Sound Wave and its Characteristics - NEET Physics Questions
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Sound Wave and its Characteristics

Question 21: easy

Assertion (A): Beats are not observed in case of light waves from two independent sources.


Reason (R): The phase difference between two light sources changes randomly.


 

1. (1) Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. (2) Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (3) (A) is true but (R) is false
4. (4) Both (A) and (R) are false
View Answer

Stable beats require coherent sources with a constant phase difference. Independent light sources have rapidly and randomly changing phase differences, making stable beats unobservable.


Thus, Assertion (A) is true, and Reason (R) is true and is the correct explanation of (A).

Question 22: easy

Assertion (A): A vibrating tuning fork sounds louder, when its stem is pressed against a desk top.


Reason (R): When a sound wave is incident on the surface of a desk, it is totally reflected.


 

1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer

Assertion (A) is true due to forced vibrations and resonance. The desk provides a larger surface area to vibrate, increasing the loudness.


Reason (R) is false because sound waves are not totally reflected; some energy is transmitted to the desk.

Question 23: easy

If speed of sound in air at 27°C is v then at what temperature speed of sound becomes 2v?

1. 1200°C
2. 927 K
3. 52°C
4. 927°C
View Answer

Since speed \(v \propto \sqrt{T}\), to double the speed, the temperature in Kelvin must quadruple: \(T_2 = 4 \times (27 + 273) = 1200\text{ K}\) or \(927^\circ\text{C}\).

Question 24: easy

If speed of sound in air at \(27^\circ\text{C}\) is \(v\) then at what temperature speed of sound becomes \(2v\)?

1. 1200°C
2. 927 K
3. 52°C
4. 927°C
View Answer

The speed of sound \(v \propto \sqrt{T}\. For the speed to double, \(T_2 = 4 T_1 = 4 \times (27 + 273) = 1200\text{ K}\. Converting to Celsius: \(1200 - 273 = 927^\circ\text{C}\).