Question

The fundamental frequency of a sonometer wire decreases by $25\%$ when the body suspended at the free end of a wire is totally immersed in water. The specific gravity of that suspended body is :

• A. $25/9$
• B. $16/7$
• C. $7/3$
• D. $4$

Answer 1

Answer: (B)

$16/7$

The natural frequency of nth mode is
$f_n=\frac{n}{2L}\sqrt{\frac{T}{\mu }}$
where
L is the length of the sonometer wire, $\mu$ is the mass per unit length of the wire, T is the tension in the string
Frequency of the first harmonic before :$f=\frac{1}{2L}\sqrt{\frac{T}{\mu }}$ .....(1)
Frequency of the first harmonic after :$f^{\prime }=\frac{1}{2L}\sqrt{\frac{T^{\prime }}{\mu }}$ .......(2)
When the body suspended at free end of wire is totally immersed in water, $T$ decreases because of buoyancy
$T-T^{\prime }=F_{buoyancy}$

$V\rho g-T^{\prime }=V{\rho }_{w}g$ where $T=V\rho g$ is the weight of the body and V is the volume of the body and $\rho$ is the density of the body.

Dividing (1) by (2)

$\frac{f}{f^{\prime }}=\sqrt{\frac{T}{T^{\prime }}}=\sqrt{\frac{V\rho g}{V\rho g-V{\rho }_{w}g}}$

$\therefore (\frac{f^{\prime }}{f}{)}^2=(\frac{0.75}{1}{)}^2=1-\frac{{\rho }_w}{\rho }$

$\Rightarrow \frac{{\rho }_w}{\rho }=1-\frac{9}{16}=\frac{7}{16}$

$\Rightarrow \frac{\rho }{{\rho }_w}=\frac{16}{7}$