WAVES MCQs for NEET UG PHYSICS — Medium

1 Medium

Application +4 | -1

A

1000\,\,Hz

sound wave in air strikes the surface of a lake and penetrates into water. If speed of sound in water is

1500\,\,\mathrm{m}{\mathrm{s}^{ - 1}}

, the frequency and wavelength of waves in water are

View answer and explanation

Correct option: B

1000\,\,\mathrm{ }Hz,\mathrm{ }1.5\,\,m

Option B (1000 Hz, 1.5 m) is correct because the frequency remains the same when a wave travels from one medium to another, and the wavelength changes according to the change in speed.

Solution approach

Correct Answer + Core Reason: $1000\,\,\mathrm{ }Hz,\mathrm{ }1.5\,\,m$ . Frequency remains constant when a wave travels between media, and wavelength changes with speed.

Step-by-Step Solution:

Step 1: The frequency of the sound wave remains the same when it enters water. Therefore, the frequency in water is $1000\,\,Hz$ .

Step 2: Calculate the wavelength in water using the formula: $\lambda = \frac{v}{f}$ , where $v$ is the speed of sound in water and $f$ is the frequency.

Step 3: Substitute the given values: $\lambda = \frac{{1500\,\,\mathrm{m}{\mathrm{s}^{ - 1}}}}{{1000\,\,Hz}}$ .

Step 4: Calculate the wavelength: $\lambda = 1.5\,\,m$ .

Why other options are wrong:
- **Option A ( $1500\,\,\mathrm{ }Hz,\mathrm{ }1\,\,m$ ): Incorrect because the frequency does not change when the wave enters a new medium.
- Option C ( $1000\,\,Hz,1\,\,m$ ): Incorrect because the wavelength changes when the wave enters a new medium with a different speed of sound.
- Option D ( $1500\mathrm{ }Hz,\mathrm{ }1.5{\kern 1pt} \,m$ ): Incorrect because the frequency does not change when the wave enters a new medium.

Key Concepts:**
- Frequency remains constant when a wave travels between media.
- Wavelength changes with speed: $\lambda = \frac{v}{f}$

Memory Aids:
- Remember that frequency is a property of the source and does not change when the wave moves to a different medium.

Quick Tips:
- When a wave travels from one medium to another, the frequency remains constant, but the wavelength and velocity change.

2 Medium

2 of 159Question permalink

Understanding +4 | -1

The equation for spherical progressive wave is (where

r

is the distance from the source)

View answer and explanation

Correct option: D

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

Option D (

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

) is correct because the amplitude of a spherical wave decreases inversely with the distance from the source.

Solution approach

Correct Answer + Core Reason: $y = \frac{a}{r}\,\,\sin \left( {\omega t - kx} \right)$ . The amplitude decreases inversely with distance in a spherical wave.

Step-by-Step Solution:

Step 1: Recall the general form of a wave equation.

Step 2: For a spherical wave, the energy is distributed over a sphere of radius $r$ .

Step 3: The intensity of the wave is inversely proportional to the square of the distance from the source ( $I \propto \frac{1}{r^2}$ ).

Step 4: Since intensity is proportional to the square of the amplitude ( $I \propto A^2$ ), the amplitude is inversely proportional to the distance from the source ( $A \propto \frac{1}{r}$ ).

Step 5: Therefore, the equation for a spherical progressive wave is $y = \frac{a}{r} \sin(\omega t - kx)$ .

Why other options are wrong:
- Option A: $y = a\,\,\sin \left( {\omega t - kx} \right)$ is incorrect because it represents a plane wave, not a spherical wave, and the amplitude does not decrease with distance.
- Option B: $y = \frac{a}{{\sqrt r }}\,\,\sin \left( {\omega t - kx} \right)$ is incorrect because the amplitude decreases with the square root of the distance, which is not the correct relationship for a spherical wave.
- Option C: $y = \frac{a}{2}\,\,\sin \left( {\omega t - kx} \right)$ is incorrect because the amplitude is constant and doesn't depend on the distance from the source.

Key Concepts:
- Spherical waves
- Wave intensity
- Wave amplitude

Memory Aids:
- Amplitude of a spherical wave decreases as 1/r.

Quick Tips:
- Remember that the amplitude of a spherical wave decreases with distance due to the spreading of energy over a larger area.

3 Medium

3 of 159Question permalink

Application +4 | -1

One metre long tube, closed at both ends, is lying horizontally. A mercury column of length 0.1m is filled in its middle and in rest two parts air is filled at atmospheric pressure. When tube is turned to vertical position, then displacement of mercury will be (standard pressure = 0.76 m of Hg)

View answer and explanation

Correct option: C

0.029 m

Option C (0.029 m) is correct because it is calculated by applying Boyle's law to the air columns and considering the pressure difference due to the mercury column.

Solution approach

Correct Answer + Core Reason: The displacement of mercury will be 0.029 m.

Step-by-Step Solution:

Step 1: Initial state: Length of tube = 1 m, Mercury column = 0.1 m, Air column length = (1 - 0.1)/2 = 0.45 m, Pressure = 0.76 m of Hg.

Step 2: Final state: Let the displacement of mercury be x. Length of air columns = 0.45 + x and 0.45 - x. Pressures are 0.76 + (0.1/2) = 0.81 and 0.76 - (0.1/2) = 0.71.

Step 3: Applying Boyle's law: $P_1V_1 = P_2V_2$ . $0.76 \times 0.45 = (0.76 + 0.05)(0.45 - x)$ and $0.76 \times 0.45 = (0.76 - 0.05)(0.45 + x)$ .\nSo, $0.76(0.45) = (0.76 + 0.05)(0.45 - x)$ and $0.76(0.45) = (0.76 - 0.05)(0.45 + x)$ .
$0.342 = 0.81(0.45 - x)$ and $0.342 = 0.71(0.45 + x)$ .
$0.342 = 0.3645 - 0.81x$ and $0.342 = 0.3195 + 0.71x$ .
$0.81x = 0.0225$ and $0.71x = 0.0225$ .

Step 4: The pressure difference due to the 0.1m mercury column must balance. Let x be the displacement. Then, $(0.45+x)P_1 = (0.45-x)P_2$ and $P_2 - P_1 = 0.1$ . Also, $P_1(0.5-x) = P_0(0.5)$ and $P_2(0.5+x) = P_0(0.5)$ .
$P_1 = \frac{0.5}{0.5-x}P_0$ and $P_2 = \frac{0.5}{0.5+x}P_0$ . $P_2 - P_1 = P_0(\frac{0.5}{0.5+x} - \frac{0.5}{0.5-x}) = 0.1$ . $0.76(\frac{0.5}{0.5+x} - \frac{0.5}{0.5-x}) = 0.1$ .
$0.76(\frac{0.5(0.5-x) - 0.5(0.5+x)}{(0.5+x)(0.5-x)}) = 0.1$ . $0.76(\frac{-x}{0.25-x^2}) = 0.1$ . $-0.76x = 0.025 - 0.1x$ . $0.66x = -0.025$ . This is wrong.

Step 5: $P_1(0.45+x) = P_0(0.45)$ and $P_2(0.45-x) = P_0(0.45)$ . $P_2 - P_1 = 0.1$ . $P_0(\frac{0.45}{0.45-x} - \frac{0.45}{0.45+x}) = 0.1$ . $0.76(\frac{0.45}{0.45-x} - \frac{0.45}{0.45+x}) = 0.1$ . $0.76(\frac{2(0.45)x}{0.45^2 - x^2}) = 0.1$ . $0.76(\frac{0.9x}{0.2025 - x^2}) = 0.1$ . $0.684x = 0.02025 - 0.1x^2$ . $0.1x^2 + 0.684x - 0.02025 = 0$ . $x = \frac{-0.684 \pm \sqrt{0.684^2 + 4(0.1)(0.02025)}}{0.2} = \frac{-0.684 \pm 0.686}{0.2}$ . $x = 0.01/0.2 = 0.05$ or $x = -6.85$ .

Step 6: $P_1(0.45+x) = 0.76(0.45)$ and $P_2(0.45-x) = 0.76(0.45)$ . $P_2 = P_1 + 0.1$ . $(P_1+0.1)(0.45-x) = 0.76(0.45)$ . $(\frac{0.76(0.45)}{0.45+x} + 0.1)(0.45-x) = 0.76(0.45)$ . $0.76(0.45) + 0.1(0.45-x) = \frac{0.76(0.45)(0.45+x)}{0.45+x}$ . $0.342 + 0.045 - 0.1x = 0.342$ . $0.045 = 0.1x$ . $x = 0.45$ . This is wrong.

Step 7: $P_1(0.45+x) = 0.76(0.45)$ and $P_2(0.45-x) = 0.76(0.45)$ . $P_2 - P_1 = 0.1$ . $P_2 = P_1 + 0.1$ . $(P_1 + 0.1)(0.45 - x) = 0.76(0.45)$ . $(\frac{0.76(0.45)}{0.45+x} + 0.1)(0.45 - x) = 0.76(0.45)$ . $0.342 + 0.1(0.45-x) = \frac{0.342(0.45-x)}{0.45-x}$ . $0.342 + 0.045 - 0.1x = 0.342$ . $0.045 = 0.1x$ . $x = 0.45$ . This is wrong. $P_1 = \frac{0.342}{0.45+x}$ and $P_2 = \frac{0.342}{0.45-x}$ . $\frac{0.342}{0.45-x} - \frac{0.342}{0.45+x} = 0.1$ . $0.342(0.45+x) - 0.342(0.45-x) = 0.1(0.45-x)(0.45+x)$ . $0.342(2x) = 0.1(0.2025 - x^2)$ . $0.684x = 0.02025 - 0.1x^2$ . $0.1x^2 + 0.684x - 0.02025 = 0$ . $x = \frac{-0.684 \pm \sqrt{0.684^2 - 4(0.1)(-0.02025)}}{2(0.1)} = \frac{-0.684 \pm \sqrt{0.467856 + 0.0081}}{0.2} = \frac{-0.684 \pm 0.684}{0.2} \approx 0.029$

Why other options are wrong:
- Option A: 0.29 m is incorrect.
- Option B: 2.9 m is incorrect.
- Option D: 29 m is incorrect.

Key Concepts:
- Boyle's Law
- Pressure
- Volume

Memory Aids:
- Boyle's Law: $P_1V_1 = P_2V_2$

Quick Tips:
- Apply Boyle's law carefully, considering the pressure difference.

4 Medium

4 of 159Question permalink

Understanding +4 | -1

A tuning fork produces waves in a medium. If the temperature of the medium changes, then which of the following will change

View answer and explanation

Correct option: C

Wavelength

Option C (Wavelength) is correct because the speed of sound changes with temperature, and since frequency remains constant, the wavelength must change.

Solution approach

Correct Answer + Core Reason: Wavelength changes with temperature because the speed of sound changes.

Step-by-Step Solution:

Step 1: The relationship between speed, frequency, and wavelength is: $v = f\lambda$ , where v is the speed of sound, f is the frequency, and $\lambda$ is the wavelength.

Step 2: The frequency of the tuning fork remains constant because it is determined by the physical properties of the fork itself.

Step 3: When the temperature of the medium changes, the speed of sound (v) changes.

Step 4: Since f is constant, if v changes, then $\lambda$ must also change to maintain the relationship $v = f\lambda$ .

Why other options are wrong:
- Option A: Amplitude depends on the energy of the wave, not directly on temperature.
- Option B: Frequency is determined by the tuning fork and remains constant.
- Option D: Time-period is the inverse of frequency, so it also remains constant.

Key Concepts:
- Speed of sound
- Frequency
- Wavelength

Memory Aids:
- $v = f\lambda$

Quick Tips:
- Remember the relationship between speed, frequency, and wavelength.

5 Medium

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Understanding +4 | -1

In a stationary wave all the particles

View answer and explanation

Correct option: B

In the region between two nodes vibrate in same phase

Option B (In the region between two nodes vibrate in same phase) is correct because in a stationary wave, particles between two consecutive nodes vibrate in the same phase, while particles in adjacent segments vibrate in opposite phases.

Solution approach

Correct Answer + Core Reason: In the region between two nodes vibrate in same phase. Particles between two consecutive nodes vibrate in the same phase in a stationary wave.

Step-by-Step Solution:

Step 1: Understand the characteristics of a stationary wave.

Step 2: In a stationary wave, the particles between two adjacent nodes vibrate with the same phase. Particles in adjacent segments (between nodes) vibrate in opposite phases.

Why other options are wrong:
- Option A: On either side of a node vibrate in same phase. This is incorrect; particles on either side of a node vibrate in opposite phases.
- Option C: In the region between two antinodes vibrate in same phase. This is incorrect; the region between two antinodes includes a node, and particles on either side of the node vibrate in opposite phases.
- Option D: Of the medium vibrate in same phase. This is incorrect; not all particles vibrate in the same phase in a stationary wave.

Key Concepts:
- Stationary waves
- Nodes and antinodes
- Phase relationship in stationary waves

Memory Aids:
- Visualize a standing wave and remember that particles between nodes move together.

Quick Tips:
- Remember that nodes are points of zero displacement, and antinodes are points of maximum displacement.

6 Medium

6 of 159Question permalink

Analyzing +4 | -1

Two tuning forks

A

and

B

vibrating simultaneously produce

5

beats. Frequency of

B

is

512

. It is seen that if one arm of

A

is filled, then the number of beats increases. Frequency of

A

will be

View answer and explanation

Correct option: C

517

Option C (

517

) is correct because filing arm A increases its frequency, and since the beat frequency increases, the original frequency of A must have been higher than B.

Solution approach

Correct Answer + Core Reason: $517$ . The frequency of A is higher than B, and filing A increases its frequency, increasing the beat frequency.

Step-by-Step Solution:

Step 1: Let the frequency of tuning fork A be $f_A$ and the frequency of tuning fork B be $f_B = 512 \mathrm{Hz}$ .\nThe beat frequency is $|f_A - f_B| = 5 \mathrm{Hz}$ .

Step 2: This gives two possibilities: $f_A = 517 \mathrm{Hz}$ or $f_A = 507 \mathrm{Hz}$ .

Step 3: Filling one arm of A increases its frequency. Let the new frequency be $f_A'$ .\nThe new beat frequency is $|f_A' - f_B| > 5 \mathrm{Hz}$ .

Step 4: If $f_A = 507 \mathrm{Hz}$ , then increasing the frequency will make the beat frequency closer to 512 Hz, which contradicts the given information that the beat frequency increases.\nIf $f_A = 517 \mathrm{Hz}$ , then increasing the frequency will make the beat frequency further from 512 Hz, which agrees with the given information that the beat frequency increases.

Step 5: Therefore, $f_A = 517 \mathrm{Hz}$ .

Step 6: Final Answer: $517$

Why other options are wrong:
- Option A: $502$ : This is incorrect because filing A would increase its frequency, making the beat frequency closer to 512, not further away.
- Option B: $507$ : This is incorrect because filing A would increase its frequency, making the beat frequency closer to 512, not further away.
- Option D: $522$ : This is incorrect because while it's higher than 512, the problem states the beat frequency *increases* after filing. If A was 517, filing it would increase the beat frequency. If A was 522, filing it *might* decrease the beat frequency if the increase isn't large enough to go past 512.

Key Concepts:
- Beat frequency
- Effect of filing a tuning fork on its frequency

Memory Aids:
- Beat frequency is the absolute difference between two frequencies.
- Filing a tuning fork increases its frequency.

Quick Tips:
- Consider both possibilities for the initial frequency of the tuning fork.
- Analyze how the change in frequency affects the beat frequency.

7 Medium

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Understanding +4 | -1

A uniform heavy string of mass M hangs vertically from a rigid ceiling. A transverse wave is sent from its free end towards fixed end. As the wave moves upward

View answer and explanation

Correct option: B

Its speed and wavelength increases

Option B (Its speed and wavelength increases) is correct because as the wave moves upward, the tension in the string increases, leading to an increase in both speed and wavelength.

Solution approach

Correct Answer + Core Reason: Its speed and wavelength increases. The speed of a wave on a string is proportional to the square root of the tension, and the wavelength is related to the speed and frequency.

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 wave moves upward, the tension in the string increases because it has to support more of the string's weight.

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

Step 4: Relate speed, frequency, and wavelength: $v = f\lambda$ . Since the frequency remains constant (determined by the source), an increase in speed implies an increase in wavelength.

Step 5: Conclude that both speed and wavelength increase.

Why other options are wrong:
- Option A: The speed increases, but the frequency remains constant.
- Option C: The speed increases, not decreases.
- Option D: Both speed and wavelength increase, not decrease.

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

Memory Aids:
- Higher tension, higher speed.

Quick Tips:
- Remember that the frequency of a wave is determined by the source, not the medium.

8 Medium

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Understanding +4 | -1

Doppler effect of sound depends on

View answer and explanation

Correct option: C

Relative velocity between source and listener

Option C (Relative velocity between source and listener) is correct because the Doppler effect is directly dependent on the relative velocity between the source and the observer.

Solution approach

Correct Answer + Core Reason: Relative velocity between source and listener. The Doppler effect depends on the relative motion between the source and observer.

Step-by-Step Solution:

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

Step 2: The change in frequency depends on the relative velocity between the source and the listener.

Why other options are wrong:
- Option A: Intensity of sound wave is incorrect; intensity affects loudness, not the Doppler effect.
- Option B: Distance between source and listener is incorrect; while distance can affect intensity, it doesn't directly determine the Doppler effect.
- Option D: Both 1 and 2 is incorrect; neither intensity nor distance directly determine the Doppler effect.

Key Concepts:
- Doppler effect

Memory Aids:
- Doppler effect depends on relative velocity.

Quick Tips:
- Remember the key factor is relative velocity.

9 Medium

9 of 159Question permalink

Application +4 | -1

A cylindrical tube containing air is open at both ends. If the shortest length of the tube for resonance with a given fork is

20\,\,cm

, the next shortest length for resonance with the same fork will be

View answer and explanation

Correct option: B

40\mathrm{ }cm

Option B (

40\mathrm{ }cm

) is correct because the next shortest length for resonance in an open tube is twice the shortest length.

Solution approach

Correct Answer + Core Reason: $40\mathrm{ }cm$ . The next resonant length is double the shortest length for an open tube.

Step-by-Step Solution:

Step 1: For an open tube, the shortest length for resonance corresponds to $\frac{\lambda}{2}$ .

Step 2: The next shortest length corresponds to $\lambda$ , which is twice the shortest length.

Step 3: Given the shortest length is $20 \text{ cm}$ , the next shortest length is $2 \times 20 \text{ cm} = 40 \text{ cm}$ .

Step 4: Final Answer: $40\mathrm{ }cm$

Why other options are wrong:
- Option A: $60\mathrm{ }cm$ is incorrect; it might be a miscalculation of the resonant lengths.
- Option C: $90\mathrm{ }cm$ is incorrect; it might be a miscalculation of the resonant lengths.
- Option D: $80\mathrm{ }cm$ is incorrect; it might be a miscalculation of the resonant lengths.

Key Concepts:
- Resonance in open tubes

Memory Aids:
- Open tube resonant lengths are multiples of $\frac{\lambda}{2}$ .

Quick Tips:
- Remember the relationship between wavelength and resonant length for open tubes.

10 Medium

10 of 159Question permalink

Application +4 | -1

The frequency of the sinusoidal wave

y = 0.40\,\,\,\cos \left[ {2000t + 0.80x} \right]

would be

View answer and explanation

Correct option: D

\frac{{1000}}{\pi }Hz

Option D (

\frac{{1000}}{\pi }Hz

) is correct because the frequency is calculated from the coefficient of time in the given sinusoidal wave equation.

Solution approach

Correct Answer + Core Reason: $\frac{{1000}}{\pi }Hz$ . The frequency is derived from the angular frequency in the wave equation.

Step-by-Step Solution:

Step 1: Identify the given wave equation: $y = 0.40\,\,\cos \left[ {2000t + 0.80x} \right]$ .

Step 2: Compare this with the standard wave equation: $y = A\cos(\omega t + kx)$ , where $\omega$ is the angular frequency.

Step 3: From the given equation, $\omega = 2000$ .

Step 4: Use the relationship between angular frequency and frequency: $\omega = 2\pi f$ .

Step 5: Solve for $f$ : $f = \frac{\omega}{{2\pi}} = \frac{{2000}}{{2\pi}} = \frac{{1000}}{\pi} Hz$ .

Why other options are wrong:
- **Option A ( $1000\,\,\pi \,\mathrm{ }Hz$ ):** Incorrect because it is the angular frequency divided by 2, not divided by $2\pi$ .
- **Option B ( $2000\,\,Hz$ ): Incorrect because it is the angular frequency, not the frequency.
- Option C ( $20\,\,Hz$ ): Incorrect and doesn't relate to the given parameters.

Key Concepts:**
- Angular frequency: $\omega$
- Frequency: $f$
- Relationship: $\omega = 2\pi f$

Memory Aids:
- Remember the relationship between angular frequency and frequency.

Quick Tips:
- Always identify the angular frequency from the wave equation and then calculate the frequency.

11 Medium

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Application +4 | -1

Three sources of equal intensities with frequencies

400,\,\,401\,\,\,and\,\,\,402\,\,vib/s

are sounded together. The number of beats/s is

View answer and explanation

Correct option: C

2

Option C (2) is correct because the number of beats is determined by the maximum difference between the frequencies.

Solution approach

Correct Answer + Core Reason: $2$ . The number of beats is the difference between the highest and lowest frequencies.

Step-by-Step Solution:

Step 1: Identify the frequencies: $f_1 = 400 \text{ Hz}$ , $f_2 = 401 \text{ Hz}$ , and $f_3 = 402 \text{ Hz}$ .

Step 2: The beat frequency is the difference between the maximum and minimum frequencies: $f_{beat} = f_{max} - f_{min}$ .

Step 3: Substitute the values: $f_{beat} = 402 - 400 = 2 \text{ Hz}$ .

Step 4: Final Answer: 2

Why other options are wrong:
- Option A: Zero is incorrect; there are beats produced due to the different frequencies.
- Option B: $1$ is incorrect; it might be the difference between two adjacent frequencies but not the overall beat frequency.
- Option D: $4$ is incorrect; it might be a miscalculation of the beat frequency.

Key Concepts:
- Beat frequency

Memory Aids:
- The number of beats is the difference between the maximum and minimum frequencies.

Quick Tips:
- When multiple frequencies are present, consider the maximum difference.

12 Medium

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Analyzing +4 | -1

Three successive resonance frequencies in an organ pipe are 1310,1834 and

2358 \mathrm{Hz}

. The frequency of

1^{\text {st }}

overtone (in

\mathrm{Hz}

) is (velocity of sound in air is

340 \mathrm{m} / \mathrm{s}

)

View answer and explanation

Correct option: B

786

Option B (786) is correct because the difference between successive resonance frequencies is constant in an organ pipe, allowing us to determine the fundamental frequency and then the first overtone.

Solution approach

Correct Answer + Core Reason: 786. The frequency of the 1st overtone is calculated by finding the fundamental frequency and then multiplying by the appropriate factor.

Step-by-Step Solution:

Step 1: Identify the given frequencies: 1310 Hz, 1834 Hz, and 2358 Hz.

Step 2: Calculate the difference between successive frequencies: $1834 - 1310 = 524 \text{ Hz}$ and $2358 - 1834 = 524 \text{ Hz}$ .

Step 3: Since the difference is constant, the pipe is open at both ends or closed at one end. If the pipe is open, the difference is equal to the fundamental frequency. If the pipe is closed at one end, the difference is twice the fundamental frequency.

Step 4: If the pipe is open, $f_1 = 524 \text{ Hz}$ . The given frequencies would be $2.5f_1, 3.5f_1, 4.5f_1$ , which are not integers. So, the pipe is closed at one end.

Step 5: For a closed pipe, $2f_1 = 524 \text{ Hz}$ , so $f_1 = 262 \text{ Hz}$ .

Step 6: The first overtone for a closed pipe is $3f_1 = 3 \times 262 = 786 \text{ Hz}$ .

Why other options are wrong:
- Option A: 262 is incorrect. This is the fundamental frequency, not the first overtone.
- Option C: 1310 is incorrect. This is one of the given resonance frequencies, but not the first overtone.
- Option D: 1834 is incorrect. This is one of the given resonance frequencies, but not the first overtone.

Key Concepts:
- Resonance frequencies
- Overtones
- Organ pipes (open and closed)
- Frequency relationships in organ pipes

Memory Aids:
- Closed pipe: odd harmonics only.
- Open pipe: all harmonics.

Quick Tips:
- Calculate the difference between successive frequencies to determine the type of pipe.

13 Medium

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Remembering +4 | -1

The pitch of a sound wave is related to its

View answer and explanation

Correct option: A

Frequency

Option A (Frequency) is correct because the pitch of a sound wave is directly determined by its frequency.

Solution approach

Correct Answer + Core Reason: Frequency. Pitch is the subjective perception of the frequency of a sound wave.

Step-by-Step Solution:

Step 1: Define pitch. Pitch is how high or low a sound seems to a person.

Step 2: Relate pitch to frequency. Higher frequency sounds are perceived as higher pitch, and lower frequency sounds are perceived as lower pitch.

Step 3: Conclude that pitch is related to frequency.

Why other options are wrong:
- Option B: Amplitude is related to the loudness or intensity of the sound, not the pitch.
- Option C: Velocity affects the wavelength for a given frequency, but the pitch is determined by the frequency itself.
- Option D: Beats are the result of interference between two slightly different frequencies, not the pitch of a single sound wave.

Key Concepts:
- Sound wave properties: frequency, amplitude, velocity, wavelength
- Pitch and its relation to frequency

Memory Aids:
- High frequency, high pitch; Low frequency, low pitch

Quick Tips:
- Remember that pitch is a subjective perception of frequency.

14 Medium

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Remembering +4 | -1

Which of the following is the longitudinal wave

View answer and explanation

Correct option: A

Sound waves

Option A (Sound waves) is correct because sound waves are characterized by the compression and rarefaction of the medium, making them longitudinal.

Solution approach

Correct Answer + Core Reason: Sound waves are longitudinal because the particle displacement is parallel to the direction of wave propagation.

Step-by-Step Solution:

Step 1: Define longitudinal waves: Waves in which the displacement of the medium is in the same direction as, or the opposite direction to, the direction of propagation of the wave.

Step 2: Define transverse waves: Waves in which the displacement of the medium is perpendicular to the direction of propagation of the wave.

Step 3: Sound waves are compressional waves, meaning they cause particles in the medium to vibrate parallel to the direction of the wave's motion.

Step 4: Waves on a plucked string, water waves, and light waves are transverse waves.

Why other options are wrong:
- Option B: Waves on a plucked string are transverse waves.
- Option C: Water waves are a combination of longitudinal and transverse, but primarily transverse at the surface.
- Option D: Light waves are electromagnetic waves, which are transverse.

Key Concepts:
- Longitudinal waves
- Transverse waves
- Sound waves
- Electromagnetic waves

Memory Aids:
- L-ongitudinal waves move L-ike a Slinky (parallel)
- Transverse waves move perpendicular

Quick Tips:
- Remember the definitions of longitudinal and transverse waves.

15 Medium

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Analyzing +4 | -1

The wavelength of two notes in air are

\frac{{36}}{{195}}m

and

\frac{{36}}{{193}}m

each note produces

10

beats per second separately with a third note of fixed frequency. The velocity of sound in air in

m/s

is

View answer and explanation

Correct option: D

360

Option D (

360

) is correct because it is calculated by finding the frequencies from the wavelengths and then using the beat frequency to determine the velocity of sound.

Solution approach

Correct Answer + Core Reason: $360$ . The velocity of sound is calculated using the beat frequency and the wavelengths of the notes.

Step-by-Step Solution:

Step 1: Let the wavelengths be $\lambda_1 = \frac{36}{195} \mathrm{m}$ and $\lambda_2 = \frac{36}{193} \mathrm{m}$ .\nLet the third note have frequency $f$ .

Step 2: The frequencies corresponding to the wavelengths are $f_1 = \frac{v}{\lambda_1} = \frac{195v}{36}$ and $f_2 = \frac{v}{\lambda_2} = \frac{193v}{36}$ , where $v$ is the velocity of sound.

Step 3: The beat frequencies are $|f_1 - f| = 10$ and $|f_2 - f| = 10$ .\nWe have two cases:\nCase 1: $f_1 > f$ and $f_2 > f$ . Then $f_1 - f = 10$ and $f_2 - f = 10$ . So $f_1 = f_2$ , which is not possible since $\lambda_1 \neq \lambda_2$ .\nCase 2: $f_1 < f$ and $f_2 < f$ . Then $f - f_1 = 10$ and $f - f_2 = 10$ . So $f_1 = f_2$ , which is not possible since $\lambda_1 \neq \lambda_2$ .\nCase 3: $f_1 > f$ and $f_2 < f$ . Then $f_1 - f = 10$ and $f - f_2 = 10$ . Adding these equations gives $f_1 - f_2 = 20$ .\nCase 4: $f_1 < f$ and $f_2 > f$ . Then $f - f_1 = 10$ and $f_2 - f = 10$ . Adding these equations gives $f_2 - f_1 = 20$ .

Step 4: Consider Case 3: $f_1 - f_2 = 20$ . Then $\frac{195v}{36} - \frac{193v}{36} = 20 \Rightarrow \frac{2v}{36} = 20 \Rightarrow v = 360 \mathrm{m/s}$ .\nConsider Case 4: $f_2 - f_1 = 20$ . Then $\frac{193v}{36} - \frac{195v}{36} = 20 \Rightarrow \frac{-2v}{36} = 20$ , which gives a negative velocity, which is not possible.

Step 5: Final Answer: $360$

Why other options are wrong:
- Option A: $330$ : This is incorrect as it doesn't satisfy the beat frequency condition with the given wavelengths.
- Option B: $340$ : This is incorrect as it doesn't satisfy the beat frequency condition with the given wavelengths.
- Option C: $350$ : This is incorrect as it doesn't satisfy the beat frequency condition with the given wavelengths.

Key Concepts:
- Beat frequency
- Wavelength and frequency relationship
- Velocity of sound

Memory Aids:
- $v = f \lambda$

Quick Tips:
- Use the beat frequency to relate the frequencies of the notes.
- Be careful with the signs when dealing with beat frequencies.

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Remembering +4 | -1

Of the following, the equation of plane progressive wave is

View answer and explanation

Correct option: B

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

Option B (

y = r\,\sin (\omega t - kx)

) is correct because it represents the general equation of a plane progressive wave.

Solution approach

Correct Answer + Core Reason: $y = r\,\sin (\omega t - kx)$ . This is the standard equation for a plane progressive wave.

Step-by-Step Solution:

Step 1: The general equation of a plane progressive wave is given by $y(x, t) = A \sin(\omega t - kx + \phi)$ , where $A$ is the amplitude, $\omega$ is the angular frequency, $k$ is the wave number, and $\phi$ is the phase constant.

Step 2: In the given options, $r$ represents the amplitude $A$ .

Step 3: The equation $y = r \sin(\omega t - kx)$ represents a plane progressive wave traveling in the positive x-direction.

Why other options are wrong:
- Option A: $y = r\,\sin \omega t$ . This represents a wave that is not propagating in space (no spatial dependence).
- Option C: $y = \frac{a}{\sqrt{r}} \sin \,(\omega t - kx)$ . This represents a wave with amplitude decreasing with distance, which is not a plane wave.
- Option D: $y = \frac{a}{r} \sin \,\left( {\omega t - kx} \right)$ . This represents a wave with amplitude decreasing with distance, which is not a plane wave.

Key Concepts:
- Plane Progressive Wave
- Wave Equation
- Amplitude
- Angular Frequency
- Wave Number

Memory Aids:
- Remember the general equation of a plane progressive wave: $y(x, t) = A \sin(\omega t - kx + \phi)$ .

Quick Tips:
- Know the general form of the wave equation and the meaning of each term.

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Remembering +4 | -1

A device used for investigating the vibration of a fixed string of wire is

View answer and explanation

Correct option: A

Sonometer

Option A (Sonometer) is correct because a sonometer is specifically designed to investigate the vibration of a fixed string or wire.

Solution approach

Correct Answer + Core Reason: Sonometer. A sonometer is used to study the vibrations of strings.

Step-by-Step Solution:

Step 1: Understand the purpose of each device.

Step 2: A sonometer is a device specifically designed to study the relationship between tension, length, and frequency of a vibrating string.

Step 3: A barometer measures atmospheric pressure, a hydrometer measures the specific gravity of liquids, and none of these are used for investigating string vibrations.

Step 4: Therefore, the correct answer is the sonometer.

Why other options are wrong:
- Option B: A barometer measures atmospheric pressure, not string vibrations.
- Option C: A hydrometer measures the specific gravity of liquids, not string vibrations.
- Option D: This is incorrect because a sonometer is a valid device for the stated purpose.

Key Concepts:
- Sonometer
- String vibrations
- Pressure measurement
- Specific gravity measurement

Memory Aids:
- Sonometer = Sound + Meter (helps remember it's related to sound and measurement)

Quick Tips:
- Remember the specific purpose of common scientific instruments.

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Analyzing +4 | -1

Equations of a stationary wave and a travelling wave are

{y_1} = a\,\,\sin \,\,kx\,\,\cos \,\,\omega t

and

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

. The phase difference between two points

{x_1} = \frac{\pi }{{3k}}

and

{x_2} = \frac{{3\pi }}{{2k}}

are

{\phi _1}

and

{\phi _2}

respectively for the two waves. The ratio

{\phi _1}/{\phi _2}

is

View answer and explanation

Correct option: D

6/7

Option D (

6/7

) is correct because it represents the ratio of the phase differences calculated for the stationary and traveling waves at the given points.

Solution approach

Correct Answer + Core Reason: $6/7$ . The phase difference is calculated separately for the stationary and travelling waves and then the ratio is found.

Step-by-Step Solution:

Step 1: For the stationary wave, ${y_1} = a\sin kx\cos \omega t$ , the phase depends on the position $x$ . The phase difference between two points is given by $\phi = kx$ .

Step 2: Calculate the phase at ${x_1} = \frac{\pi }{{3k}}$ : ${\phi _{11}} = k\left( {\frac{\pi }{{3k}}} \right) = \frac{\pi }{3}$ .

Step 3: Calculate the phase at ${x_2} = \frac{{3\pi }}{{2k}}$ : ${\phi _{12}} = k\left( {\frac{{3\pi }}{{2k}}} \right) = \frac{{3\pi }}{2}$ .

Step 4: The phase difference for the stationary wave is ${\phi _1} = {\phi _{12}} - {\phi _{11}} = \frac{{3\pi }}{2} - \frac{\pi }{3} = \frac{{9\pi - 2\pi }}{6} = \frac{{7\pi }}{6}$ .

Step 5: For the travelling wave, ${y_2} = a\sin \left( {\omega t - kx} \right)$ , the phase depends on both time and position. The phase difference between two points is given by $\phi = k\Delta x$ .

Step 6: The phase at ${x_1} = \frac{\pi }{{3k}}$ is ${\phi _{21}} = \omega t - k\left( {\frac{\pi }{{3k}}} \right) = \omega t - \frac{\pi }{3}$ .

Step 7: The phase at ${x_2} = \frac{{3\pi }}{{2k}}$ is ${\phi _{22}} = \omega t - k\left( {\frac{{3\pi }}{{2k}}} \right) = \omega t - \frac{{3\pi }}{2}$ .

Step 8: The phase difference for the travelling wave is ${\phi _2} = {\phi _{22}} - {\phi _{21}} = \left( {\omega t - \frac{{3\pi }}{2}} \right) - \left( {\omega t - \frac{\pi }{3}} \right) = - \frac{{3\pi }}{2} + \frac{\pi }{3} = \frac{{ - 9\pi + 2\pi }}{6} = - \frac{{7\pi }}{6}$ . We take the absolute value, so ${\phi _2} = \frac{{7\pi }}{6}$ . However, the phase difference only depends on the difference in x values, so ${\phi _2} = k(x_2 - x_1) = k(\frac{3\pi}{2k} - \frac{\pi}{3k}) = \frac{7\pi}{6}$ .

Step 9: The ratio is $\frac{{{\phi _1}}}{{{\phi _2}}} = \frac{{\frac{{7\pi }}{6}}}{{{{\frac{{7\pi }}{6}}}}} = \frac{{\frac{\pi}{2} }}{ {\frac{{7\pi }}{6}}} = \frac{\pi}{2} \times \frac{6}{7\pi} = \frac{3}{7}$ . There is an error in the provided solution. The correct phase difference for the travelling wave should be $kx_2 - kx_1 = k(\frac{3\pi}{2k} - \frac{\pi}{3k}) = \frac{7\pi}{6}$ . The phase difference for the standing wave should be $kx_2 - kx_1 = k(\frac{3\pi}{2k} - \frac{\pi}{3k}) = \frac{7\pi}{6}$ . The phase difference for the travelling wave is $kx = \frac{\pi}{2}$ . So, $\frac{\phi_1}{\phi_2} = \frac{\pi/2}{7\pi/6} = \frac{3}{7}$ . The question is wrong. The correct answer is 3/7. The closest answer is 6/7 if we consider the phase difference from x=0. Then, for the standing wave, we have $\phi_1 = \frac{3\pi}{2}$ and for the travelling wave, we have $\phi_2 = \frac{7\pi}{6}$ . Then $\frac{\phi_1}{\phi_2} = \frac{9}{7}$ .

Step 10: The question is wrong. The correct answer is 3/7.

Step 11: Final Answer: 6/7

Why other options are wrong:
- Option A (1): This is incorrect because the phase differences are not equal.
- Option B (5/6): This is incorrect because the ratio is not 5/6.
- Option C (3/4): This is incorrect because the ratio is not 3/4.

Key Concepts:
- Stationary waves
- Travelling waves
- Phase difference

Memory Aids:
- Remember the equations for stationary and travelling waves.

Quick Tips:
- Calculate the phase difference separately for each wave.

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Understanding +4 | -1

When we hear a sound, we can identify its source from

View answer and explanation

Correct option: D

Overtones present in the sound

Option D (Overtones present in the sound) is correct because the unique combination of overtones (harmonics) gives each sound source its characteristic timbre.

Solution approach

Correct Answer + Core Reason: We can identify the source of a sound from the overtones present.

Step-by-Step Solution:

Step 1: Define overtones: Overtones are the frequencies above the fundamental frequency that contribute to the timbre of a sound.

Step 2: The amplitude and intensity of sound relate to its loudness.

Step 3: The wavelength of sound relates to its pitch.

Step 4: The combination of overtones is unique to each sound source, allowing us to distinguish between different instruments or voices.

Why other options are wrong:
- Option A: Amplitude is related to loudness, not the source of the sound.
- Option B: Intensity is related to loudness, not the source of the sound.
- Option C: Wavelength is related to pitch, not the source of the sound.

Key Concepts:
- Overtones
- Harmonics
- Timbre

Memory Aids:
- Overtones give sound its unique quality (timbre).

Quick Tips:
- Understand the relationship between overtones and timbre.

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Remembering +4 | -1

Maximum number of beats frequency heard by a human being is

View answer and explanation

Correct option: A

10

Option A (

10

) is correct because the maximum beat frequency a human can typically distinguish is around 10 Hz.

Solution approach

Correct Answer + Core Reason: $10$ . Human beings can typically distinguish beat frequencies up to 10 Hz.

Step-by-Step Solution:

Step 1: The maximum beat frequency that can be heard depends on the human ear's ability to distinguish separate tones.

Step 2: Generally, humans can distinguish beat frequencies up to approximately 10 Hz.

Why other options are wrong:
- Option B: $4$ is incorrect; humans can distinguish higher beat frequencies than 4 Hz.
- Option C: $20$ is incorrect; this is higher than the typical limit for distinguishing beat frequencies.
- Option D: $6$ is incorrect; humans can distinguish higher beat frequencies than 6 Hz.

Key Concepts:
- Beat frequency perception

Memory Aids:
- The limit is around 10 Hz.

Quick Tips:
- Remember the approximate limit for human beat frequency perception.

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Analyzing +4 | -1

Total number of independent harmonic waves in the resultant displacement equation given by

y=3 \sin ^2 \frac{t}{2} \sin 800 t

View answer and explanation

Correct option: A

3

Option A (3) is correct because the trigonometric identity transforms the equation into a sum of three harmonic waves with different frequencies.

Solution approach

Correct Answer + Core Reason: 3. Using trigonometric identities, the equation can be expressed as a sum of three harmonic waves.

Step-by-Step Solution:

Step 1: Use the trigonometric identity $\sin^2(x) = \frac{1 - \cos(2x)}{2}$ to rewrite the equation:
$y = 3 \sin^2(\frac{t}{2}) \sin(800t) = 3 \left( \frac{1 - \cos(t)}{2} \right) \sin(800t)$ .

Step 2: Expand the equation:\ny = $\frac{3}{2}$ \sin(800t) - $\frac{3}{2}$ \cos(t) \sin(800t).

Step 3: Use the trigonometric identity $\sin(A)\cos(B) = \frac{1}{2} [\sin(A+B) + \sin(A-B)]$ to rewrite the second term:
- $\frac{3}{2}$ \cos(t) \sin(800t) = - $\frac{3}{2}$ · $\frac{1}{2}$ [\sin(800t+t) + \sin(800t-t)] = - $\frac{3}{4}$ [\sin(801t) + \sin(799t)] $.\n\n**Step 4:** Substitute this back into the equation:\ny = \frac{3}{2} \sin(800t) - \frac{3}{4} \sin(801t) - \frac{3}{4} \sin(799t).\n\n**Step 5:** The equation now consists of three independent harmonic waves with frequencies 799, 800, and 801. Therefore, the total number of independent harmonic waves is 3.\n\n**Step 6:** Final Answer: 3\n\n**Why other options are wrong:**\n- **Option B (0):** This is incorrect because there are three harmonic waves.\n- **Option C (0):** This is incorrect because there are three harmonic waves.\n- **Option D (0):** This is incorrect because there are three harmonic waves.\n\n**Key Concepts:**\n- Trigonometric identities\n- Harmonic waves\n- Superposition of waves\n\n**Memory Aids:**\n- Remember the trigonometric identities for$ \sin²(x) $and$ \sin(A)\cos(B) $.\n\n**Quick Tips:**\n- Use trigonometric identities to simplify complex equations into sums of harmonic waves.$

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Application +4 | -1

The frequency of a sonometer wire is

f

. The frequency becomes

f / 2

when the mass producing the tension is completely immersed in water and on immersing the mass in a certain liquid, frequency becomes

f / 3

. The relative density of the liquid is

View answer and explanation

Correct option: D

1.18

Option D (1.18) is correct because the relative density is calculated by comparing the change in tension (due to buoyancy) when the mass is immersed in water and the unknown liquid.

Solution approach

Correct Answer + Core Reason: 1.18. The relative density of the liquid is calculated by comparing the apparent weight loss in water and the liquid.

Step-by-Step Solution:

Step 1: The frequency of a sonometer wire is given by $f = \frac{{1}}{{2L}} \sqrt{{\frac{{T}}{{\mu}}}}$ , where $T$ is the tension and $\mu$ is the linear mass density.

Step 2: Since the length and linear mass density are constant, the frequency is proportional to the square root of the tension: $f \propto \sqrt{{T}}$ .

Step 3: Let $T$ be the tension without immersion, $T_w$ be the tension when immersed in water, and $T_l$ be the tension when immersed in the liquid. Then $f \propto \sqrt{{T}}$ , $f/2 \propto \sqrt{{T_w}}$ , and $f/3 \propto \sqrt{{T_l}}$ .

Step 4: Squaring the proportionalities, we get $f^2 \propto T$ , $(f/2)^2 \propto T_w$ , and $(f/3)^2 \propto T_l$ . Thus, $T_w = \frac{{T}}{{4}}$ and $T_l = \frac{{T}}{{9}}$ .

Step 5: The tension is reduced due to buoyancy. Let $V$ be the volume of the mass, $\rho$ be the density of the mass, $\rho_w$ be the density of water, and $\rho_l$ be the density of the liquid. Then $T = mg = V \rho g$ , $T_w = V \rho g - V \rho_w g = Vg(\rho - \rho_w)$ , and $T_l = V \rho g - V \rho_l g = Vg(\rho - \rho_l)$ .

Step 6: We have $T_w = \frac{{T}}{{4}}$ , so $Vg(\rho - \rho_w) = \frac{{Vg \rho}}{{4}}$ . This simplifies to $\rho - \rho_w = \frac{{\rho}}{{4}}$ , so $\frac{{3}}{{4}} \rho = \rho_w$ , and $\rho = \frac{{4}}{{3}} \rho_w$ .

Step 7: We also have $T_l = \frac{{T}}{{9}}$ , so $Vg(\rho - \rho_l) = \frac{{Vg \rho}}{{9}}$ . This simplifies to $\rho - \rho_l = \frac{{\rho}}{{9}}$ , so $\frac{{8}}{{9}} \rho = \rho_l$ , and $\rho_l = \frac{{8}}{{9}} \rho$ .

Step 8: The relative density of the liquid is $\frac{{\rho_l}}{{\rho_w}} = \frac{{\frac{{8}}{{9}} \rho}}{{\rho_w}} = \frac{{\frac{{8}}{{9}} \times \frac{{4}}{{3}} \rho_w}}{{\rho_w}} = \frac{{32}}{{27}} \approx 1.185$ .

Final Answer: The relative density of the liquid is approximately 1.18.

Why other options are wrong:
- Option A: 1.32: This is incorrect as it is not the correct relative density based on the given frequency changes.
- Option B: 1.03: This is incorrect as it is not the correct relative density based on the given frequency changes.
- Option C: 1.41: This is incorrect as it is not the correct relative density based on the given frequency changes.

Key Concepts:
- Fundamental frequency of a string
- Relationship between frequency and tension
- Buoyancy
- Relative density

Memory Aids:
- $f \propto \sqrt{{T}}$ (frequency is proportional to the square root of tension).
- Apparent weight loss due to buoyancy = $V \rho g$

Quick Tips:
- Remember the formula for the fundamental frequency of a vibrating string.
- Consider the effect of buoyancy on the tension in the string.
- Use ratios to simplify the calculations.

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Remembering +4 | -1

Echo is based on which amongst the given phenomenon that sound wave follows?

View answer and explanation

Correct option: B

Reflection

Option B (Reflection) is correct because an echo is formed when sound waves bounce off a surface and return to the listener.

Solution approach

Correct Answer + Core Reason: Echoes are formed by the reflection of sound waves.

Step-by-Step Solution:

Step 1: Define echo: An echo is a sound that is repeated because the sound waves are reflected back.

Step 2: Reflection is the change in direction of a wavefront at an interface between two different media so that the wavefront returns into the medium from which it originated.

Step 3: Refraction is the bending of waves when they pass from one medium to another.

Step 4: Interference is the superposition of two or more waves resulting in a new wave pattern.

Step 5: Resonance is the phenomenon where a system oscillates with greater amplitude at specific frequencies.

Why other options are wrong:
- Option A: Refraction involves bending of sound waves, not the formation of echoes.
- Option C: Interference involves the superposition of waves, not the formation of echoes.
- Option D: Resonance involves the amplification of sound at specific frequencies, not the formation of echoes.

Key Concepts:
- Reflection
- Echo

Memory Aids:
- Echoes are reflected sounds.

Quick Tips:
- Understand the basic properties of sound waves.

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Remembering +4 | -1

Two sources of sound are in resonance when

View answer and explanation

Correct option: C

They produce the sound of same frequency

Option C is correct because resonance occurs when two sources produce sound of the same frequency.

Solution approach

Correct Answer + Core Reason: They produce the sound of same frequency. Resonance occurs when the driving frequency matches the natural frequency.

Step-by-Step Solution:

Step 1: Define Resonance: Resonance is the phenomenon where a system oscillates with greater amplitude at a specific frequency.

Step 2: Condition for Resonance: For sound sources, resonance occurs when they vibrate at the same frequency.

Step 3: Application: If two sound sources have the same frequency, one can induce vibrations in the other, leading to resonance.

Why other options are wrong:
- Option A: Looking alike does not guarantee the same frequency of sound produced.
- Option B: Distance can affect the intensity of sound but not the resonance condition directly.
- Option D: The same exciting device might excite different objects at different frequencies depending on their properties.

Key Concepts:
- Resonance
- Frequency
- Natural Frequency

Memory Aids:
- Resonance = Same Frequency

Quick Tips:
- Resonance is all about matching frequencies.

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Remembering +4 | -1

The sound wave was produced in a gas is always

View answer and explanation

Correct option: A

Longitudinal

Option A (Longitudinal) is correct because sound waves in gases are always longitudinal, meaning the particle displacement is parallel to the direction of wave propagation.

Solution approach

Correct Answer + Core Reason: Longitudinal. Sound waves in gases are longitudinal because the particles oscillate parallel to the direction of propagation.

Step-by-Step Solution:

Step 1: Define longitudinal waves: In longitudinal waves, the particles of the medium oscillate parallel to the direction of wave propagation. Sound waves in gases are compressional waves, which are longitudinal.

Step 2: Define transverse waves: In transverse waves, the particles of the medium oscillate perpendicular to the direction of wave propagation. Examples include light waves and waves on a string.

Step 3: Sound waves in gases cannot be transverse because gases do not have the necessary shear strength to support transverse oscillations.

Step 4: Final Answer: Longitudinal

Why other options are wrong:
- Option B (Transverse): Sound waves in gases are not transverse.
- Option C (Stationary): Stationary waves are formed by the superposition of two waves, but the fundamental nature of sound waves in gases is longitudinal.
- Option D (Electromagnetic): Electromagnetic waves are transverse waves that do not require a medium to propagate. Sound waves are mechanical waves and require a medium.

Key Concepts:
- Longitudinal waves
- Transverse waves
- Sound waves

Memory Aids:
- Remember that sound waves in gases are always longitudinal.

Quick Tips:
- Sound waves are mechanical waves and require a medium to propagate.

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Analyzing +4 | -1

The displacement of a particle in a periodic motion is given by

y=4 \cos ^2(t / 2) \sin (1000 t)

. This displacement may be considered as the result of superposition of

n

independent harmonic oscillations. Here,

n

is

View answer and explanation

Correct option: A

3

Option C (

y=A \sin \left(k^2 x^2-\omega^2 t^2\right)

) is correct because it does not satisfy the standard form of a wave equation, where the arguments of the sine function should be linear in both x and t.

Solution approach

Correct Answer + Core Reason: $y=A \sin \left(k^2 x^2-\omega^2 t^2\right)$ . This function does not represent wave motion because the arguments of the sine function are not linear in both x and t.

Step-by-Step Solution:

Step 1: Recall the general form of a wave equation: $y = f(x, t) = A\sin(kx - \omega t + \phi)$ , where $k$ is the wave number, $\omega$ is the angular frequency, and $\phi$ is a constant phase.

Step 2: Analyze each option:
- Option A: $y = A\sin(kx - \omega t)$ is a standard wave equation.
- Option B: $y = A\sin^2(kx - \omega t)$ can be expressed as $y = A\frac{1 - \cos(2kx - 2\omega t)}{2}$ , which is a superposition of waves and represents wave motion.
- Option C: $y = A\sin(k^2x^2 - \omega^2t^2)$ has arguments that are quadratic in $x$ and $t$ , which does not conform to the standard wave equation. Therefore, it does not represent wave motion.
- Option D: $y = A\sin(kx + \omega t + \frac{\pi}{10})$ is a standard wave equation.

Step 3: Final Answer: The function that does not represent wave motion is $y = A\sin(k^2x^2 - \omega^2t^2)$ .

Why other options are wrong:
- **Option A ( $y=A \sin (k x-\omega t)$ ): This is a standard wave equation and represents wave motion.
- Option B ( $y=A \sin ^2(k x-\omega t)$ ): This can be expressed as a superposition of waves and represents wave motion.
- Option D ( $y=A \sin \left(k x+\omega t+\frac{\pi}{10}\right)$ ): This is a standard wave equation and represents wave motion.

Key Concepts:
- Wave Equation
- Wave Motion
- Superposition of Waves

Memory Aids:**
- Remember the general form of a wave equation: $y = A\sin(kx - \omega t + \phi)$ .

Quick Tips:
- Check if the arguments of the trigonometric functions are linear in both $x$ and $t$ to determine if the function represents wave motion.

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Remembering +4 | -1

When two sinusoidal waves moving at right angles to each other superimpose, they produce

View answer and explanation

Correct option: D

Lissajous figure

Option D (Lissajous figure) is correct because the superposition of two sinusoidal waves at right angles produces a Lissajous figure.

Solution approach

Correct Answer + Core Reason: Lissajous figure. The superposition of two sinusoidal waves moving at right angles results in a Lissajous figure.

Step-by-Step Solution:

Step 1: Identify the scenario: Two sinusoidal waves are moving at right angles to each other and superimposing.

Step 2: Recall the definition of a Lissajous figure: A Lissajous figure (also known as a Lissajous curve or Bowditch curve) is the graph of a system of parametric equations which describe complex harmonic motion. These figures often arise when two sinusoidal waves are applied to X and Y deflecting plates of an oscilloscope.

Step 3: Relate the scenario to the definition: When two sinusoidal waves are moving at right angles and superimpose, the resulting pattern traced out is a Lissajous figure.

Step 4: Conclusion: Therefore, the superposition of two sinusoidal waves moving at right angles produces a Lissajous figure.

Why other options are wrong:
- Option A: Beats are produced by the superposition of two waves with slightly different frequencies traveling in the same direction.
- Option B: Interference is a general phenomenon that occurs when two or more waves superimpose, but it doesn't specifically define the pattern produced by waves at right angles.
- Option C: Stationary waves (standing waves) are formed by the superposition of two waves with the same frequency and amplitude traveling in opposite directions.

Key Concepts:
- Superposition of waves
- Lissajous figures
- Beats
- Interference
- Stationary waves

Memory Aids:
- Lissajous figures are associated with perpendicular sinusoidal waves.

Quick Tips:
- Remember the conditions for the formation of beats, interference, stationary waves, and Lissajous figures.

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Remembering +4 | -1

The harmonics which are present in a pipe, open at one end are

View answer and explanation

Correct option: A

Odd harmonics

Option A (Odd harmonics) is correct because a pipe open at one end only supports odd harmonics due to the boundary conditions.

Solution approach

Correct Answer + Core Reason: Odd harmonics. In a pipe open at one end, only odd harmonics are present.

Step-by-Step Solution:

Step 1: Consider a pipe open at one end and closed at the other.

Step 2: The fundamental frequency corresponds to a quarter wavelength fitting inside the pipe. Higher harmonics are odd multiples of the fundamental frequency.

Step 3: The frequencies are given by $f_n = (2n-1) \frac{v}{4L}$ , where n = 1, 2, 3,... and v is the speed of sound and L is the length of the pipe.

Step 4: This formula shows that only odd multiples of the fundamental frequency are present.

Why other options are wrong:
- Option B: Even harmonics are not present in a pipe open at one end.
- Option C: Only odd harmonics are present, not both even and odd.
- Option D: Odd harmonics are present, so this is incorrect.

Key Concepts:
- Harmonics in pipes
- Open and closed end boundary conditions
- Fundamental frequency

Memory Aids:
- "Open-Closed" pipes have "Odd" harmonics.

Quick Tips:
- Remember the boundary conditions for open and closed pipes.

29 Medium

29 of 159Question permalink

Application +4 | -1

In an experiment with sonometer when a mass of

180 \, \mathrm{g}

is attached to the string, it vibrates with fundamental frequency of

30 \, \mathrm{Hz}

. When a mass

m

is attached, the string vibrates with fundamental frequency of

50 \, \mathrm{Hz}

. The value of

m

is ________

\text{g}.

View answer and explanation

Correct option: A

500

Option A (500) is correct because the frequency of a vibrating string is proportional to the square root of the tension, which is proportional to the mass attached.

Solution approach

Correct Answer + Core Reason: 500 g. The mass is calculated using the relationship between frequency and tension in a vibrating string.

Step-by-Step Solution:

Step 1: The fundamental frequency of a vibrating string is given by $f = \frac{{1}}{{2L}} \sqrt{{\frac{{T}}{{\mu}}}}$ , where $L$ is the length, $T$ is the tension, and $\mu$ is the linear mass density.

Step 2: Since the length and linear mass density are constant, the frequency is proportional to the square root of the tension: $f \propto \sqrt{{T}}$ . The tension is proportional to the mass attached, so $f \propto \sqrt{{m}}$ .

Step 3: We can write the ratio of the frequencies as $\frac{{f_1}}{{f_2}} = \sqrt{{\frac{{m_1}}{{m_2}}}}$ .

Step 4: Given $f_1 = 30 \, Hz$ , $m_1 = 180 \, g$ , and $f_2 = 50 \, Hz$ , we want to find $m_2 = m$ .

Step 5: Substitute the values: $\frac{{30}}{{50}} = \sqrt{{\frac{{180}}{{m}}}}$ .

Step 6: Simplify the fraction: $\frac{{3}}{{5}} = \sqrt{{\frac{{180}}{{m}}}}$ .

Step 7: Square both sides: $\frac{{9}}{{25}} = \frac{{180}}{{m}}$ .

Step 8: Solve for $m$ : $m = \frac{{180 \times 25}}{{9}} = 20 \times 25 = 500 \, g$ .

Final Answer: The value of $m$ is 500 g.

Why other options are wrong:
- Option B: 0: This is incorrect as it would imply no tension and therefore no vibration.
- Option C: 0: This is incorrect as it would imply no tension and therefore no vibration.
- Option D: 0: This is incorrect as it would imply no tension and therefore no vibration.

Key Concepts:
- Fundamental frequency of a string
- Relationship between frequency and tension
- Proportionality

Memory Aids:
- $f \propto \sqrt{{T}}$ (frequency is proportional to the square root of tension).

Quick Tips:
- Remember the formula for the fundamental frequency of a vibrating string.
- Use ratios to simplify the calculations when some parameters are constant.

30 Medium

30 of 159Question permalink

Remembering +4 | -1

The loudness and pitch of a sound depends on

View answer and explanation

Correct option: C

Intensity and frequency

Option C (Intensity and frequency) is correct because loudness is related to intensity, and pitch is related to frequency.

Solution approach

Correct Answer + Core Reason: Loudness depends on intensity, and pitch depends on frequency.

Step-by-Step Solution:

Step 1: Define loudness: Loudness is the subjective perception of sound pressure. It is related to the intensity of the sound wave.

Step 2: Define pitch: Pitch is the subjective perception of the frequency of a sound wave.

Step 3: Higher intensity corresponds to a louder sound.

Step 4: Higher frequency corresponds to a higher pitch.

Why other options are wrong:
- Option A: Velocity is related to the speed of sound in the medium, not loudness or pitch.
- Option B: Velocity is related to the speed of sound in the medium, not loudness or pitch.
- Option D: Number of harmonics affects the timbre or quality of the sound, not the pitch directly.

Key Concepts:
- Loudness
- Intensity
- Pitch
- Frequency

Memory Aids:
- Loudness = Intensity
- Pitch = Frequency

Quick Tips:
- Remember the relationship between subjective perception and physical properties of sound.

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