JEE Main Physics question
Two springs A and B are stretched by applying forces of equal magnitude at the four ends. If the spring constant of A is 2 times greater than that of B, and the energy stored in A is E, then that in B is
Answer: (B)
Solution
For the same force, E = F²/(2k) ∝ 1/k. / = / = 2 ⟹ = 2E.