Question

The frequency of sonometer wire is $f$, but when the weights producing the tensions are completely immersed in water, the frequency becomes $f / 2$ and on immersing the weights in a certain liquid, the frequency becomes $f / 3$. The specific gravity of the liquid is

• A. 4/3
• B. 16/9
• C. 15/12
• D. 32/27

Answer 1

Answer: (D)

$\because f \propto \sqrt{T}$
$\frac{ f _{\text {air }}}{ f _{\text {water }}}=\sqrt{\frac{W_{\text {air }}}{W_{\text {water }}}}=\sqrt{\frac{V \rho g}{V \rho g-V \rho_{w} g}}$
$=\frac{f}{\frac{f}{2}}=\sqrt{\frac{\rho}{\rho-\rho_{w}}}$
$\Rightarrow 2 =\sqrt{\frac{\rho}{\rho-\rho_{w}}}$
$\Rightarrow 4 \rho-4 \rho_{w} =\rho$
$\Rightarrow \rho =\frac{4}{3} \rho_{w}$
Similarly in second case,
$\frac{f}{\frac{f}{3}}=\sqrt{\frac{\rho}{\rho-\rho_{L}}}$
$\Rightarrow 3=\sqrt{\frac{\frac{4}{3} \rho_{w}}{4-\frac{3 \rho_{L}}{\rho_{w}}}}=\sqrt{\frac{4}{4-\frac{3 \rho_{L}}{\rho_{w}}}}$
Here, $ \frac{\rho_{L}}{\rho_{w}}=$ specific gravity (say $S$ )
$\Rightarrow 9=\frac{4}{4-3 S}$
$\Rightarrow 36-27 S=4 $
$\Rightarrow S=\frac{32}{27}$