Question

A sonometer wire is stretched by a hanging metal bob. Its fundamental frequency is $f_1$. When the bob is completely immersed in water, the frequency becomes $f_2$. The relative density of the metal is

• A. $\frac{f_1^2}{f_1^2-f_2^2}$
• B. $\frac{f_2^2}{f_1^2-f_2^2}$
• C. $\frac{f_1}{f_1-f_2}$
• D. $\frac{f_2}{f_1-f_2}$

Answer 1

The correct option is A $\frac{f_1^2}{f_1^2-f_2^2}$

Let $W_1$ be the weight of the bob in air and $W_2$ when it is immersed in water.
Given that, $f_1=\frac{1}{2L}\sqrt{\frac{W_1}{\mu }}$ and $f_2=\frac{1}{2L}\sqrt{\frac{W_2}{\mu }}$
$\therefore \frac{W_1}{W_2}=\frac{f_1^2}{f_2^2}$
Relative density $=\frac{Weightinair}{Lossofweightinwater}$
$=\frac{W_1}{W_1-W_2}=\frac{f_1^2}{f_1^2-f_2^2}$
Hence, the correct choice is (a).