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

Question

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

Dr. Abubacker · Accepted Answer

(c) $Y = \frac{\frac{F}{A}}{\frac{l}{L}} = \frac{F 	imes L}{A 	imes l}$
(where Y is Young’s modulus of elasticity Since, Y, L and A remain same.
$F \propto l$
$\frac{F_{1}}{F_{2}} = \frac{l_{1}}{l_{2}}$
⟹ $\frac{F}{F_{2}} = \frac{2 	imes 10^{- 3}}{4 	imes 10^{- 3}}$
$F_{2} = 2F$

Q. 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

Explanation

Reviewed by Dr. Abubacker

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