A spring, which is initially in its unstretched condition, is first stretched by a length $x$ and then again by a further length $x$ . The work is done in the first case $W_{1}$ and in the second case $W_{2}$ . Find $\frac{W_{2}}{W_{1}}$ .

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Force constant of spring one is K and another spring is 2K. When both spring are stretched through same distance, then the work done

The correct option is B W2=2W1Work done = - Δ(Potentialenergy) = −12kx2Both springs are stretched by the same length , let it be l .Then ,W1 = −12kl2W2 = −122kl2Therefore , W2 = 2W1

Force constant of spring one is $K$ and another spring is $2 K$ . When both spring are stretched through same distance, then the work done

The correct option is B $W_{2} = 2 W_{1}$
Work done = - $Δ (P o t e n t i a l e n e r g y)$
= $- \frac{1}{2} k x^{2}$
Both springs are stretched by the same length , let it be $l$ .
Then ,
$W_{1}$ = $- \frac{1}{2} k l^{2}$
$W_{2}$ = $- \frac{1}{2} 2 k l^{2}$
Therefore , $W_{2}$ = $2$ $W_{1}$