A string 25 cm long and having a mass of 2.5 g is under tension. A pipe closed at one end is 40 cm long. When the string is set vibrating in its first overtone and the air in the pipe in its fundamental frequency, 8 beats per second are heard. It is observed that decreasing the tension in the string decreases the beat frequency. If the speed of sound in air is 320 m/s, find the tension in the string

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$37.01 N$

$25.03 N$

$20.02 N$

$27.04 N$

answer is B.

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Detailed Solution

Frequency of fundamental mode of closed pipe
$f_{1} = \frac{V}{l} = 200 Hz$
Decreasing the tension in string decrease the beat frequency.

Hence the first overtone frequency of the string should be 208 Hz (not 192 Hz)
$\begin{array}{l} 208 = \frac{1}{l} \sqrt{\frac{T}{μ}} \Rightarrow \frac{T}{μ} = {(208 l)}^{2} \\ T = (208 \times 0.25)^{2} [\frac{2.5 \times 10^{- 3}}{0.25}] = 27.04 N \end{array}$