In a resonance tube the first resonance with a tunning fork occurs at 16 ~cm and second at 49 ~cm . If the velocity of sound is 330 ~m / s , the frequency of tunning fork is

Correct answer: (A) 500. For closed pipe l_1=v/(4 n) and l_2=(3 v)/(4 n) v =2 n (l_2-l_1 ) n =fracv2 (l_2-l_1 ) =330/(2 ×(0.49-0.16))=500 ~Hz

Q121 / 150 · MHT CET 2011 · Physics

In a resonance tube the first resonance with a tunning fork occurs at $16 \mathrm{~cm}$ and second at $49 \mathrm{~cm} .$ If the velocity of sound is $330 \mathrm{~m} / \mathrm{s}$, the frequency of tunning fork is

Correct answer
A
500
B
300
C
330
D
165

Solution

For closed pipe $l_{1}=\frac{v}{4 n}$ and $l_{2}=\frac{3 v}{4 n}$ $$ \begin{aligned} v &=2 n\left(l_{2}-l_{1}\right) \\ n &=\frac{v}{2\left(l_{2}-l_{1}\right)} \\ &=\frac{330}{2 \times(0.49-0.16)}=500 \mathrm{~Hz} \end{aligned} $$

Hence, the correct answer is option: 'A'.

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