Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
$K$ is the force constant of a spring. The work done in increasing its extension from $I_1$ to $I_2$ will be
PhysicsMechanical Properties of SolidsJEE Main
Options:
  • A $K\left(I_2-I_1\right)$
  • B $\frac{K}{2}\left(I_2+I_1\right)$
  • C $K\left(I_2^2-I_1^2\right)$
  • D $\frac{K}{2}\left(I_2^2-I_1^2\right)$
Solution:
1702 Upvotes Verified Answer
The correct answer is: $\frac{K}{2}\left(I_2^2-I_1^2\right)$
At extension $l_1$, the stored energy $=\frac{1}{2} K l_1^2$
At extension $l_2$, the stored energy $=\frac{1}{2} \mathrm{Nl}_2^2$
Work done in increasing its extension from $l_1$ to $l_2=\frac{1}{2} K\left(l_2^2-l_1^2\right)$

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.