mg(Rcosθ−Rsinϕ)=21mV2.......(1)
On losing contact N=0⇒mgsinϕ=RmV2......(2)

⇒mgR(cosθ−sinϕ)=21mgRsinϕ
⇒2cosθ=3sinϕ
JEE Main 2014 — Physics Mechanics

A particle is released on a vertical smooth semicircular track from point X so that, OX makes angle θ from the vertical (see figure). The normal reaction of the track on the particle vanishes at the point Y where OY makes an angle ϕ with the horizontal. Then
Held on 19 Apr 2014 · Verified 6 Jul 2026.
sinϕ=32cosθ
sinϕ=43cosθ
sinϕ=21cosθ
sinϕ=cosθ
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