Question

A particular force $F$ applied on a wire increases its length by $2\times {10}^{-3}\text{m}$. If we were to increase the wire's length by $4\times {10}^{-3}\text{m}$, then the applied force should be

• A. $F$
• B. $2F$
• C. $\frac{F}{2}$
• D. $4F$

Answer 1

The correct option is B $2F$

As we know,
$Y=\frac{\frac{F}{A}}{\frac{l}{L}}=\frac{F\times L}{A\times l}$
[Where, $l$ = Change in length , $L$ = Original length, $Y$ =Young's modulus of elasticity, $A=$ area of cross-section]

Since $Y,L$ and $A$ remain same.
$\therefore F\propto l$
$\Rightarrow \frac{F_1}{F_2}=\frac{l_1}{l_2}$
$\Rightarrow \frac{F_1}{F_2}=\frac{2\times {10}^{-3}}{4\times {10}^{-3}}$
$\therefore F_2=2F_1=2F$