
mgsinθ−ma0cosθ=ma
a=gsinθ−a0cosθ
Now using,
S=ut+21adownt2
cosθL=21(gsinθ−a0cosθ)t2
t=gsinθcosθ−a0cos2θ2L
t=gsin2θ−a0(1+cos2θ)4L
JEE Main 2026 — Physics Mechanics
A small block of mass m slides down from the top of a frictionless inclined surface, while the inclined plane is moving towards left with constant acceleration a0. The angle between the inclined plane and ground is θ and its base length is L. Assuming that initially the small block is at the top of the inclined plane, the time it takes to reach the lowest point of the inclined plane is ____.

Held on 28 Jan 2026 · Verified 6 Jul 2026.
gsinθ−a0cosθ2L
gcos2θ−a0sinθcosθ4L
gsin2θ−a0(1+cos2θ)2L
gsin2θ−a0(1+cos2θ)4L
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