Question

A steel wire is $1\text{m}$ long and $1{\text{mm}}^2$ in area of cross section. If it takes $200\text{N}$ to stretch this wire by $1\text{mm}$. How much force will be required to stretch a wire of same material as well as same cross-section area from its normal length of $2\text{m}$ by a length of $20\text{mm}$?

• A. $200\text{N}$
• B. $2000\text{N}$
• C. $4000\text{N}$
• D. $400\text{N}$

Answer 1

The correct option is B $2000\text{N}$

We know, Young Modulus $\left(Y\right)=\frac{F/A}{\Delta l/l}$
Where, $\frac{F}{A}=$ Normal stress
$\Delta l=$ Change in length
$l=$ Original length
$\Rightarrow \Delta l=\frac{Fl}{AY}$
Since, $A,Y$ remain constant in this case.
$\Delta l\propto Fl$
So, $\frac{\Delta l_1}{\Delta l_2}=\frac{F_1{l}_1}{F_2{l}_2}$
Substitute the values, $F_1=200\text{N}$
$l_1=1\text{m}$
$\Delta l_1=1\text{mm}$
$l_2=2\text{m}$
$\Delta l_2=20\text{mm}$
$\frac{1\times {10}^{-3}}{20\times {10}^{-3}}=\frac{200\times 1}{F_2\times 2}$
$\Rightarrow F_2=2000\text{N}$