Question

A particular force $(F)$ applied on a wire increases its length by $2\times 10^{-3}m$. To increase the wire's length by $4\times 10^{-3}m$ the applied force will be

• A. 4 F
• B. 3 F
• C. 2 F
• D. F

Answer 1

Answer: (C)

$Y=\frac{F/A}{\Delta l/l}=\frac{F\times l}{A\times \Delta l}$
(where $Y$ is Young's modulus of elasticity)
Since, $Y,I$ and $A$ remain same.
$F\propto \Delta l$
$\frac{F_1}{F_2}=\frac{\Delta l_1}{\Delta l_2}$
$\Rightarrow \frac{F}{F_2}=\frac{2\times 10^{-3}}{4\times 10^{-3}}$
$F_2=2F$