Question

A load of mass $M$ is suspended from a steel wire of length $2$ and radius $1.0\text{mm}$ in Searle's apparatus experiment. The increase in length produced in the wire is $4.0\text{mm}$. Now the load is fully immersed in a liquid of relative density $2$. The relative density of the material of load is $8$. The new value of increase in length of the steel wire is:

• A. $\text{5.0 mm}$
• B. $\text{4.0 mm}$
• C. $\text{3.0 mm}$
• D. Zero

Answer 1

The correct option is C $\text{3.0 mm}$


Using $\frac{F}{A}=Y\cdot \frac{\Delta l}{l}$

$\Delta l\propto F---\left(1\right)$
When the load is suspended,
$T=Mg$
When the load is fully immersed in the liquid,
$T^{\prime }=Mg-f_B=Mg-\frac{M}{{\rho }_o}\cdot {\rho }_l\cdot g$
$\Rightarrow T^{\prime }=(1-\frac{{\rho }_l}{{\rho }_o})Mg$
$\Rightarrow T^{\prime }=(1-\frac{2}{8})Mg$
$\Rightarrow T^{\prime }=\frac{3}{4}Mg$
From equation $\left(1\right)$
$\frac{\Delta l^{\prime }}{\Delta l}=\frac{T^{\prime }}{T}=\frac{3}{4}$

$[\text{Given:}\Delta l=4\text{mm}]$
$\therefore \Delta l^{\prime }=\frac{3}{4}\cdot \Delta l=\frac{3}{4}\times 4=3\text{mm}$

Hence, option $\left(A\right)$ is correct.