Question

A wire under a load has a frequency $n_1$. When the load is completely immersed in water, its frequency is $n_2$. The relative density of the material of the load is

• A. $\frac{n_1^2}{(n_1^2-n_2^2)}$
• B. $\frac{n_2^2}{(n_1^2-n_2^2)}$
• C. $\frac{n_1}{(n_1+n_2)}$
• D. $\frac{n_2}{(n_1+n_2)}$

Answer 1

The correct option is A $\frac{n_1^2}{(n_1^2-n_2^2)}$

$n_1=\frac{v_1}{2l}=\frac{1}{2}\sqrt{\frac{T_1}{ml}}$
$=\frac{1}{2}\sqrt{\frac{{\rho }_{m}vg}{ml}}$

$n_2=\frac{v_2}{2l}=\frac{1}{2}\sqrt{\frac{T_2}{ml}}$
$=\frac{1}{2}\sqrt{\frac{({\rho }_m-{\rho }_{\omega })vg}{ml}}$

So, $\frac{n_1}{n_2}=\sqrt{\frac{{\rho }_m}{{\rho }_m-{\rho }_{\omega }}}$
$\frac{{n_1}^2}{{n_2}^2}=\frac{{\rho }_m}{{\rho }_m-{\rho }_{\omega }}$
${n_1}^2{\rho }_m-{\rho }_{\omega }{n_1}^2={n_2}^2{\rho }_m$
$({n_1}^2-{n_2}^2){\rho }_m={n_1}^2{\rho }_{\omega }$
relative density $=\frac{{\rho }_m}{{\rho }_{\omega }}=\frac{{n_1}^2}{{n_1}^2-{n_2}^2}$