
Loss in the gravitational potential energy of the ball= gain in the potential energy by the spring
−ΔPEgravitational=ΔPEspring
mg(h+2h)=21k(2h)2⇒mg23h=8kh2
⇒k=h12mg=0.112×0.1×10=120Nm−1
JEE Main 2022 — Physics Mechanics
A ball of mass 100g is dropped from a height h=10cm on a platform fixed at the top of a vertical spring (as shown in figure). The ball stays on the platform and the platform is depressed by a distance 2h. The spring constant is _____ Nm−1
(Use g=10ms−2)

Held on 24 Jun 2022 · Verified 6 Jul 2026.
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