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Two similar springs $P$ and $Q$ have spring constants $K_P$ and $K_Q$, such that $K_P > K_Q .$ They are stretched first by the same amount (case $a$), then by the same force (case $b$). The work done by the springs $W_P$ and $W_Q$ are related as, in case $(a)$ and case $(b)$ respectively
$W_P=W_Q ,W_P >W_Q$
$W_P=W_Q , W_P =W_Q$
$W_P >W_Q , W_Q > W_P$
$W_P < W_Q , W_Q < W_P$
Solution
$\begin{array}{l}
\,\,\,\,\,\,\,\,Here,\,{K_p} > {K_Q}\\
Case\,\left( a \right)\,:\,Elongation\,\left( x \right)\,in\,each\,spring\\
is\,same.\\
{W_p} = \frac{1}{2}{K_p}{x^2},\,{W_Q} = \frac{1}{2}{K_Q}{x^2}\therefore {W_p} > {W_Q}\\
Case\,\left( b \right)\,:\,Force\,of\,elongation\,is\,same.
\end{array}$
$\begin{array}{l}
So,\,{x_1} = \frac{F}{{{K_p}}}and\,{x_2} = \frac{F}{{{K_Q}}}\\
{W_p} = \frac{1}{2}{K_p}x_1^2 = \frac{1}{2}\,\frac{{{F^2}}}{{{K_p}}}\\
{W_Q} = \frac{1}{2}\,{K_Q}x_2^2 = \frac{1}{2}\,\frac{{{F^2}}}{{{K_Q}}}\,\,\,\,\,\,\therefore \,{W_p} < {W_Q}
\end{array}$
