Question

The volume of a wire remains constant, when subjected to tensile stress is applied on it. If the percentage change in lateral strain is $2$ %, then percentage change in longitudinal strain is

• A. $2$ %
• B. $4$ %
• C. $6$ %
• D. $8$ %

Answer 1

The correct option is B $4$ %

Let us suppose length of wire is $L$ and radius is $r$ then it's volume is
$V=\pi r^{2}L$
On differentiating both side
$\Delta V=\pi \left(2r\Delta r\right)L+\pi r^2\Delta L$
As, $V=$ constant, then $\Delta V=0$
$\Rightarrow 2\pi rL\Delta r+\pi r^2\Delta L=0$
$\Rightarrow 2\pi rL\Delta r=-\pi r^2\Delta L$
$\Rightarrow \frac{\Delta r}{r}=\frac{-1}{2}\times \frac{\Delta L}{L}$
$\Rightarrow \frac{\Delta L}{L}=\frac{-2\Delta r}{r}$
$\Rightarrow \frac{\Delta L}{L}\times 100=\frac{-2\Delta r}{r}\times 100$


$\Rightarrow \frac{\Delta L}{L}\times 100=-2\times -2$
$[\frac{\Delta r}{r}\times 100=-2$%, given, Negative sign in percentage lateral strain shows radius will decrease due to tensile stress.$]$
$\Rightarrow \frac{\Delta L}{L}\times 100=+4$ %
Percentage change in longitudnal strain is $+4$ %