Ncert Exercises & Solutions

A tuning fork vibrating with a frequency of 512 Hz is kept close to the open end of a tube filled with water (figure). The water level in the tube is gradually lowered. When the water level is 17 cm below the open end, the maximum intensity of sound is heard. If the room temperature is $20 ° C$ , calculate;

1. Speed of sound in air at room temperature.

2. Speed of sound in air at $0 ° C$ .

3. If the water in the tube is replaced with mercury, will there be any difference in your observations?

Hint:

Consider the diagram frequency of tuning fork

ν

=512 Hz

For observation of first maxima of intensity

1. Step 1: Find the speed of sound in air at room temperature.

L = \frac{λ}{4} \Rightarrow λ = 4 L (for closed pipe)

ν = νλ = 512 \times 4 \times 17 \times 10^{- 2} = 348.16 m / s

2. Step 2: Find the speed of sound in the air at

0 ° C

We know that

v \propto \sqrt{T}

where temperature (T) is in kelvin.

\frac{v_{20}}{v_{0}} = \sqrt{\frac{273 + 20}{273 + 0}} = \sqrt{\frac{293}{273}}

\frac{v_{20}}{v_{0}} = \sqrt{1.073} = 1.03

v_{0} = \frac{v_{20}}{1.03} = \frac{348.16}{1.03} = 338 m / s

3. Step 3: Find the length of the air column for which the resonance occurs in the case of mercury.

Resonance will be observed at 17 cm length of the air column, only the intensity of sound heard may be greater due to more complete reflection of the sound waves at the mercury surface because mercury is denser than water.

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