Question

Two wires are made of the same material and have the same volume. The first wire has a cross-sectional area $1{\text{mm}}^2$ and the second wire has cross-sectional area $4{\text{mm}}^2$. If the length of the first wire is increased by $\Delta L$ on applying a force of $10\text{N}$, how much force is needed to stretch the second wire by the same amount?

• A. $160\text{N}$
• B. $100\text{N}$
• C. $90\text{N}$
• D. $80\text{N}$

Answer 1

The correct option is A $160\text{N}$

To stretch the second wire through same length $\left(\Delta L\right)$, let force needed be $F^{\prime }$
$\Rightarrow$Young's modulus should be same for both cases due to same material of wire
$Y=\frac{\left(F/A\right)}{\left(\Delta L/L\right)}$
$\Rightarrow \frac{F/A}{\Delta L/L}=\frac{F^{\prime }/A^{\prime }}{\Delta L/L^{\prime }}...\left(i\right)$
Since,volume is same
$\therefore A\times L=A^{\prime }{L}^{\prime }$
$\Rightarrow \frac{L}{L^{\prime }}=\frac{A^{\prime }}{A}=\frac{4}{1}...\left(ii\right)$
From Eq. $\left(i\right)\&\left(ii\right)$,
$\Rightarrow \frac{FL}{A}=\frac{F^{\prime }{L}^{\prime }}{A^{\prime }}$
$\Rightarrow F^{\prime }=\frac{A^{\prime }}{A}\times \frac{L}{L^{\prime }}\times F$
$\Rightarrow F^{\prime }={\left(\frac{4}{1}\right)}^2\times F$
$\therefore F^{\prime }=16\times 10=160\text{N}$