Question

Two exactly similar wires of steel and copper are stretched by equal forces. If the difference in their elongation is $0.5cm$, then elongation of the copper wire is $[Y_{st}=2\times {10}^{12}dynec{m}^{-2}\&Y_{cu}=12\times {10}^{11}dyne/c{m}^2]$

• A. $0.75cm$
• B. $0.15cm$
• C. $1.5cm$
• D. $1.25cm$

Answer 1

The correct option is D $1.25cm$

Given
$\Delta l_{cu}$-$\Delta l_s=0.5cm$ ....(i)
Here $\Delta l_{cu}$ is extension of copper wire
$\Delta l_s$ is extension of steel wire
From definition of Youngs Modulus

$Y_s=\frac{F}{A}\times \frac{l}{\Delta l_s}{Y}_{cu}=\frac{F}{A}\times \frac{l}{\Delta l_{cu}}$
As Force and area for both wires are same

$\therefore \frac{Y_s}{Y_{cu}}=\frac{\Delta l_{cu}}{\Delta l_s}$
$\Delta l_{cu}=\Delta l_s\times \frac{Y_s}{Y_{cu}}$
$\Delta l_{cu}=\frac{5}{3}\Delta l_s$

using eq. (i),
$\Delta l_s+0.5=1.67\Delta l_s$
$\Delta l_s=0.75cm$
so, $\Delta l_{cu}=0.5+0.75=1.25cm$