Question

A hollow metallic tube of length L is closed at one end. The resonance occurs with a fork of frequency n. The tube is then heated up uniformly and its length changes by l . If the change in velocity V of the sound is v, then the resonance will now be produced by a fork of frequency

• A. $\frac{V+v}{4(L+l)}$
• B. $4(L-l)$
• C. $\frac{V+v}{4(L-l)}$
• D. $\frac{V-v}{4(L-l)}$

Answer 1

The correct option is A $\frac{V+v}{4(L+l)}$

Pipe closed at one end

$\Rightarrow \frac{\lambda }{4}=l$

$f=\frac{V}{\lambda }=\frac{V}{4l}$

$\therefore f_1=\frac{V+v}{4(L+l)}$