Question

A stone hangs from the free end of a sonometer wire whose vibrating length, when tuned to a tuning fork, is $40\text{cm}$. When the stone hangs wholly immersed in water, the vibrating length is reduced to $30\text{cm}$. The relative density of the stone is

• A. $\frac{16}{9}$
• B. $\frac{16}{7}$
• C. $\frac{16}{5}$
• D. $\frac{16}{3}$

Answer 1

The correct option is B $\frac{16}{7}$

Frequency of waves in sonometer wire
$f=\frac{1}{2l}\sqrt{\frac{Mg}{\mu }}$
where $M=$ mass of stone
$l=$ vibrating length of wire
$\mu =$ mass per unit length of wire (constant)

If $\rho$ is the density of the stone and $V$ its volume, then $M=\rho V$. When the stone is wholly immersed in water of density ${\rho }^{\prime }$, the effective weight of the stone
$M^{{}^{\prime }}g=(M-V{\rho }^{{}^{\prime }})g=(V\rho -V{\rho }^{{}^{\prime }})g=V{\rho }^{{}^{\prime }}(\frac{\rho }{{\rho }^{{}^{\prime }}}-1)g$

Now, $f=\frac{1}{2l}\sqrt{\frac{\rho Vg}{\mu }}$
and $f^{{}^{\prime }}=\frac{1}{2l^{{}^{\prime }}}\sqrt{\frac{V{\rho }^{{}^{\prime }}(\frac{\rho }{{\rho }^{{}^{\prime }}}-1)g}{\mu }}$

Given, $l=40\text{cm}$ and $l^{\prime }=30\text{cm}$.
Also $f=f^{{}^{\prime }}$ (same tuning fork)

which gives $\frac{\sqrt{\rho }}{l}=\frac{\sqrt{\rho -{\rho }^{{}^{\prime }}}}{l^{\prime }}$
$\Rightarrow \frac{\rho }{{\rho }^{{}^{\prime }}}=\frac{16}{7}$
Hence, the correct choice is (b).