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.
$F \propto \Delta l$
$\frac{F_{1}}{F_{2}}=\frac{\Delta l_{1}}{\Delta l_{2}}$
$\Rightarrow \frac{F}{F_{2}}=\frac{2 \times 10^{-3}}{4 \times 10^{-3}}$
$F_{2} =2\, F$