A wire of length L and radius r is clamped at | vedclass

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(D) Young's modulus $Y$ is given by the formula: $Y = \frac{	ext{Stress}}{	ext{Strain}} = \frac{F/A}{\ell/L} = \frac{FL}{A\ell}$.Since $A = \pi r^2$

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(D) Young's modulus $Y$ is given by the formula: $Y = \frac{	ext{Stress}}{	ext{Strain}} = \frac{F/A}{\ell/L} = \frac{FL}{A\ell}$.Since $A = \pi r^2$,we have $Y = \frac{FL}{\pi r^2 \ell}$,which implies $\ell = \frac{FL}{Y \pi r^2}$.In the initial state,the extension is $\ell = \frac{FL}{Y \pi r^2}$.In the new state,the force $F' = F/2$ and the radius $r' = r/2$. The length $L$ remains constant.The new extension $\ell'$ is given by: $\ell' = \frac{F' L}{Y \pi (r')^2} = \frac{(F/2) L}{Y \pi (r/2)^2}$.Simplifying this,we get: $\ell' = \frac{(F/2) L}{Y \pi (r^2/4)} = \frac{FL}{2 Y \pi (r^2/4)} = \frac{FL}{Y \pi r^2 / 2} = 2 	imes \frac{FL}{Y \pi r^2}$.Substituting the initial expression for $\ell$,we get $\ell' = 2\ell$.Therefore,the increase in length becomes $2$ times the original value.

$A$ wire of length $L$ and radius $r$ is clamped at one end. If its other end is pulled by a force $F$,its length increases by $l$. If the radius of the wire and the applied force both are reduced to half of their original values keeping original length constant,the increase in length will become.

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