Question

A stretched sonometer wire is in unison with a tuning fork$.$ When the length of the wire is increased by $2\%$ the number of beats heard per second is $5$ then$,$ the frequency of the fork is

• A. 255$Hz$
• B. 450$Hz$
• C. 5$Hz$
• D. 0.5$Hz$

Answer 1

The correct option is A 255$Hz$

Let the Length of wire is L$,$

Length is increased by $2\%$ then number of beats heard per second is $5$

Means $n₁-n₂=$ $5sec⁻¹$ or, $5Hz$ -----$\left(1\right)$

Here n₁ and n₂ are the frequencies of initial and final$.$

we know$,n=$ here n is frequency $,$ l is the length of wire , $m$ is the mass per unit length and $T$ is the tension in string$.$

so$,$ here it is clear that frequency is inversely proportional to length of wire.

$\therefore n₁l₁=n₂l₂$

$n₁L=n₂1.02L$ { wire is increased by $2$, so $L₂$ $=1.02L$ }

$50n₁=51n₂$

$n₂=50/51n₁$ ----------$\left(2\right)$

from equations $\left(1\right)$ and $\left(2\right)$

$n₁-50/51n₁=5$

$\Rightarrow (51-50)/51n₁=5$

$\Rightarrow 1/51n₁=5$

$\Rightarrow n₁=51\times 5Hz=255Hz$

So$,$ frequency of turning fork is $255Hz$

Hence,

option $\left(A\right)$ is correct answer.