A tuning fork vibrating with a frequency of 512Hz is kept close to the open end of a tube filled with water (Fig. 15.4). The water level in the tube is gradually lowered. When the water level is 17cm below the open end, maximum intensity of sound is heard. If the room temperature is 20° C, calculate

(a) Speed of sound in air at room temperature

(b) Speed of sound in air at 0° C

(c) If the water in the tube is replaced with mercury, will there be any difference in your observations?

4 Views|Posted 7 months ago
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7 months ago

This is a short answer type question as classified in NCERT Exemplar

consider the frequency of tuning fork= 512Hz

(a) L= λ 4

λ = 4 L

V= v λ = 512 * 4 * 17 * 10 - 2

= 348.16m/s

(b) we know that v T

V 20 V 0 = 273 + 20 273 = 293 273

V 20 V 0 = 1.073 = 1.03

Vo= V 20 1.03 = 348.16 1.03 = 338 m / s

(c) Resonance will be observed at 17cm length of air column only intensity of sound heard may be greater due to more com

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Physics NCERT Exemplar Solutions Class 11th Chapter Fifteen 2025

Physics NCERT Exemplar Solutions Class 11th Chapter Fifteen 2025

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