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CGP EDU Academic Team
Published on: September 12, 2026
K is the force constant of a spring. The work done in increasing its extension from $l_1$ to $l_{2}$ will be
$(a) K(l_{2} - l_{1}) \quad (b) \frac{K}{2}(l_{2} + l_{1}) \\ (c) K(l_{2}^{2} - l_{1}^{2}) \quad (d) \frac{K}{2}(l_{2}^{2} - l_{1}^{2})$
Text Solution
Verified by ExpertsThe correct answer is:
D
At extension $l_1$, the stored energy $= \frac{1}{2} K l_1^2$ At extension $l_2$, the stored energy $= \frac{1}{2} K l_2^2$ Work done in increasing its extension from $l_1$ to $l_2$ $= \frac{1}{2} K (l_2^2 - l_1^2)$
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