Amplitude is zero at all the points

A progressive wave ${\rm{y = a}}\,{\rm{sin}}\left[ {\left( {{\rm{kx}} - {\rm{\omega t}}} \right)} \right]$ is reflected by a rigid wall at ${\rm{x}} = 0$. Then the reflected wave can be represented by

$\begin{array}{*{20}{l}} {{\rm{y = }} - {\rm{asin}}\left( {{\rm{kx + \omega t}}} \right)} \end{array}$

${\rm{y = a}}\,{\rm{sin}}\left( {{\rm{kx + \omega t}}} \right)$

$\begin{array}{*{20}{l}} {{\rm{y = a}}\,{\rm{cos}}\left( {{\rm{kx + \omega t}}} \right)} \end{array}$

$\begin{array}{*{20}{l}} {{\rm{y}} = - {\rm{asin}}\left( {{\rm{kx}} - {\rm{\omega t}}} \right)} \end{array}$

Two Cu wires of radii ${{\rm{R}}_1}\,\,{\rm{and}}\,\,{{\rm{R}}_2}$ such that $\left( {{R_1} > {R_2}} \right)$. Then which of the following is true?

Transverse wave travels after in thicker wire

Travels with the same speed in both the wires

WAVES MCQs for NEET UG PHYSICS — Medium, Page 5

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Remembering +4 | -1

The speed of sound in gas of density

ρ

at a pressure

p

is proportional to

View answer and explanation

Correct option: D

\sqrt {\frac{p}{\mathrm{\rho }}}

Option D (

\sqrt {\frac{p}{\mathrm{\rho }}}

) is correct because the speed of sound in a gas is proportional to the square root of the pressure divided by the density.

Solution approach

Correct Answer + Core Reason: $\sqrt {\frac{p}{\mathrm{\rho }}}$ . The speed of sound is proportional to the square root of pressure divided by density.

Step-by-Step Solution:

Step 1: Recall the formula for the speed of sound in a gas.

Step 2: The speed of sound in a gas is given by the formula $v = \sqrt{\frac{{\gamma p}}{\rho}}$ , where $v$ is the speed of sound, $p$ is the pressure, $\rho$ is the density, and $\gamma$ is the adiabatic index (a constant for a given gas).

Step 3: Since $\gamma$ is a constant, the speed of sound $v$ is proportional to $\sqrt{\frac{p}{\rho}}$ .

Step 4: Therefore, the correct answer is $\sqrt{\frac{p}{\rho}}$ .

Why other options are wrong:
- Option A: ${\left( {\frac{p}{\mathrm{\rho }}} \right)^2}$ is incorrect because the speed of sound is proportional to the square root, not the square, of the ratio.
- Option B: ${\left( {\frac{p}{\mathrm{\rho }}} \right)^{\frac{3}{2}}}$ is incorrect because the speed of sound is proportional to the square root, not the 3/2 power, of the ratio.
- Option C: $\sqrt {\frac{\mathrm{\rho }}{\mathrm{p}}}$ is incorrect because the speed of sound is proportional to the square root of pressure divided by density, not the inverse.

Key Concepts:
- Speed of sound in a gas
- Pressure
- Density
- Adiabatic index

Memory Aids:
- Remember the formula: Speed of Sound = Square Root of (Pressure / Density).

Quick Tips:
- Always remember the correct formula for the speed of sound in a gas.

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A pulse or a wave train travels along a stretched string and reaches the fixed end of the string. It will be reflected back with

View answer and explanation

Correct option: D

A phase change of

180^\circ

with velocity reversed

Option D (A phase change of

180^\circ

with velocity reversed) is correct because when a pulse reflects from a fixed end, it undergoes a phase change of 180 degrees and the velocity is reversed.

Solution approach

Correct Answer + Core Reason: A phase change of $180^\circ$ with velocity reversed. Reflection from a fixed end results in a phase change of 180 degrees and a reversal of velocity.

Step-by-Step Solution:

Step 1: Recall the behavior of a wave pulse at a fixed end.

Step 2: When a wave pulse encounters a fixed end, it is reflected. The reflection causes a phase change of 180 degrees (or $\pi$ radians). This is because the fixed end cannot move, so the reflected wave must be inverted to cancel out the incident wave at that point.

Step 3: The velocity of the reflected pulse is also reversed in direction.

Why other options are wrong:
- Option A: The same phase as the incident pulse but with velocity reversed. This is incorrect because there is a phase change of 180 degrees.
- Option B: A phase change of $180^\circ$ with no reversal of velocity. This is incorrect because the velocity is also reversed.
- Option C: The same phase as the incident pulse with no reversal of velocity. This is incorrect because there is a phase change of 180 degrees and the velocity is reversed.

Key Concepts:
- Reflection of waves
- Phase change upon reflection
- Fixed end reflection

Memory Aids:
- Remember that a fixed end inverts the wave (180-degree phase change) and reverses the velocity.

Quick Tips:
- Distinguish between reflection from a fixed end and a free end.

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If wave

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

is moving along

x - axis

, the shape of pulse at

t = 0\,\,and\,\,t = 2\,s

View answer and explanation

Correct option: B

Are same

Option B (Are same) is correct because the shape of the pulse is determined by the cosine function, and the cosine function repeats its shape after a certain time period if no changes are made to the wave.

Solution approach

Correct Answer + Core Reason: Are same. The shape of the pulse is determined by the cosine function, which remains the same if no parameters change.

Step-by-Step Solution:

Step 1: The wave equation is given by $y = a \cos(\omega t + kx)$ .

Step 2: At $t = 0$ , the shape of the pulse is $y = a \cos(kx)$ .

Step 3: At $t = 2s$ , the shape of the pulse is $y = a \cos(2\omega + kx)$ .

Step 4: Since the wave is moving along the x-axis and no parameters have changed, the shape of the pulse remains the same, only its position changes.

Why other options are wrong:
- Option A: Are different is incorrect because the shape of the pulse is determined by the cosine function, which remains the same if no parameters change.
- Option C: May not be same is incorrect because the shape will be same as the wave is progressing without any change in its properties.
- Option D: None of these is incorrect because the shape is the same.

Key Concepts:
- Wave propagation
- Wave shape

Memory Aids:
- The shape of a wave remains the same as it propagates, unless there are changes in the medium or wave parameters.

Quick Tips:
- Visualize the wave as it moves along the x-axis. The shape of the wave remains constant.

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Two tuning forks of frequency 250 Hz and 256 Hz produce beats. Then beat frequency is

View answer and explanation

Correct option: A

6 Hz

Option A (6 Hz) is correct because the beat frequency is the absolute difference between the two frequencies.

Solution approach

Correct Answer + Core Reason: 6 Hz. The beat frequency is the difference between the two frequencies.

Step-by-Step Solution:

Step 1: Identify the frequencies: $f_1 = 250 \text{ Hz}$ and $f_2 = 256 \text{ Hz}$ .

Step 2: Calculate the beat frequency using the formula: $f_{beat} = |f_1 - f_2|$ .

Step 3: Substitute the values: $f_{beat} = |250 - 256| = 6 \text{ Hz}$ .

Step 4: Final Answer: 6 Hz

Why other options are wrong:
- Option B: 12 Hz is incorrect; it might be a miscalculation of the frequencies or their difference.
- Option C: 18 Hz is incorrect; it might be a miscalculation of the frequencies or their difference.
- Option D: 21 Hz is incorrect; it might be a miscalculation of the frequencies or their difference.

Key Concepts:
- Beat frequency: $f_{beat} = |f_1 - f_2|$

Memory Aids:
- Beat frequency is always the absolute difference.

Quick Tips:
- Remember to take the absolute difference.

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Analyzing +4 | -1

If

x=a \sin [\omega t+(\pi / 3)]

and

x^{\prime}=a \cos \omega t

, then what is the phase difference between two waves?

View answer and explanation

Correct option: B

\quad \pi / 6

Option B (

\quad \pi / 6

) is correct because it represents the phase difference between the two given sinusoidal waves after converting cosine to sine.

Solution approach

Correct Answer + Core Reason: $\quad \pi / 6$ . The phase difference is calculated by converting cosine to sine and comparing the phase constants.

Step-by-Step Solution:

Step 1: Given $x = a \sin [\omega t + (\pi / 3)]$

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Assertion (A): When a travelling wave is sent along a particular string by oscillating one end, the speed of the wave will increase if we increase the frequency of oscillations.
Reason (R): If a travelling wave sent along a particular string by oscillating its one end, it is the wave length of the wave that decreases.

View answer and explanation

Correct option: D

Assertion is false but Reason is true

Assertion is false but Reason is true. The speed of a wave on a string depends on the tension and mass per unit length, not the frequency. The wavelength decreases when the frequency increases, keeping the speed constant.

Solution approach

Correct Answer + Core Reason: Assertion is false but Reason is true. The speed of a wave depends on the medium, not the frequency.

Step-by-Step Solution:

Step 1: The speed of a 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 2: The speed of the wave depends only on the properties of the string (tension and mass per unit length) and not on the frequency of the oscillations.

Step 3: Therefore, if we increase the frequency of oscillations, the speed of the wave will not increase. The assertion is false.

Step 4: The relationship between speed, frequency, and wavelength is given by $v = f\lambda$ , where $f$ is the frequency and $\lambda$ is the wavelength.

Step 5: If the speed remains constant and the frequency increases, the wavelength must decrease to maintain the relationship $v = f\lambda$ .

Step 6: Therefore, the wavelength of the wave decreases when the frequency increases. The reason is true.

Why other options are wrong:
- Option A (Both Assertion and Reason are true and Reason is the correct explanation of Assertion): This is incorrect because the assertion is false.
- Option B (Both Assertion and Reason are true but Reason is not the correct explanation of Assertion): This is incorrect because the assertion is false.
- Option C (Assertion is true but Reason is false): This is incorrect because the assertion is false and the reason is true.

Key Concepts:
- Wave speed on a string
- Frequency, wavelength, and speed relationship

Memory Aids:
- Remember the formula for wave speed on a string: $v = \sqrt{\frac{T}{\mu}}$ .
- Remember the relationship between speed, frequency, and wavelength: $v = f\lambda$ .

Quick Tips:
- The speed of a wave on a string depends only on the properties of the string, not the frequency.

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Application +4 | -1

y = 5\,\,\sin \left( {30\pi t - \frac{\pi }{7} + 30^\circ } \right)y \to mm,t \to s,x \to m

. for given progressive wave equation, phase difference between two vibrating particle having path difference

3.5\mathrm{ }m

would be

View answer and explanation

Correct option: D

\pi /2

Option D (

\pi /2

) is correct because the phase difference is calculated using the path difference and the wave number.

Solution approach

Correct Answer + Core Reason: $\pi /2$ . The phase difference is calculated using the path difference and the wave number.

Step-by-Step Solution:

Step 1: Identify the given information: The equation of the wave is $y = 5\,\sin(30\pi t - \frac{\pi}{7} + 30^\circ)$ , and the path difference is $\Delta x = 3.5\mathrm{ }\,m$ .

Step 2: The given equation doesn't directly provide the wave number $k$ . We need more information to find $k$ or the wavelength $\lambda$ .\nHowever, the question seems to have an error. The term $\frac{\pi}{7}$ is a constant phase, and $30^\circ$ is also a constant phase. The path difference is given as 3.5 m, but there's no 'x' term in the equation to relate it to. We can assume there's a missing 'kx' term in the equation. Let's assume the equation is meant to be $y = 5\,\sin(30\pi t - kx - \frac{\pi}{7} + 30^\circ)$ .

Step 3: Without the 'kx' term, we cannot directly relate the path difference to the phase difference. Let's assume the question meant to provide the wave number or wavelength indirectly. If we assume $k = \frac{\pi}{7}$ , then $\Delta \phi = k \Delta x = \frac{\pi}{7} \times 3.5 = \frac{\pi}{7} \times \frac{7}{2} = \frac{\pi}{2}$ .

Step 4: Final Answer: The phase difference is $\frac{\pi}{2}$ .

Why other options are wrong:
- **Option A ( $\pi /4$ ): This is incorrect because it might be calculated using an incorrect value for the wave number or an incorrect relationship between path difference and phase difference.
- Option B ( $\pi$ ): This is incorrect because it might be calculated using an incorrect value for the wave number or an incorrect relationship between path difference and phase difference.
- Option C ( $\pi /3$ ): This is incorrect because it might be calculated using an incorrect value for the wave number or an incorrect relationship between path difference and phase difference.

Key Concepts:
- Progressive Wave Equation
- Phase Difference
- Path Difference
- Wave Number

Memory Aids:**
- Remember the relationship: $\Delta \phi = k \Delta x$

Quick Tips:
- Always ensure the wave equation is complete and includes the 'kx' term to relate path difference to phase difference.
- If the wave number is not directly given, try to derive it from other information provided.

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Statement-I: In transverse wave particle velocity is perpendicular to wave velocity.
Statement-II: In a longitudinal wave particle velocity is along the direction of propagation of wave.

View answer and explanation

Correct option: C

Both statement-I

\&

II are correct

Option C (Both statement-I

\& II are correct) is correct because particle velocity is perpendicular to wave velocity in transverse waves and parallel in longitudinal waves.

Solution approach

Correct Answer + Core Reason: Both statement-I $\& II are correct. This accurately describes the relationship between particle and wave velocity in transverse and longitudinal waves.\n\n**Step-by-Step Solution:**\n\n**Step 1:** Recall the definitions of transverse and longitudinal waves.\n\n**Step 2:** In a transverse wave, the particles of the medium vibrate perpendicular to the direction of wave propagation. Therefore, particle velocity is perpendicular to wave velocity.\n\n**Step 3:** In a longitudinal wave, the particles of the medium vibrate parallel to the direction of wave propagation. Therefore, particle velocity is along the direction of propagation of the wave.\n\n**Step 4:** Both statements are correct based on the definitions of transverse and longitudinal waves.\n\n**Why other options are wrong:**\n- **Option A:** Statement-I is correct is incorrect because Statement-II is also correct.\n- **Option B:** Statement-II is correct is incorrect because Statement-I is also correct.\n- **Option D:** Both statement-I$ \& II are incorrect is incorrect because both statements are correct.

Key Concepts:
- Transverse waves
- Longitudinal waves
- Particle velocity
- Wave velocity

Memory Aids:
- Transverse: Perpendicular; Longitudinal: Parallel.

Quick Tips:
- Visualize the motion of particles in both types of waves to understand the relationship between particle and wave velocity.

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Analyzing +4 | -1

A string is stretched between fixed points separated by 75.0 cm. It is observed to have resonant frequencies of 420 Hz and 315 Hz. There are no other resonant frequency for this string is:

View answer and explanation

Correct option: A

105 Hz

Option A (105 Hz) is correct because it is the greatest common divisor of the two resonant frequencies, representing the fundamental frequency.

Solution approach

Correct Answer + Core Reason: 105 Hz. The fundamental frequency is the greatest common divisor of the given resonant frequencies.

Step-by-Step Solution:

Step 1: The resonant frequencies of a string fixed at both ends are given by $f_n = n f_1$ , where $f_1$ is the fundamental frequency and $n$ is an integer.

Step 2: We are given two resonant frequencies: 420 Hz and 315 Hz.

Step 3: Let $420 = n_1 f_1$ and $315 = n_2 f_1$ , where $n_1$ and $n_2$ are integers.

Step 4: We need to find the greatest common divisor (GCD) of 420 and 315, which will be the fundamental frequency $f_1$ .

Step 5: Prime factorization of 420: $420 = 2^2 \times 3 \times 5 \times 7$ .\nPrime factorization of 315: $315 = 3^2 \times 5 \times 7$ .

Step 6: The GCD is $3 \times 5 \times 7 = 105$ .

Step 7: Therefore, the fundamental frequency is 105 Hz.

Why other options are wrong:
- Option B (155 Hz): This is incorrect because it is not a common divisor of 420 and 315.
- Option C (205 Hz): This is incorrect because it is not a common divisor of 420 and 315.
- Option D (245 Hz): This is incorrect because it is not a common divisor of 420 and 315.

Key Concepts:
- Resonant frequencies
- Fundamental frequency
- Greatest common divisor (GCD)

Memory Aids:
- Remember that resonant frequencies are integer multiples of the fundamental frequency.

Quick Tips:
- Find the GCD of the given frequencies to determine the fundamental frequency.

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Remembering +4 | -1

The waves in which the particles of the medium vibrate in a direction perpendicular to the direction of wave motion is known as

View answer and explanation

Correct option: A

Transverse wave

Option A (Transverse wave) is correct because transverse waves are defined by particle vibration perpendicular to the wave's direction of motion.

Solution approach

Correct Answer + Core Reason: Transverse wave. The particles vibrate perpendicular to the wave direction.

Step-by-Step Solution:

Step 1: Define the terms. A transverse wave is a wave where the particle displacement is perpendicular to the direction of wave propagation.

Step 2: Consider the definition of a longitudinal wave. In longitudinal waves, the particle displacement is parallel to the direction of wave propagation.

Step 3: Relate the definition to the question. The question specifically asks for waves where particles vibrate perpendicularly.

Step 4: Conclude that the answer is a transverse wave.

Why other options are wrong:
- Option B: Longitudinal waves involve particle vibration parallel to the wave direction, not perpendicular.
- Option C: Propagated waves simply refer to waves that travel, and do not specify the direction of particle vibration relative to wave motion. All waves propagate, so this is not specific enough.
- Option D: Since option A is correct, this option is incorrect.

Key Concepts:
- Transverse waves
- Longitudinal waves
- Wave propagation

Memory Aids:
- Transverse: Think of \"across\" or \"perpendicular\"
- Longitudinal: Think of \"long\" or \"parallel\"

Quick Tips:
- Remember the definitions of transverse and longitudinal waves. Visualize the particle motion in each type.

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Understanding +4 | -1

In stationary wave

View answer and explanation

Correct option: A

Strain is maximum at nodes

Option A (Strain is maximum at nodes) is correct because in a stationary wave, the displacement is minimum at the nodes, but the strain (change in displacement with respect to position) is maximum at the nodes.

Solution approach

Correct Answer + Core Reason: Strain is maximum at nodes. In a stationary wave, strain is maximum where displacement is minimum (nodes).

Step-by-Step Solution:

Step 1: Understand the concept of strain in a wave.

Step 2: Strain is defined as the change in displacement with respect to position, i.e., the derivative of displacement with respect to position. In a stationary wave, the displacement is minimum at the nodes.

Step 3: At the nodes, the particles are momentarily at rest, but the surrounding particles are experiencing maximum compression or extension, leading to maximum strain.

Why other options are wrong:
- Option B: Strain is maximum at antinodes. This is incorrect because displacement is maximum at antinodes, and strain is minimum.
- Option C: Strain is minimum at nodes. This is incorrect because strain is maximum at nodes.
- Option D: Amplitude is zero at all the points. This is incorrect because amplitude is zero only at the nodes.

Key Concepts:
- Stationary waves
- Nodes and antinodes
- Strain in a wave

Memory Aids:
- Remember that strain is related to the derivative of displacement.

Quick Tips:
- Visualize the displacement and strain in a stationary wave.

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Remembering +4 | -1

The number of waves contained in unit length of the medium is called

View answer and explanation

Correct option: B

Wave number

Option B (Wave number) is correct because wave number is defined as the number of waves per unit length.

Solution approach

Correct Answer + Core Reason: Wave number. Wave number is the number of waves contained in a unit length.

Step-by-Step Solution:

Step 1: Define wave number (k).

Step 2: Wave number is defined as $k = \frac{2\pi}{\lambda}$ , where $\lambda$ is the wavelength.

Step 3: Since wavelength is the length of one wave, the number of waves in unit length is $1/\lambda$ , which is proportional to the wave number.

Why other options are wrong:
- Option A: Elastic wave is a type of wave, not the number of waves per unit length.
- Option C: Wave pulse is a short burst of a wave, not the number of waves per unit length.
- Option D: Electromagnetic wave is a type of wave, not the number of waves per unit length.

Key Concepts:
- Wave number
- Wavelength

Memory Aids:
- Wave number is related to spatial frequency.

Quick Tips:
- Remember the formula for wave number: $k = \frac{2\pi}{\lambda}$

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Analyzing +4 | -1

Given that

y = A\sin \left[ {\left( {\frac{{2\pi }}{\lambda }\left( {ct - x} \right)} \right)} \right]

, where

y

and

x

are measured in metres. Which of the following statements is true

View answer and explanation

Correct option: A

The unit of

{\lambda ^{ - 1}}

is same as that of

\frac{{2\pi }}{\lambda }

Option A (The unit of

{\lambda ^{ - 1}}

is same as that of

\frac{{2\pi }}{\lambda }

) is correct because both terms represent inverse lengths and therefore have the same units.

Solution approach

Correct Answer + Core Reason: The unit of ${\lambda ^{ - 1}}$ is same as that of $\frac{{2\pi }}{\lambda }$ . Both represent inverse lengths.

Step-by-Step Solution:

Step 1: Analyze the given equation: $y = A\sin \left[ {\left( {\frac{{2\pi }}{\lambda }\left( {ct - x} \right)} \right)} \right]$

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Understanding +4 | -1

A progressive wave 

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

is reflected by a rigid wall at

\mathrm{x} = 0

. Then the reflected wave can be represented by

View answer and explanation

Correct option: D

\begin{array}{llllllllllllllllllll} {\mathrm{y = } - \mathrm{asin}\left( {\mathrm{kx + \omega t}} \right)} \end{array}

Option D (

\begin{array}{llllllllllllllllllll}{\mathrm{y = } - \mathrm{asin}\left( {\mathrm{kx + \omega t}} \right)}\end{array}

) is correct because reflection from a rigid wall introduces a phase change of

\pi

, and the wave travels in the opposite direction.

Solution approach

Correct Answer + Core Reason: $\begin{array}{llllllllllllllllllll}{\mathrm{y = } - \mathrm{asin}\left( {\mathrm{kx + \omega t}} \right)}\end{array}$ . Reflection at a rigid wall introduces a phase change of $\pi$ and reverses the direction of propagation.

Step-by-Step Solution:

Step 1: The incident wave is given by $\mathrm{y = a}\,\mathrm{sin}\left[ {\left( {\mathrm{kx} - \mathrm{\omega t}} \right)} \right]$ .

Step 2: Reflection from a rigid wall at $x=0$ introduces a phase change of $\pi$ , which is equivalent to multiplying the wave function by -1.

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

Step 4: Therefore, the reflected wave is $y = -a \sin(-kx - \omega t) = -a \sin(kx + \omega t)$ .

Why other options are wrong:
- Option A: $\mathrm{y = a}\,\mathrm{sin}\left( {\mathrm{kx + \omega t}} \right)$ is incorrect because it doesn't account for the phase change of $\pi$ upon reflection from a rigid wall.
- Option B: $\begin{array}{llllllllllllllllllll}{\mathrm{y = a}\,\mathrm{cos}\left( {\mathrm{kx + \omega t}} \right)}\end{array}$ is incorrect because it changes the sine function to a cosine function, which is not a result of reflection.
- Option C: $\begin{array}{llllllllllllllllllll}{\mathrm{y} = - \mathrm{asin}\left( {\mathrm{kx} - \mathrm{\omega t}} \right)}\end{array}$ is incorrect because it doesn't account for the change in direction of the wave after reflection (x becomes -x).

Key Concepts:
- Wave reflection
- Phase change upon reflection
- Direction of wave propagation

Memory Aids:
- Reflection from a rigid wall: phase change of $\pi$ and reversal of direction.

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

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Application +4 | -1

A tuning fork of frequency

512 \mathrm{~Hz}

is vibrated with a sonometer wire and 6 beats per second are heard. The beat frequency reduces if the tension in the string is slightly increased. The original frequency of vibration of the string is

View answer and explanation

Correct option: A

506 \mathrm{~Hz}

Option A (

506 \mathrm{~Hz}

) is correct because when the tension increases, the frequency increases, meaning the original frequency was lower than the tuning fork's frequency.

Solution approach

Correct Answer + Core Reason: $506 \mathrm{~Hz}$ . The beat frequency decreases when tension increases, implying the original frequency was lower.

Step-by-Step Solution:

Step 1: Identify the given information: Tuning fork frequency $f_t = 512 \mathrm{~Hz}$ , beat frequency $= 6 \mathrm{~Hz}$ .

Step 2: The possible frequencies of the string are $f_s = f_t \pm \text{beat frequency} = 512 \pm 6 \mathrm{~Hz}$ . So, $f_s$ could be $506 \mathrm{~Hz}$ or $518 \mathrm{~Hz}$ .

Step 3: When the tension in the string is increased, the frequency of the string increases. The beat frequency reduces. This means that the original frequency of the string was less than the frequency of the tuning fork.

Step 4: If the original frequency was $518 \mathrm{~Hz}$ , increasing the tension would increase the frequency further away from $512 \mathrm{~Hz}$ , increasing the beat frequency. This contradicts the given information.

Step 5: Therefore, the original frequency must have been $506 \mathrm{~Hz}$ . Increasing the tension would increase the frequency towards $512 \mathrm{~Hz}$ , reducing the beat frequency.

Why other options are wrong:
- **Option B ( $512 \mathrm{~Hz}$ ): If the string frequency was the same as the tuning fork, there would be no beats.
- Option C ( $518 \mathrm{~Hz}$ ): Increasing tension would increase the frequency and thus the beat frequency, contradicting the problem statement.
- Option D ( $524 \mathrm{~Hz}$ ): Increasing tension would increase the frequency and thus the beat frequency, contradicting the problem statement.

Key Concepts:**
- Beat frequency: $|f_1 - f_2|$
- Frequency of a string: $f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$ , where $T$ is tension and $\mu$ is linear mass density.

Memory Aids:
- Beats occur when two frequencies are close but not identical.
- Increasing tension increases frequency.

Quick Tips:
- Always consider both possible frequencies when dealing with beats.
- Think about how changing tension affects frequency.

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Remembering +4 | -1

The speed of sound in a gas of density

\rho

at a pressure

P

is proportional to

View answer and explanation

Correct option: D

\sqrt {\frac{P}{\rho }}

Option D is correct because the speed of sound in a gas is directly proportional to the square root of the pressure divided by the density, as given by the formula

v = \sqrt{{\frac{{P}}{{\rho}}}}

Solution approach

Correct Answer + Core Reason: $\sqrt {\frac{P}{\rho }}$ . The speed of sound in a gas is proportional to the square root of pressure divided by density.

Step-by-Step Solution:

Step 1: Recall the formula for the speed of sound in a gas.

Step 2: The speed of sound in a gas is given by $v = \sqrt{{\frac{{\gamma P}}{{\rho}}}}$ , where $\gamma$ is the adiabatic index, $P$ is the pressure, and $\rho$ is the density.

Step 3: Since $\gamma$ is a constant for a given gas, the speed of sound $v$ is proportional to $\sqrt{{\frac{P}{\rho}}}$ .

Step 4: Therefore, the speed of sound is proportional to $\sqrt{{\frac{P}{\rho}}}$ .

Why other options are wrong:
- Option A: ${\left( {\frac{p}{\rho }} \right)^2}$ This is incorrect because the speed of sound is proportional to the square root, not the square, of the pressure-density ratio.
- Option B: ${\left( {\frac{P}{\rho }} \right)^{3/2}}$ This is incorrect because the speed of sound is proportional to the square root, not the 3/2 power, of the pressure-density ratio.
- Option C: $\sqrt {\frac{\rho }{P}}$ This is incorrect because the speed of sound is proportional to the square root of pressure divided by density, not the inverse.

Key Concepts:
- Speed of sound in a gas

Memory Aids:
- Remember the formula: $v = \sqrt{{\frac{{\gamma P}}{{\rho}}}}$

Quick Tips:
- Always remember the correct formula for the speed of sound in a gas.

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Analyzing +4 | -1

A tuning fork

A

produces

4\,\,beats/\sec

with another tuning fork

B

of frequency

320\,\,Hz

. On filing the fork

A

,

4\,\,beats/\sec

are again heard. The frequency of fork

A

, after filing is

View answer and explanation

Correct option: A

324\,\,Hz

Option A (

324{\mathrm{ }}\,Hz

) is correct because filing the fork increases its frequency, and since the beat frequency remains the same, the original frequency must have been lower than B.

Solution approach

Correct Answer + Core Reason: $324\mathrm{ }\,Hz$ . Filing the fork increases its frequency, and the beat frequency remains the same, indicating the original frequency was lower.

Step-by-Step Solution:

Step 1: Let the frequency of tuning fork A be $f_A$ and the frequency of tuning fork B be $f_B = 320 \mathrm{Hz}$ .\nThe beat frequency is $|f_A - f_B| = 4 \mathrm{Hz}$ .

Step 2: This gives two possibilities: $f_A = 324 \mathrm{Hz}$ or $f_A = 316 \mathrm{Hz}$ .

Step 3: Filing the fork A increases its frequency. Let the new frequency be $f_A'$ .\nThe new beat frequency is $|f_A' - f_B| = 4 \mathrm{Hz}$ .

Step 4: If $f_A = 316 \mathrm{Hz}$ , then filing the fork increases its frequency, so $f_A' > 316 \mathrm{Hz}$ . If the new beat frequency is 4 Hz, then $f_A' = 324 \mathrm{Hz}$ .\nIf $f_A = 324 \mathrm{Hz}$ , then filing the fork increases its frequency, so $f_A' > 324 \mathrm{Hz}$ . But the new beat frequency is 4 Hz, so $|f_A' - 320| = 4$ . This would mean $f_A' = 324 \mathrm{Hz}$ or $f_A' = 316 \mathrm{Hz}$ . Since filing increases the frequency, this case is not possible.

Step 5: Therefore, the original frequency was 316 Hz, and after filing, it became 324 Hz.\nHowever, the question asks for the frequency *after* filing, which is 324 Hz.

Step 6: Final Answer: $324\mathrm{ }\,Hz$

Why other options are wrong:
- Option B: $320\mathrm{ }\,Hz$ : This is incorrect because the beat frequency is 4 Hz, so the frequency of A cannot be 320 Hz.
- Option C: $316\mathrm{ }\,Hz$ : This is the original frequency of A, not the frequency after filing.
- Option D: $314\mathrm{ }\,Hz$ : This is incorrect because filing A would increase its frequency, and the beat frequency remains the same.

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.

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Application +4 | -1

If the distance between successive compressions and rarefraction in a sound wave is 2m and velocity of sound is 360 m/s, then the frequency is

View answer and explanation

Correct option: D

90 Hz

Option D (90 Hz) is correct because the frequency is calculated using the wavelength (twice the distance between compression and rarefaction) and the velocity of sound.

Solution approach

Correct Answer + Core Reason: 90 Hz. The frequency is calculated using the wavelength and velocity of sound.

Step-by-Step Solution:

Step 1: Identify the given information: Distance between successive compressions and rarefactions is 2 m, and the velocity of sound is 360 m/s.

Step 2: Determine the wavelength. The distance between successive compressions and rarefactions is half the wavelength, so $\frac{\lambda}{2} = 2 \text{ m}$ . Therefore, the wavelength $\lambda = 4 \text{ m}$ .

Step 3: Use the formula relating velocity, frequency, and wavelength: $v = f\lambda$ .

Step 4: Solve for the frequency: $f = \frac{v}{\lambda} = \frac{360 \text{ m/s}}{4 \text{ m}} = 90 \text{ Hz}$ .

Why other options are wrong:
- Option A: 180 Hz is incorrect. It might be a result of not dividing the wavelength by 2 or using an incorrect formula.
- Option B: 45 Hz is incorrect. It might be a result of dividing the velocity by twice the wavelength.
- Option C: 120 Hz is incorrect. It might be a result of using a wrong value for the wavelength.

Key Concepts:
- Wavelength
- Frequency
- Velocity of sound
- Relationship between velocity, frequency, and wavelength: $v = f\lambda$

Memory Aids:
- Remember that the distance between successive compressions and rarefactions is half the wavelength.

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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Analyzing +4 | -1

The frequency of fundamental note in an organ pipe is

240\,\,Hz

. On blowing air, frequencies

720\,\,Hz

and

1200\,\,Hz

are heard. This indicates that organ pipe is

View answer and explanation

Correct option: A

A pipe closed at one end

Option A (A pipe closed at one end) is correct because the given frequencies are odd multiples of the fundamental frequency.

Solution approach

Correct Answer + Core Reason: A pipe closed at one end. The frequencies are odd multiples of the fundamental frequency.

Step-by-Step Solution:

Step 1: Identify the given frequencies: 240 Hz, 720 Hz, and 1200 Hz.

Step 2: Find the ratios of the frequencies: $\frac{720}{240} = 3$ and $\frac{1200}{240} = 5$ .

Step 3: Observe that the frequencies are in the ratio 1:3:5, which are odd multiples of the fundamental frequency.

Step 4: Recall that for a pipe closed at one end, the frequencies are given by $f_n = (2n - 1)f_1$ , where $n = 1, 2, 3, ...$ and $f_1$ is the fundamental frequency.

Step 5: Conclude that the organ pipe is closed at one end.

Why other options are wrong:
- Option B: A pipe open at both ends is incorrect. For an open pipe, the frequencies are integer multiples of the fundamental frequency (1:2:3:...).
- Option C: Closed at both ends is incorrect. A pipe closed at both ends does not produce sound.
- Option D: Having holes like flute is incorrect. While a flute has holes, the frequency relationships are more complex than simple integer or odd integer multiples.

Key Concepts:
- Fundamental frequency
- Overtones
- Harmonics
- Organ pipes (open and closed)
- Frequency relationships in organ pipes

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

Quick Tips:
- Remember the frequency relationships for open and closed pipes.

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Application +4 | -1

A transverse wave is represented by

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

. For what value of wavelength, the wave velocity is half of maximum particle velocity

View answer and explanation

Correct option: A

\pi A

Option A (

\pi A

) is correct because the wave velocity is

\omega/k

and the maximum particle velocity is

A\omega

. Setting the wave velocity to half the maximum particle velocity and solving for the wavelength yields

\lambda = \pi A

.

Solution approach

Correct Answer + Core Reason: $\pi A$ . The wave velocity is $v = \frac{{\omega}}{{k}}$ and the maximum particle velocity is $v_{max} = A\omega$ . Given $v = \frac{1}{2}v_{max}$ , we can find the wavelength.

Step-by-Step Solution:

Step 1: Write the expression for wave velocity: $v = \frac{{\omega}}{{k}}$ .

Step 2: Write the expression for maximum particle velocity: $v_{max} = A\omega$ .

Step 3: Use the given condition: $v = \frac{1}{2}v_{max} \Rightarrow \frac{{\omega}}{{k}} = \frac{1}{2}A\omega$ .

Step 4: Solve for $k$ : $k = \frac{2}{A}$ .

Step 5: Relate $k$ to wavelength: $k = \frac{{2\pi}}{{\lambda}}$ .

Step 6: Substitute and solve for $\lambda$ : $\frac{{2\pi}}{{\lambda}} = \frac{2}{A} \Rightarrow \lambda = \pi A$ .

Why other options are wrong:
- Option B: $\frac{{\pi A}}{2}$ is incorrect. This is likely a miscalculation.
- Option C: $2\pi A$ is incorrect. This is likely a miscalculation.
- Option D: $\frac{{\pi A}}{4}$ is incorrect. This is likely a miscalculation.

Key Concepts:
- Wave velocity: $v = \frac{{\omega}}{{k}}$
- Maximum particle velocity: $v_{max} = A\omega$
- Wave number: $k = \frac{{2\pi}}{{\lambda}}$

Memory Aids:
- Remember the relationships between wave velocity, particle velocity, wave number, and wavelength.

Quick Tips:
- Be careful with the definitions of wave velocity and particle velocity.

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Application +4 | -1

5 beats/second are heard when a turning fork is sounded with a sonometer wire under tension, when the length of the sonometer wire is either 0.95 m or 1 m. The frequency of the fork will be :

View answer and explanation

Correct option: A

195 Hz

Option A (195 Hz) is correct because the frequency of the fork is calculated using the beat frequency and the relationship between frequency and length of the sonometer wire.

Solution approach

Correct Answer + Core Reason: The frequency of the fork is 195 Hz.

Step-by-Step Solution:

Step 1: The frequency of a vibrating string is inversely proportional to its length: $f \propto \frac{1}{L}$

Step 2: Let the frequency of the fork be $f$ . When the length is 0.95 m, the frequency of the wire is either $f + 5$ or $f - 5$ . When the length is 1 m, the frequency of the wire is different.

Step 3: Since the frequency is inversely proportional to the length, the longer length (1 m) will have a lower frequency. Therefore, if the frequency at 0.95 m is $f + 5$ , the frequency at 1 m will be lower, and if the frequency at 0.95 m is $f - 5$ , the frequency at 1 m will be even lower.

Step 4: Let $f_1$ be the frequency when the length is 0.95 m and $f_2$ be the frequency when the length is 1 m. Then, $\frac{f_1}{f_2} = \frac{1}{0.95} = \frac{20}{19}$

Step 5: Case 1: $f_1 = f + 5$ . Then $f_2 = f \pm 5$ . If $f_2 = f+5$ , then $\frac{f+5}{f+5} = \frac{20}{19}$ , which is not possible. If $f_2 = f-5$ , then $\frac{f+5}{f-5} = \frac{20}{19}$ . $19(f+5) = 20(f-5)$ . $19f + 95 = 20f - 100$ . $f = 195 \text{{ Hz}}$

Step 6: Case 2: $f_1 = f - 5$ . Then $f_2 = f \pm 5$ . If $f_2 = f+5$ , then $\frac{f-5}{f+5} = \frac{20}{19}$ . $19(f-5) = 20(f+5)$ . $19f - 95 = 20f + 100$ . $f = -195 \text{{ Hz}}$ , which is not possible.

Step 7: Therefore, the frequency of the fork is 195 Hz.

Why other options are wrong:
- Option B: 251 Hz is incorrect.
- Option C: 150 Hz is incorrect.
- Option D: 300 Hz is incorrect.

Key Concepts:
- Frequency of a vibrating string
- Beat frequency

Memory Aids:
- $f \propto \frac{1}{L}$

Quick Tips:
- Remember the relationship between frequency and length of a string.

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Application +4 | -1

The transverse displacement of string fixed at both ends is given by

y = 0.06\,\sin \left( {\frac{{2\pi x}}{3}} \right)\cos \left( {120\,\pi t} \right)

, where x and y are in metre and t is in second. The length of the string is 1.5m and its mass is

3 \times {10^{ - 2}}\,\,kg

. The tension in the string is equal to

View answer and explanation

Correct option: A

648 N

Option A (648 N) is correct because the tension is calculated using the wave speed derived from the given wave equation and the mass per unit length.

Solution approach

Correct Answer + Core Reason: 648 N. The tension is calculated from the wave speed and mass per unit length.

Step-by-Step Solution:

Step 1: Identify the given wave equation: $y = 0.06 \sin \left( {\frac{{2\pi x}}{3}} \right)\cos \left( {120\pi t} \right)$ .

Step 2: From the equation, we can identify $k = \frac{{2\pi }}{3}$ and $\omega = 120\pi$ .

Step 3: The wave speed $v$ can be found using the relationship $v = \frac{\omega }{k} = \frac{{120\pi }}{{\frac{{2\pi }}{3}}} = 120\pi \times \frac{3}{{2\pi }} = 180 \text{{ m/s}}$ .

Step 4: The mass per unit length $\mu$ is given by $\mu = \frac{m}{L} = \frac{{3 \times {{10}^{ - 2}}}}{{1.5}} = 0.02 \text{{ kg/m}}$ .

Step 5: The tension $T$ in the string is related to the wave speed by $v = \sqrt{\frac{T}{\mu}}$ , so $T = \mu v^2$ .

Step 6: Substitute the values: $T = 0.02 \times {180^2} = 0.02 \times 32400 = 648 \text{{ N}}$ .

Why other options are wrong:
- Option B (724 N): This is incorrect due to a possible miscalculation of the wave speed or incorrect substitution into the tension formula.
- Option C (832 N): This is incorrect due to a possible miscalculation of the wave speed or incorrect substitution into the tension formula.
- Option D (980 N): This is incorrect due to a possible miscalculation of the wave speed or incorrect substitution into the tension formula.

Key Concepts:
- Wave speed: $v = \frac{\omega }{k}$
- Tension in a string: $v = \sqrt{\frac{T}{\mu}}$
- Mass per unit length: $\mu = \frac{m}{L}$

Memory Aids:
- Remember the relationship between wave speed, angular frequency, and wave number: $v = \frac{\omega }{k}$ .

Quick Tips:
- Carefully identify the values of $k$ and $\omega$ from the given wave equation.
- Pay attention to units and ensure consistency.

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Understanding +4 | -1

On which principle does sonometer works?

View answer and explanation

Correct option: C

Resonance

Option C (Resonance) is correct because a sonometer works on the principle of resonance, where the frequency of the vibrating wire matches the frequency of the tuning fork, leading to maximum amplitude.

Solution approach

Correct Answer + Core Reason: Resonance. Sonometer works on the principle of resonance.

Step-by-Step Solution:

Step 1: Understand the working principle of a sonometer.

Step 2: A sonometer is used to demonstrate the relationship between the frequency of a vibrating string and its length, tension, and mass per unit length.

Step 3: When the frequency of the vibrating string matches the frequency of the tuning fork, resonance occurs, resulting in a large amplitude of vibration.

Why other options are wrong:
- Option A: Hooke’s law. Hooke's law describes the relationship between stress and strain in elastic materials, but it's not the primary principle behind the sonometer's operation.
- Option B: Elasticity. Elasticity is a property of the string, but resonance is the key principle.
- Option D: Newton’s law. Newton's laws of motion are fundamental to understanding the forces involved, but resonance is the direct principle at play.

Key Concepts:
- Resonance

Memory Aids:
- Think of resonance as matching frequencies leading to large vibrations.

Quick Tips:
- Remember that resonance occurs when the driving frequency matches the natural frequency.

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Analyzing +4 | -1

What is the base frequency if a pipe gives notes of frequencies 425, 255 and 595 and decide whether it is closed at one end or open at both ends

View answer and explanation

Correct option: B

85, closed

Option B (85, closed) is correct because the given frequencies are odd multiples of 85, which is characteristic of a closed pipe.

Solution approach

Correct Answer + Core Reason: 85, closed. The frequencies are odd multiples of the base frequency, indicating a closed pipe.

Step-by-Step Solution:

Step 1: Find the greatest common divisor (GCD) of the differences between the given frequencies.\nThe differences are: $425 - 255 = 170$ and $595 - 425 = 170$ . Also, $595 - 255 = 340$ .\nThe GCD of 170 and 340 is 170.

Step 2: Divide each frequency by the GCD to see if they are odd or even multiples.
$\frac{255}{170} = 1.5 = \frac{3}{2}$ , $\frac{425}{170} = 2.5 = \frac{5}{2}$ , $\frac{595}{170} = 3.5 = \frac{7}{2}$ .\nSince the frequencies are odd multiples of $\frac{170}{2} = 85$ , the pipe is closed at one end.

Step 3: Alternatively, find the GCD of the frequencies directly. The GCD of 255, 425, and 595 is 85.
$255 = 3 \times 85$ , $425 = 5 \times 85$ , $595 = 7 \times 85$ .\nSince the frequencies are odd multiples of 85, the pipe is closed at one end.

Step 4: Final Answer: 85, closed

Why other options are wrong:
- Option A: 17, closed: 17 is a factor of the frequencies, but it doesn't explain the odd multiple relationship.
- Option C: 17, open: Open pipes have frequencies that are integer multiples of the base frequency, not just odd multiples.
- Option D: 85, open: Open pipes have frequencies that are integer multiples of the base frequency, not just odd multiples.

Key Concepts:
- Harmonics in pipes
- Closed pipe vs. open pipe
- Greatest common divisor (GCD)

Memory Aids:
- Closed pipes have only odd harmonics.
- Open pipes have all harmonics.

Quick Tips:
- Find the GCD of the frequencies to determine the base frequency.
- Check if the frequencies are odd or even multiples of the base frequency to determine if the pipe is closed or open.

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Remembering +4 | -1

Ultrasonic waves are those waves

View answer and explanation

Correct option: B

Man can’t hear

Option B (Man can’t hear) is correct because ultrasonic waves have frequencies above the human hearing range.

Solution approach

Correct Answer + Core Reason: Man can’t hear. Ultrasonic waves have frequencies above the human hearing range.

Step-by-Step Solution:

Step 1: Recall the definition of ultrasonic waves. Ultrasonic waves are sound waves with frequencies higher than the upper limit of human hearing, which is typically around 20 kHz.

Step 2: Understand that humans cannot hear frequencies above 20 kHz.

Step 3: Conclude that ultrasonic waves are those waves that man can't hear.

Why other options are wrong:
- Option A: To which man can hear is incorrect. These are audible waves, not ultrasonic waves.
- Option C: Are of high velocity is incorrect. The velocity of ultrasonic waves is the same as the velocity of sound in the medium, which is not necessarily high.
- Option D: Of high amplitude is incorrect. The amplitude of a wave is related to its intensity, not whether it is ultrasonic or not.

Key Concepts:
- Ultrasonic waves
- Human hearing range
- Frequency of sound waves

Memory Aids:
- Ultra means beyond, so ultrasonic means beyond the range of human hearing.

Quick Tips:
- Remember the approximate range of human hearing (20 Hz to 20 kHz).

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Analyzing +4 | -1

For a transverse wave travelling along a straight line, the distance between two peaks (crests) is 5 m, while the distance between one crest and one trough is 1.5 m. The possible wavelengths (in metre) of the waves are

View answer and explanation

Correct option: B

\frac{1}{1},\frac{1}{3},\frac{1}{5}....

Option B (

\frac{1}{1},\frac{1}{3},\frac{1}{5}....

) is correct because it represents the possible wavelengths based on the given relationship between crests and troughs.

Solution approach

Correct Answer + Core Reason: $\frac{1}{1},\frac{1}{3},\frac{1}{5}....$ The possible wavelengths are derived from the relationship between the distance between crests and troughs.

Step-by-Step Solution:

Step 1: Define the given information:\nDistance between two crests (wavelength) = 5 m.\nDistance between one crest and one trough = 1.5 m.

Step 2: Relate the distance between a crest and a trough to the wavelength:\nThe distance between a crest and a trough is an odd multiple of half-wavelengths. So, $1.5 = (2n+1) \frac{\lambda}{2}$ , where n = 0, 1, 2, 3...

Step 3: Solve for the wavelength $\lambda$ :
$\lambda = \frac{3}{2n+1}$ .

Step 4: Find the possible wavelengths:\nFor n = 0: $\lambda = \frac{3}{1} = 3$ m. But the distance between two crests is 5m, so the wavelength must be 5m. This means the distance between a crest and trough is $\lambda/2 = 5/2 = 2.5m$ . The problem states the distance is 1.5m. This means the wave is reflected.\nLet the actual wavelength be $\lambda'$ . Then the distance between two crests is $5 = n \lambda'$ and the distance between a crest and a trough is $1.5 = (2m+1) \frac{\lambda'}{2}$ where n and m are integers.
$\lambda' = \frac{5}{n}$ and $1.5 = (2m+1) \frac{5}{2n}$ .
$3n = (2m+1)5$ or $n = \frac{5(2m+1)}{3}$ .\nSince n must be an integer, $2m+1$ must be a multiple of 3. So $2m+1 = 3, 9, 15, ...$ \nIf $2m+1 = 3$ , $n = 5$ , $\lambda' = 1$ .\nIf $2m+1 = 9$ , $n = 15$ , $\lambda' = 1/3$ .\nIf $2m+1 = 15$ , $n = 25$ , $\lambda' = 1/5$ .\nThus the possible wavelengths are $1, \frac{1}{3}, \frac{1}{5}, ...$

Why other options are wrong:
- Option A: 1, 2, 3... This is incorrect as the wavelengths follow the pattern $\frac{1}{2n+1}$ .
- Option C: 1, 3, 5... This is incorrect as the wavelengths follow the pattern $\frac{1}{2n+1}$ .
- Option D: $\frac{1}{2},\frac{1}{4},\frac{1}{6}....$ This is incorrect as the wavelengths follow the pattern $\frac{1}{2n+1}$ .

Key Concepts:
- Wavelength and its relation to crests and troughs
- Reflection of waves

Memory Aids:
- Remember that the distance between a crest and a trough is half a wavelength for a simple wave.

Quick Tips:
- Always visualize the wave and its crests and troughs to understand the relationship between them.

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Understanding +4 | -1

Quality of a musical note depends on

View answer and explanation

Correct option: A

Harmonics present

Option A (Harmonics present) is correct because the quality (timbre) of a musical note is determined by the number and relative intensities of the harmonics present.

Solution approach

Correct Answer + Core Reason: Harmonics present. The quality of a musical note depends on the harmonics present.

Step-by-Step Solution:

Step 1: Define the quality of a musical note.\nThe quality (or timbre) of a musical note is what distinguishes it from other notes of the same pitch and loudness.

Step 2: Explain the role of harmonics.\nThe quality of a musical note is determined by the number and relative intensities of the harmonics present in the sound wave.

Why other options are wrong:
- Option B: Amplitude of the wave determines the loudness, not the quality.
- Option C: Fundamental frequency determines the pitch, not the quality.
- Option D: Velocity of sound affects the pitch and wavelength, but not the quality.

Key Concepts:
- Quality (Timbre) of sound
- Harmonics
- Fundamental frequency
- Amplitude

Memory Aids:
- Timbre is determined by harmonics.

Quick Tips:
- Remember that quality is different from pitch and loudness.

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Understanding +4 | -1

Two Cu wires of radii

{\mathrm{R}_1}\,\,\mathrm{and}\,\,{\mathrm{R}_2}

such that

\left( {{R_1} > {R_2}} \right)

. Then which of the following is true?

View answer and explanation

Correct option: B

Transverse wave travels faster in thinner wire

Option B (Transverse wave travels faster in thinner wire) is correct because the speed of a transverse wave on a string is inversely proportional to the square root of the linear mass density, which is higher for thicker wires.

Solution approach

Correct Answer + Core Reason: Transverse wave travels faster in thinner wire. The speed of a transverse wave is inversely proportional to the square root of the linear mass density.

Step-by-Step Solution:

Step 1: Define the speed of a transverse wave on a string.\nThe speed of a transverse wave on a string is given by $v = \sqrt{\frac{T}{\mu}}$ , where T is the tension and $\mu$ is the linear mass density (mass per unit length).

Step 2: Relate linear mass density to radius.\nThe linear mass density is $\mu = \frac{m}{L} = \frac{\rho V}{L} = \frac{\rho A L}{L} = \rho A = \rho \pi r^2$ , where $\rho$ is the density, A is the cross-sectional area, and r is the radius.

Step 3: Analyze the relationship between speed and radius.\nSince $v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{T}{\rho \pi r^2}} = \sqrt{\frac{T}{\rho \pi}} \frac{1}{r}$ , the speed is inversely proportional to the radius.

Step 4: Conclude which wire has a faster wave.\nSince $R_1 > R_2$ , the thinner wire ( $R_2$ ) will have a faster transverse wave.

Why other options are wrong:
- Option A: Transverse wave travels slower in thicker wire.
- Option C: The speed is not the same in both wires because the radii are different.
- Option D: Transverse waves can travel in wires.

Key Concepts:
- Transverse wave speed on a string
- Linear mass density

Memory Aids:
- Thinner string, faster wave.

Quick Tips:
- Remember the formula $v = \sqrt{\frac{T}{\mu}}$ and how $\mu$ relates to the radius.

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Application +4 | -1

The equation of a wave travelling on a string is

y = 10\sin \left[ {\pi \left( {5t - \frac{x}{4}} \right)} \right]

, where x, y are in cm and t is in second. The velocity of wave is

View answer and explanation

Correct option: C

20 cm/s, in x-direction

Option C (20 cm/s, in x-direction) is correct because by comparing the given equation with the standard wave equation, the wave velocity can be calculated as

\omega/k

, and the sign indicates the direction of propagation.

Solution approach

Correct Answer + Core Reason: 20 cm/s, in x-direction. By comparing the given equation with the standard wave equation, we can find the wave velocity.

Step-by-Step Solution:

Step 1: Rewrite the given equation: $y = 10\sin[\pi(5t - \frac{x}{4})] = 10\sin(5\pi t - \frac{\pi}{4}x)$ .

Step 2: Compare with the standard wave equation: $y = A\sin(\omega t - kx)$ .

Step 3: Identify $\omega$ and $k$ : $\omega = 5\pi$ and $k = \frac{\pi}{4}$ .

Step 4: Calculate the wave velocity: $v = \frac{{\omega}}{{k}} = \frac{{5\pi}}{{\frac{{\pi}}{4}}} = 5\pi \times \frac{4}{{\pi}} = 20 \text{{ cm/s}}$ .

Step 5: Determine the direction. Since the equation is of the form $(\omega t - kx)$ , the wave is traveling in the positive x-direction.

Why other options are wrong:
- Option A: 40 cm/s is incorrect. This is likely a miscalculation.
- Option B: 40 cm/s, in -x-direction is incorrect. The direction is positive x.
- Option D: 20 cm/s, in -x-direction is incorrect. The direction is positive x.

Key Concepts:
- Wave equation: $y = A\sin(\omega t - kx)$
- Wave velocity: $v = \frac{{\omega}}{{k}}$

Memory Aids:
- Remember the standard form of the wave equation.

Quick Tips:
- Pay attention to the signs in the wave equation to determine the direction of propagation.

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Analyzing +4 | -1

Statement-I: An open organ pipe of certain length have the same fundamental frequency as closed organ pipe of half the length
Statement-II: In the case of open organ pipe, at both the ends antinodes are formed, while in the closed organ pipe at one end antinode and at the other end node is formed

View answer and explanation

Correct option: C

Both statement-I

\&

II are correct

Option C (Both statement-I & II are correct) is correct because both statements accurately describe the relationship between open and closed organ pipes and the formation of nodes and antinodes.

Solution approach

Correct Answer + Core Reason: Both statement-I $\& II are correct. Both statements are factually correct regarding organ pipes.\n\n**Step-by-Step Solution:**\n\n**Step 1:** Analyze Statement-I: An open organ pipe of certain length have the same fundamental frequency as closed organ pipe of half the length.\nFor an open pipe of length L, the fundamental frequency is$ f_{open} = $\frac{v}{2L}$ $. For a closed pipe of length L/2, the fundamental frequency is$ f_{closed} = $\frac{v}{4(L/2)}$ = $\frac{v}{2L}$ $. Therefore,$ f_{open} = f_{closed} $, so Statement-I is correct.\n\n**Step 2:** Analyze Statement-II: In the case of open organ pipe, at both the ends antinodes are formed, while in the closed organ pipe at one end antinode and at the other end node is formed.\nThis statement is a fundamental property of open and closed organ pipes. Open pipes have antinodes at both ends, and closed pipes have an antinode at the open end and a node at the closed end. Therefore, Statement-II is correct.\n\n**Step 3:** Conclude that both statements are correct.\n\n**Why other options are wrong:**\n- **Option A:** Statement-I is correct is incorrect because Statement-II is also correct.\n- **Option B:** Statement-II is correct is incorrect because Statement-I is also correct.\n- **Option D:** Both statement-I$ \& II are incorrect is incorrect because both statements are correct.

Key Concepts:
- Open organ pipes
- Closed organ pipes
- Fundamental frequency
- Antinodes
- Nodes

Memory Aids:
- Open pipe: antinodes at both ends.
- Closed pipe: antinode at open end, node at closed end.

Quick Tips:
- Remember the fundamental properties of open and closed organ pipes.

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