Question

A wire is found to have a length $L$ when it is loaded with a block of mass $M$ and relative density $n$. When the block is immersed in water, the length of the wire reduces by $x$. Then, which of the following is/are correct?

• A. Weight of water displaced when the block is immersed in water is $\frac{Mg}{n}$.
• B. The apparent loss of weight due to immersion is $Mg(1-\frac{1}{n})$.
• C. The original length of wire before it was loaded is $L_1=L-nx$.
• D. The original length of the wire before it was loaded is $L_1=L-\frac{x}{n}$.

Answer 1

Answer: (A), (C)

The original length of wire before it was loaded is $L_1=L-nx$.

Volume of block is $V=\frac{M}{\rho }$,
where $\rho =$ density of block,

${\rho }_w=$ density of water.
& Volume of water displaced $=V$
Weight of water displaced is $w^{{}^{\prime }}={\rho }_{w}Vg=\frac{Mg}{n}[\because n=\frac{\rho }{{\rho }_w}]$
and Apparent loss of weight $=\frac{Mg}{n}$
Hence, apparent weight of block $=Mg-\frac{Mg}{n}=Mg(1-\frac{1}{n})$

Let $L_1$ be the original length of unloaded wire.
Then, Young's modulus $Y=\frac{F{L}_1}{A\Delta L}$
Before the block is immersed in water,
$\Delta L=(L-L_1),F=Mg$
Hence, $Y=\frac{Mg{L}_1}{A(L-L_1)}\dots \dots \left(i\right)$
After the block is immersed in water, $\Delta L=(L-L_1-x)$
& $F=Mg(1-\frac{1}{n})$
Hence, $Y=\frac{Mg(1-\frac{1}{n})L_1}{A(L-L_1-x)}\dots \dots \left(ii\right)$

From (i) and (ii), we get
$\frac{1}{L-L_1}=\frac{n-1}{n(L-L_1-x)}$
$\Rightarrow -nx=L_1-L$
$\Rightarrow L_1=L-nx$
Options (a) and (c) are correct.