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Wires $A$ and $B$ are made from the same material. $A$ has twice the diameter and three times the length of $B$ . If the elastic limits are not reached, When each is stretched by the same tension, the ratio of energy stored in $A$ to that in $B$ is :

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$3 : 4$

$2 : 3$

$6 : 1$

$3 : 2$

answer is B.

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Detailed Solution

$W = \frac{1}{2} {(\frac{F}{A})}^{2} . \frac{V o l u m e}{Y}$

$W α \frac{f^{2}}{r^{2}} l$

$W α \frac{l}{r^{2}}$

$\frac{w_{1}}{w_{2}} = \frac{3}{4}$

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