Question

An additional bridge is kept below a sonometer wire so that it is divided into two segments of lengths in the ratio $2:3$, and $n$${}_1$, $n$${}_2$ are their respective fundamental frequencies. If the additional bridge is removed then the fundamental frequency of that sonometer wire is $n$. The ratio of $n,n$${}_1$, $n$${}_2$ is:

• A. $2:3:5$
• B. $2:5:3$
• C. $4:9:25$
• D. $6:15:10$

Answer 1

Answer: (D)

$6:15:10$

Given
$\frac{l_1}{l_2}=\frac{2}{3}$
But, $\frac{n_1}{n_2}=\frac{l_2}{l_1}=\frac{3}{2}$ ......(1) [$\because n\propto \frac{1}{l}$]
Given, $l_1+l_2=l$
Dividing by $l_1$
$\therefore 1+\frac{l_2}{l_1}=\frac{l}{l_1}$
$\Rightarrow 1+\frac{3}{2}=\frac{l}{l_1}$
$\Rightarrow \frac{5}{2}=\frac{l}{l_1}$
$\Rightarrow \frac{l}{l_1}=\frac{n_1}{n}=\frac{5}{2}$ ....(2)
Combining (1) and (2) we get, $n:n_1:n_2=6:15:10$