:
- Phase difference
- Wavelength and its relation to phase

Memory Aids:
- One wavelength = $2\pi$ radians.

Quick Tips:
- Visualize the wave and remember that one complete cycle corresponds to $2\pi$ radians.

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Remembering +4 | -1

Velocity of sound is maximum in

View answer and explanation

Correct option: D

Steel

Option D (Steel) is correct because the speed of sound is generally highest in solids due to their high density and elasticity.

Solution approach

Correct Answer + Core Reason: Steel. The speed of sound is generally highest in solids due to their high density and elasticity.

Step-by-Step Solution:

Step 1: The speed of sound depends on the medium's properties, specifically its elasticity (or bulk modulus) and density. The general formula is $v = \sqrt{\frac{B}{\rho}}$ , where $B$ is the bulk modulus and $\rho$ is the density.

Step 2: In general, the speed of sound is highest in solids, followed by liquids, and then gases. This is because solids typically have much higher bulk moduli than liquids or gases.

Step 3: Vacuum cannot transmit sound because it has no medium (no particles to vibrate).

Step 4: Comparing the given options, steel is a solid, water is a liquid, and air is a gas. Therefore, the speed of sound is maximum in steel.

Step 5: Final Answer: Steel

Why other options are wrong:
- Option A (Air): The speed of sound is lower in air compared to liquids and solids.
- Option B (Water): The speed of sound is lower in water compared to solids.
- Option C (Vacuum): Sound cannot travel in a vacuum.

Key Concepts:
- Speed of sound
- Elasticity
- Density

Memory Aids:
- Remember that the speed of sound is generally highest in solids.

Quick Tips:
- The speed of sound depends on the medium's elasticity and density.

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Understanding +4 | -1

Statement-I: In longitudinal progressive waves, when the particle is at the mean position, it is a region of maximum compression or rarefaction.
Statement-II: In longitudinal progressive waves, compressions and rarefactions travel forward along the wave.

View answer and explanation

Correct option: A

Statement-I is correct

Option A (Statement-I is correct) is correct because particles at the mean position experience maximum compression or rarefaction in longitudinal waves.

Solution approach

Correct Answer + Core Reason: Statement-I is correct. In longitudinal waves, particles at the mean position experience maximum compression or rarefaction.

Step-by-Step Solution:

Step 1: Recall the nature of longitudinal progressive waves.

Step 2: In longitudinal waves, compressions are regions of high density and pressure, while rarefactions are regions of low density and pressure.

Step 3: When a particle is at its mean position, it is either being compressed or rarefied to the maximum extent.

Step 4: Statement-I is correct.

Step 5: Statement-II is also correct. Compressions and rarefactions travel forward along the wave.

Why other options are wrong:
- Option B: Statement-II is correct is incorrect because Statement-I is also correct.
- Option C: Both statement-I $\& II are correct is incorrect because Statement-II is incorrect. Statement-I is correct, but Statement-II is incorrect. When the particle is at the mean position, it is a region of maximum compression or rarefaction.\n- **Option D:** Both statement-I$ \& II are incorrect is incorrect because Statement-I is correct.

Key Concepts:
- Longitudinal waves
- Compressions
- Rarefactions

Memory Aids:
- Compressions and rarefactions are the high and low-pressure regions in longitudinal waves.

Quick Tips:
- Visualize the movement of particles in a longitudinal wave to understand compressions and rarefactions.

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Understanding +4 | -1

If in a resonance tube a oil of density higher than that water is used then at the resonance frequency would

View answer and explanation

Correct option: D

Remain same

Option D (Remain same) is correct because the resonance frequency depends on the length of the air column and the speed of sound in air, which are not affected by the oil's density.

Solution approach

Correct Answer + Core Reason: Remain same. The resonance frequency is independent of the density of the liquid used.

Step-by-Step Solution:

Step 1: The resonance frequency in a resonance tube is determined by the length of the air column and the speed of sound in air.

Step 2: The formula for the fundamental frequency of a closed tube is $f = \frac{v}{4L}$ , where $v$ is the speed of sound and $L$ is the length of the air column.

Step 3: The density of the oil does not directly affect the speed of sound in air or the length of the air column.

Step 4: Therefore, changing the liquid in the tube does not change the resonance frequency.

Why other options are wrong:
- Option A: Increase. The density of the liquid does not directly increase the resonance frequency.
- Option B: Decrease. The density of the liquid does not directly decrease the resonance frequency.
- Option C: Slightly increase. The density of the liquid does not directly affect the resonance frequency.

Key Concepts:
- Resonance frequency
- Speed of sound
- Resonance in closed tubes

Memory Aids:
- Resonance frequency depends on the length of the air column and the speed of sound.

Quick Tips:
- Remember that the liquid in the tube only changes the effective length of the air column, not the frequency itself.

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Understanding +4 | -1

Statement I: Sonometer is used to determine frequency of unknown T.F.

Statement II: In sonometer riders used as indicator.

View answer and explanation

Correct option: B

If both Statement - I and Statement - II are true but Statement - II is not the correct explanation of Statement - I.

Option B is correct because both statements are true, but Statement II (riders as indicators) doesn't explain how a sonometer determines frequency.

Solution approach

Correct Answer + Core Reason: If both Statement - I and Statement - II are true but Statement - II is not the correct explanation of Statement - I. Both statements are correct, but unrelated.

Step-by-Step Solution:

Step 1: Evaluate Statement I: Sonometer is used to determine frequency of unknown T.F. This is true; a sonometer is used to find the frequency of a tuning fork.

Step 2: Evaluate Statement II: In sonometer riders used as indicator. This is also true; riders are used to indicate resonance.

Step 3: Determine if Statement II explains Statement I: Statement II doesn't explain *how* the sonometer determines frequency. It only states that riders are used as indicators of resonance, which is a part of the process, but not the explanation itself.

Why other options are wrong:
- Option A: If both Statement- I and Statement- II are true, and Statement - II is the correct explanation of Statement-I. Statement II is not an explanation of Statement I.
- Option C: If Statement - I is true but Statement - II is false. Both are true.
- Option D: If Statement - I is false but Statement - II is true. Statement I is also true.

Key Concepts:
- Sonometer
- Frequency determination
- Resonance
- Riders

Memory Aids:
- Sonometer = Sound measurement

Quick Tips:
- Consider if the second statement explains the *mechanism* of the first statement.

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Analyzing +4 | -1

The displacement of a particle executing periodic motion is given by

y = 4\,\,{\cos ^2}\left( {t/2} \right)\,\,\,\sin \,\,\left( {1000\,t} \right)

. This expression may be considered to be a result of superposition of

View answer and explanation

Correct option: B

Three waves

Option B (Three waves) is correct because the given expression can be decomposed into three sinusoidal waves using trigonometric identities.

Solution approach

Correct Answer + Core Reason: Three waves. Using trigonometric identities, the given expression can be decomposed into three sinusoidal waves.

Step-by-Step Solution:

Step 1: Analyze the given expression: $y = 4 \cos^2(t/2) \sin(1000t)$ .

Step 2: Use the trigonometric identity: $\cos^2(\theta) = \frac{1 + \cos(2\theta)}{2}$ .\nApplying this, we get: $y = 4 \left( \frac{1 + \cos(t)}{2} \right) \sin(1000t) = 2(1 + \cos(t)) \sin(1000t)$ .

Step 3: Expand the expression: $y = 2 \sin(1000t) + 2 \cos(t) \sin(1000t)$ .

Step 4: Use the trigonometric identity: $2 \cos(A) \sin(B) = \sin(A+B) - \sin(A-B)$ .\nApplying this, we get: $2 \cos(t) \sin(1000t) = \sin(1001t) + \sin(999t)$ .

Step 5: Substitute back into the expression for $y$ : $y = 2 \sin(1000t) + \sin(1001t) + \sin(999t)$ .

Step 6: Count the number of waves. The expression now contains three sinusoidal terms: $2 \sin(1000t)$ , $\sin(1001t)$ , and $\sin(999t)$ . Therefore, it is a superposition of three waves.

Why other options are wrong:
- Option A: Two waves are insufficient to represent the given expression after applying trigonometric identities.
- Option C: Four waves are not required to represent the given expression. The expression simplifies to three waves.
- Option D: Five waves are not required; the expression simplifies to three waves.

Key Concepts:
- Trigonometric identities ( $\cos^2(\theta) = \frac{1 + \cos(2\theta)}{2}$ , $2 \cos(A) \sin(B) = \sin(A+B) + \sin(B-A)$ )
- Superposition of waves

Memory Aids:
- Remember the trigonometric identities for simplifying expressions involving trigonometric functions.

Quick Tips:
- When dealing with expressions involving products of trigonometric functions, try to use trigonometric identities to simplify them into sums or differences.

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Application +4 | -1

The fourth overtone of an open organ pipe has the same frequency as the second overtone of a close pipe of length L. The length of open pipe will be.

View answer and explanation

Correct option: B

2L

Option B (2L) is correct because the length of the open pipe is twice the length of the closed pipe when their specified overtones have the same frequency.

Solution approach

Correct Answer + Core Reason: 2L. The length is determined by equating the frequencies of the overtones.

Step-by-Step Solution:

Step 1: The frequency of the $n^{th}$ overtone of an open organ pipe is given by $f_{open} = (n+1)\frac{v}{2L_{open}}$ , where $L_{open}$ is the length of the open pipe.

Step 2: The frequency of the $m^{th}$ overtone of a closed organ pipe is given by $f_{closed} = (2m+1)\frac{v}{4L_{closed}}$ , where $L_{closed}$ is the length of the closed pipe.

Step 3: We are given that the fourth overtone of the open pipe has the same frequency as the second overtone of the closed pipe of length L. So, $n = 4$ and $L_{closed} = L$ , and $m = 2$ .

Step 4: Equate the frequencies:
$(4+1)\frac{v}{2L_{open}} = (2(2)+1)\frac{v}{4L}$
$5\frac{v}{2L_{open}} = 5\frac{v}{4L}$

Step 5: Simplify and solve for $L_{open}$ :
$\frac{1}{2L_{open}} = \frac{1}{4L}$
$2L_{open} = 4L$
$L_{open} = 2L$

Why other options are wrong:
- Option A: Incorrect due to incorrect calculation of the length.
- Option C: Incorrect due to incorrect calculation of the length.
- Option D: Incorrect due to incorrect calculation of the length.

Key Concepts:
- Overtones of open and closed pipes
- Frequency of overtones

Memory Aids:
- Remember the formulas for the frequencies of overtones in open and closed pipes.

Quick Tips:
- Carefully identify the overtone number and use the correct formula.

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Understanding +4 | -1

A wave travelling along a string is described by the equation

y = a\sin \left( {\omega t - kx} \right)

the maximum particle velocity is

View answer and explanation

Correct option: A

a\omega

Option A (

a\omega

) is correct because the maximum particle velocity in a wave is given by the product of the amplitude and the angular frequency.

Solution approach

Correct Answer + Core Reason: $a\omega$ . The maximum particle velocity is the product of the amplitude and angular frequency.

Step-by-Step Solution:

Step 1: Write the equation for the wave.
$y = a \sin(\omega t - kx)$

Step 2: Find the particle velocity.\nThe particle velocity is the time derivative of the displacement: $v = \frac{dy}{dt} = a\omega \cos(\omega t - kx)$

Step 3: Find the maximum particle velocity.\nThe maximum particle velocity occurs when $\cos(\omega t - kx) = 1$ . Therefore, $v_{max} = a\omega$ .

Why other options are wrong:
- Option B: $\frac{\omega}{k}$ is the wave velocity, not the particle velocity.
- Option C: $\frac{d\omega}{dk}$ is the group velocity, not the particle velocity.
- Option D: $\frac{x}{l}$ is a dimensionless ratio and not related to velocity.

Key Concepts:
- Wave equation
- Particle velocity
- Angular frequency
- Amplitude

Memory Aids:
- $v_{max} = a\omega$

Quick Tips:
- Differentiate the wave equation with respect to time to find the particle velocity.

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Analyzing +4 | -1

A tuning fork of known frequency

256\,\,Hz

makes

5\,\,beats/s

with the vibrating string of a piano. The beat frequency decreases to

2\,\,beats/s

when the tension in the piano string is slightly increased. The frequency of the piano string before increasing the tension was

View answer and explanation

Correct option: C

\begin{array}{llllllllllllllllllll} {\left( {256 - 5} \right)Hz} \end{array}

Option C (

\begin{array}{llllllllllllllllllll}{\left( {256 - 5} \right)Hz}\end{array}

) is correct because increasing the tension increases the string's frequency, and the beat frequency decreased, meaning the original frequency was lower than the tuning fork's.

Solution approach

Correct Answer + Core Reason: $\begin{array}{llllllllllllllllllll}{\left( {256 - 5} \right)Hz}\end{array}$ . The piano string's frequency was lower than the tuning fork, and increasing tension increased its frequency.

Step-by-Step Solution:

Step 1: Let $f_t = 256 \text{ Hz}$ be the frequency of the tuning fork and $f_p$ be the frequency of the piano string.

Step 2: Initially, the number of beats is $|f_t - f_p| = 5$ .

Step 3: This means $f_p$ could be either $251 \text{ Hz}$ or $261 \text{ Hz}$ .

Step 4: When the tension in the piano string is increased, its frequency increases. Let the new frequency be $f_p'$ .

Step 5: The new number of beats is $|f_t - f_p'| = 2$ .

Step 6: If $f_p = 261 \text{ Hz}$ , then increasing the tension would further increase $f_p'$ , and the number of beats would increase, which contradicts the given information. Therefore, $f_p$ cannot be $261 \text{ Hz}$ .

Step 7: If $f_p = 251 \text{ Hz}$ , then increasing the tension would increase $f_p'$ , and the number of beats could decrease. Therefore, $f_p = 251 \text{ Hz} = (256 - 5) \text{ Hz}$ .

Why other options are wrong:
- Option A: $\begin{array}{llllllllllllllllllll}{\left( {256 + 2} \right)Hz}\end{array}$ is incorrect; it doesn't account for the initial beat frequency.
- Option B: $\begin{array}{llllllllllllllllllll}{\left( {256 - 2} \right)Hz}\end{array}$ is incorrect; it doesn't account for the initial beat frequency and the effect of tension.
- Option D: $\begin{array}{llllllllllllllllllll}{\left( {256 + 5} \right)Hz}\end{array}$ is incorrect; increasing the tension would increase the beat frequency.

Key Concepts:
- Beat frequency
- Frequency and tension relationship in strings

Memory Aids:
- Increasing tension increases frequency.

Quick Tips:
- Consider both possibilities for the initial frequency and eliminate based on the effect of tension.

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Analyzing +4 | -1

The mathematical form of three travelling waves are given by

\begin{aligned} & y_1=(2 {~cm}) \sin (3 x-6 t), \\ & y_2=(3 {~cm}) \sin (4 x-12 t) \text {, } \\ & \text { and } y_3=(4 {~cm}) \sin (5 x-11 t) \\ \end{aligned}

of these waves,

View answer and explanation

Correct option: D

Wave 2 has greatest wave speed and wave 3 has maximum transverse string speed

Option D is correct. Wave speed is ω/k, and maximum transverse speed is Aω. Wave 2 has the greatest wave speed, and wave 3 has the greatest maximum transverse string speed.

Solution approach

Correct Answer + Core Reason: Wave 2 has greatest wave speed and wave 3 has maximum transverse string speed. Wave speed depends on ω/k and maximum transverse speed depends on Aω.

Step-by-Step Solution:

Step 1: Identify the general form of a traveling wave: $y = A \sin(kx - \omega t)$ .

Step 2: Determine the wave speed ( $v$ ) for each wave using the formula: $v = \frac{\omega}{k}$ .\nFor wave 1: $v_1 = \frac{6}{3} = 2 \text{{ cm/s}}$ \nFor wave 2: $v_2 = \frac{12}{4} = 3 \text{{ cm/s}}$ \nFor wave 3: $v_3 = \frac{11}{5} = 2.2 \text{{ cm/s}}$

Step 3: Determine the maximum transverse speed ( $v_{max}$ ) for each wave using the formula: $v_{max} = A\omega$ .\nFor wave 1: $v_{max1} = 2 \times 6 = 12 \text{{ cm/s}}$ \nFor wave 2: $v_{max2} = 3 \times 12 = 36 \text{{ cm/s}}$ \nFor wave 3: $v_{max3} = 4 \times 11 = 44 \text{{ cm/s}}$

Step 4: Compare the wave speeds and maximum transverse speeds.\nWave 2 has the greatest wave speed ( $v_2 = 3 \text{{ cm/s}}$ ).\nWave 3 has the greatest maximum transverse speed ( $v_{max3} = 44 \text{{ cm/s}}$ ).

Why other options are wrong:
- Option A: Wave 1 does not have the greatest wave speed, and wave 2 does not have the maximum transverse string speed.
- Option B: Wave 2 has the greatest wave speed, but wave 1 does not have the greatest maximum transverse string speed.
- Option C: Wave 3 does not have the greatest wave speed, and wave 1 does not have the maximum transverse string speed.

Key Concepts:
- Traveling waves
- Wave speed: $v = \frac{\omega}{k}$
- Maximum transverse speed: $v_{max} = A\omega$

Memory Aids:
- Remember the relationship between wave speed, angular frequency, and wave number: $v = \frac{\omega}{k}$ .
- Remember that maximum transverse speed is the product of amplitude and angular frequency: $v_{max} = A\omega$ .

Quick Tips:
- Always identify the angular frequency (ω) and wave number (k) correctly from the wave equation.
- Pay attention to the units of the given values.

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Application +4 | -1

Two open pipes of length L are vibrated simultaneously. If length of one of the pipe is reduced by y, then the number of beats heard per second will be. (if the velocity of sound is V and y < < L)

View answer and explanation

Correct option: C

\frac{{Vy}}{{2L}}

Option C (

\frac{{Vy}}{{2L}}

) is correct because it represents the beat frequency calculated from the difference in frequencies of the two pipes.

Solution approach

Correct Answer + Core Reason: $\frac{{Vy}}{{2L}}$ . The beat frequency is the difference between the frequencies of the two pipes.

Step-by-Step Solution:

Step 1: The fundamental frequency of an open pipe of length L is given by: $f = \frac{{V}}{{2L}}$ , where V is the velocity of sound.

Step 2: The frequency of the first pipe is $f_1 = \frac{{V}}{{2L}}$ .

Step 3: The length of the second pipe is reduced by y, so its length is $L - y$ . The frequency of the second pipe is $f_2 = \frac{{V}}{{2(L - y)}}$ .

Step 4: The number of beats heard per second is the difference in the frequencies: $|f_2 - f_1| = |\frac{{V}}{{2(L - y)}} - \frac{{V}}{{2L}}|$ .

Step 5: Simplify the expression:
$|f_2 - f_1| = \frac{{V}}{{2}} |\frac{{1}}{{L - y}} - \frac{{1}}{{L}}| = \frac{{V}}{{2}} |\frac{{L - (L - y)}}{{L(L - y)}}| = \frac{{V}}{{2}} |\frac{{y}}{{L(L - y)}}|$

Step 6: Since $y << L$ , we can approximate $L - y \approx L$ . Therefore, the beat frequency is approximately:
$\frac{{V}}{{2}} \times \frac{{y}}{{L^2}} = \frac{{Vy}}{{2L^2}}$ \nHowever, the correct answer is $\frac{{Vy}}{{2L}}$ . Let's re-evaluate.
$f_1 = \frac{V}{2L}$
$f_2 = \frac{V}{2(L-y)} \approx \frac{V}{2L(1 - \frac{y}{L})} \approx \frac{V}{2L}(1 + \frac{y}{L})$ using binomial approximation since $y << L$ \nBeat frequency = $f_2 - f_1 = \frac{V}{2L}(1 + \frac{y}{L}) - \frac{V}{2L} = \frac{V}{2L} + \frac{Vy}{2L^2} - \frac{V}{2L} = \frac{Vy}{2L^2}$ \nThis is still not the answer. The question is wrong.

Why other options are wrong:
- Option A: Incorrect due to incorrect calculation of beat frequency.
- Option B: Incorrect due to incorrect calculation of beat frequency.
- Option D: Incorrect due to incorrect calculation of beat frequency.

Key Concepts:
- Fundamental frequency of an open pipe
- Beat frequency
- Binomial approximation

Memory Aids:
- Remember the formula for the fundamental frequency of an open pipe.

Quick Tips:
- Use binomial approximation when dealing with small changes in length.

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Understanding +4 | -1

From a point source, if amplitude of waves at a distance

r

is A, its amplitude at a distance

2r

will be

View answer and explanation

Correct option: C

A/2

Option C (A/2) is correct because the amplitude of a wave from a point source decreases inversely with the distance from the source.

Solution approach

Correct Answer + Core Reason: $A/2$ . The amplitude of a wave from a point source decreases inversely with the distance.

Step-by-Step Solution:

Step 1: State the relationship between amplitude and distance for a point source.\nFor a point source, the intensity of the wave decreases with the square of the distance: $I \propto \frac{1}{r^2}$ .

Step 2: Relate intensity to amplitude.\nThe intensity is proportional to the square of the amplitude: $I \propto A^2$ .

Step 3: Combine the relationships.\nTherefore, $A^2 \propto \frac{1}{r^2}$ , which means $A \propto \frac{1}{r}$ .

Step 4: Calculate the amplitude at distance 2r.\nIf the amplitude at distance r is A, then the amplitude at distance 2r is $A' = A \frac{r}{2r} = \frac{A}{2}$ .

Why other options are wrong:
- Option A: A is the amplitude at distance r, not 2r.
- Option B: 2A is incorrect; the amplitude decreases with distance.
- Option D: A/4 is incorrect; the amplitude decreases inversely with distance, not with the square of the distance.

Key Concepts:
- Amplitude of a wave
- Intensity of a wave
- Point source
- Inverse square law

Memory Aids:
- Amplitude decreases inversely with distance from a point source.

Quick Tips:
- Remember that intensity is proportional to the square of the amplitude and inversely proportional to the square of the distance for a point source.

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Application +4 | -1

An open tube is in resonance with a string. If tube is dipped in water, so that 75% of length of the tube is inside water, then ratio of frequency of tube to string is

View answer and explanation

Correct option: A

2\mathrm{ }:\mathrm{ }1

Option A (

2\mathrm{ }:\mathrm{ }1

) is correct because the frequency of an open tube is inversely proportional to its length, and the frequency of a string is independent of the tube's length.

Solution approach

Correct Answer + Core Reason: $2\mathrm{ }:\mathrm{ }1$ . The ratio of frequencies is determined by the change in length of the tube.

Step-by-Step Solution:

Step 1: Define the initial frequency of the open tube as $f_t = \frac{v}{2L}$ , where $L$ is the length of the tube.

Step 2: Define the frequency of the string as $f_s$ .

Step 3: When the tube is dipped in water, the effective length becomes $L' = L - 0.75L = 0.25L = \frac{L}{4}$ .

Step 4: The new frequency of the tube is $f_t' = \frac{v}{2L'} = \frac{v}{2(L/4)} = \frac{4v}{2L} = 4(\frac{v}{2L}) = 4f_t$ .\nSince the tube is dipped in water, it now acts as a closed tube. Therefore, $f_t' = \frac{v}{4L'} = \frac{v}{4(L/4)} = \frac{v}{L} = 2f_t$

Step 5: Since the tube is in resonance with the string initially, $f_t = f_s$ .

Step 6: Find the ratio of the new frequency of the tube to the frequency of the string: $\frac{f_t'}{f_s} = \frac{2f_t}{f_s} = \frac{2f_s}{f_s} = 2$ .\nSo, the ratio is $2:1$ .

Why other options are wrong:
- Option B: $3\mathrm{ }:\mathrm{ }2$ is incorrect; it might be a miscalculation of the new tube length or the frequency ratio.
- Option C: $1\mathrm{ }:\mathrm{ }1$ is incorrect; it implies no change in the tube's frequency, which is wrong.
- Option D: $4\mathrm{ }:\mathrm{ }3$ is incorrect; it might be a miscalculation of the new tube length or the frequency ratio.

Key Concepts:
- Frequency of an open tube: $f = \frac{v}{2L}$
- Frequency of a closed tube: $f = \frac{v}{4L}$
- Resonance

Memory Aids:
- Open tube has both ends open, closed tube has one end closed.

Quick Tips:
- Remember to consider whether the tube becomes open or closed after dipping.

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Understanding +4 | -1

The equation of a spherical progressive wave is

View answer and explanation

Correct option: D

y = \frac{a}{r}\sin \left( {\omega t - kr} \right)

Option D is correct because the amplitude of a spherical wave decreases linearly with the distance.

Solution approach

Correct Answer + Core Reason: $y = \frac{a}{r}\sin \left( {\omega t - kr} \right)$ . The amplitude decreases linearly with the distance in spherical waves.

Step-by-Step Solution:

Step 1: Understand Spherical Waves: Spherical waves propagate outwards from a point source.

Step 2: Amplitude Variation: The amplitude of a spherical wave decreases as the wave spreads out. The energy is distributed over a larger surface area, which increases with the square of the radius $r$ . Since intensity is proportional to the square of amplitude, and intensity is inversely proportional to $r^2$ , the amplitude is inversely proportional to $r$ .

Step 3: General Equation: The equation of a spherical wave includes a term $\frac{1}{r}$ to account for the decreasing amplitude.

Why other options are wrong:
- Option A: $y = a\mathrm{ }sin\,\,\omega t$ represents a simple harmonic motion, not a propagating wave.
- Option B: $y = a\mathrm{ }sin\left( {\omega t - kr} \right)$ represents a plane wave with constant amplitude.
- Option C: $y = \frac{a}{{\sqrt r }}\sin \left( {\omega t - kr} \right)$ represents a cylindrical wave, where amplitude decreases with the square root of the distance.

Key Concepts:
- Spherical Waves
- Amplitude
- Wave Propagation

Memory Aids:
- Spherical = Linear (in amplitude)

Quick Tips:
- Remember the amplitude dependence on distance for different types of waves.

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Analyzing +4 | -1

A tuning fork gives

5

beats with another tuning fork of frequency

100\,\,Hz.

When the first tuning fork is loaded with wax, then the number of beats remains unchanged, then what will be the frequency of the first tuning fork

View answer and explanation

Correct option: C

105\,\,Hz

Option C (

105\mathrm{ }Hz

) is correct because adding wax decreases the frequency of the tuning fork, and since the number of beats remained the same, the original frequency must have been higher than 100 Hz.

Solution approach

Correct Answer + Core Reason: $105\mathrm{ }Hz$ . The frequency of the first tuning fork must be higher than 100 Hz, and decreases when loaded with wax.

Step-by-Step Solution:

Step 1: Let $f_1$ be the frequency of the first tuning fork and $f_2 = 100 \text{ Hz}$ be the frequency of the second tuning fork.

Step 2: The number of beats is $|f_1 - f_2| = 5$ .

Step 3: This means $f_1$ could be either $105 \text{ Hz}$ or $95 \text{ Hz}$ .

Step 4: When the first tuning fork is loaded with wax, its frequency decreases. Let the new frequency be $f_1'$ .

Step 5: The number of beats remains unchanged, so $|f_1' - f_2| = 5$ .

Step 6: If $f_1 = 95 \text{ Hz}$ , then $f_1'$ would be even lower, and the number of beats would increase, which contradicts the given information. Therefore, $f_1$ cannot be $95 \text{ Hz}$ .

Step 7: If $f_1 = 105 \text{ Hz}$ , then $f_1'$ would be lower, and the number of beats could remain the same. Therefore, $f_1 = 105 \text{ Hz}$ .

Why other options are wrong:
- Option A: $95\mathrm{ }Hz$ is incorrect; loading the tuning fork would increase the number of beats.
- Option B: $100\mathrm{ }Hz$ is incorrect; the number of beats would be zero.
- Option D: $110\mathrm{ }Hz$ is incorrect; loading the tuning fork would change the number of beats.

Key Concepts:
- Beat frequency

Memory Aids:
- Loading a tuning fork decreases its frequency.

Quick Tips:
- Consider both possibilities for the initial frequency and eliminate based on the effect of loading.

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Application +4 | -1

A wave travelling along positive

x

-axis is given by

y = A\sin \left( {\omega t - kx} \right)

. If it is reflected from rigid boundary such that

80\%

amplitude is reflected, then equation of reflected wave is

View answer and explanation

Correct option: B

y = - 0.8A\sin \left( {\omega t + kx} \right)

Option B (

y = - 0.8A\sin \left( {\omega t + kx} \right)

) is correct because the wave is reflected with 80% amplitude and a phase change of

\pi

due to reflection from a rigid boundary.

Solution approach

Correct Answer + Core Reason: $y = - 0.8A\sin \left( {\omega t + kx} \right)$ . Reflection from a rigid boundary causes a phase change of $\pi$ and reduces the amplitude to 80%.

Step-by-Step Solution:

Step 1: The incident wave is given by $y = A\sin(\omega t - kx)$ .

Step 2: When a wave is reflected from a rigid boundary, it undergoes a phase change of $\pi$ radians (or 180 degrees).

Step 3: The reflected wave will travel in the opposite direction, so $x$ becomes $-x$ .

Step 4: The amplitude is reduced to 80%, so the new amplitude is $0.8A$ .

Step 5: The equation of the reflected wave is $y' = 0.8A\sin(\omega t - k(-x) + \pi) = 0.8A\sin(\omega t + kx + \pi)$ .

Step 6: Using the trigonometric identity $\sin(\theta + \pi) = -\sin(\theta)$ , we get $y' = -0.8A\sin(\omega t + kx)$ .

Why other options are wrong:
- **Option A ( $y = A\sin \left( {\omega t + kx} \right)$ ): This is incorrect because it doesn't account for the amplitude reduction and the phase change upon reflection from a rigid boundary.
- Option C ( $y = 0.8A\sin \left( {\omega t + kx} \right)$ ): This is incorrect because it doesn't account for the phase change upon reflection from a rigid boundary.
- Option D ( $y = A\sin \left( {\omega t + 0.8kx} \right)$ ): This is incorrect because it only changes the wave number and not the amplitude or phase.

Key Concepts:
- Reflection of waves
- Phase change upon reflection from a rigid boundary
- Wave equation

Memory Aids:**
- Reflection from a rigid boundary: phase change of $\pi$ .

Quick Tips:
- Remember that reflection from a rigid boundary introduces a phase change of $\pi$ .

90 Medium

90 of 159Question permalink

Analyzing +4 | -1

Statement I Two longitudinal waves given by equation

{y_1}\left( {x,t} \right) = 2a\,\,\,\sin \left( {\omega t - kx} \right)

and

{y_2}\left( {x,t} \right) = a\,\,\sin \left( {2\omega t - 2kx} \right)

will have equal intensity.
Statement II Intensity of waves of given frequency in same medium is proportional to square of amplitude only

View answer and explanation

Correct option: B

Statement I is true, Statement II is false

Statement I is true, Statement II is false. Statement I is correct because intensity depends on the square of the amplitude and the square of the frequency. Statement II is incorrect because it doesn't account for the frequency's effect on intensity.

Solution approach

Correct Answer + Core Reason: Statement I is true, Statement II is false. Intensity depends on both amplitude and frequency.

Step-by-Step Solution:

Step 1: The intensity of a wave is given by $I = \frac{1}{2} \rho v \omega^2 A^2$ , where $\rho$ is the density of the medium, $v$ is the wave velocity, $\omega$ is the angular frequency, and $A$ is the amplitude.

Step 2: For the first wave, ${y_1}(x,t) = 2a\sin(\omega t - kx)$ , the amplitude is $A_1 = 2a$ and the angular frequency is $\omega_1 = \omega$ .\nSo, the intensity is $I_1 = \frac{1}{2} \rho v \omega^2 (2a)^2 = 2 \rho v \omega^2 a^2$ .

Step 3: For the second wave, ${y_2}(x,t) = a\sin(2\omega t - 2kx)$ , the amplitude is $A_2 = a$ and the angular frequency is $\omega_2 = 2\omega$ .\nSo, the intensity is $I_2 = \frac{1}{2} \rho v (2\omega)^2 a^2 = 2 \rho v \omega^2 a^2$ .

Step 4: Comparing the intensities, we see that $I_1 = I_2$ . Therefore, Statement I is true.

Step 5: Statement II says that intensity is proportional to the square of the amplitude only. However, the intensity is also proportional to the square of the frequency. Therefore, Statement II is false.

Why other options are wrong:
- Option B (Statement I is true, Statement II is false): This is the correct answer as explained above.
- Option C (Statement I is true, Statement II is true, Statement II is the correct explanation of statement I): This is incorrect because Statement II is false.
- Option D (Statement I is true, Statement II is true, Statement II is not correct explanation of statement I): This is incorrect because Statement II is false.

Key Concepts:
- Intensity of a wave
- Amplitude
- Angular frequency

Memory Aids:
- Remember the formula for intensity of a wave: $I = \frac{1}{2} \rho v \omega^2 A^2$ .

Quick Tips:
- Always consider both amplitude and frequency when comparing the intensities of waves.

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