a=(1+I/mR2)g sin θ
aS=(1+2/5)g sin θs
aC=(1+1/2)gsinθC
Now, aS = aC
(57)gsinθS=(23)gsinθC
75 sin θS=32 sin θC
∴sinθSsinθC=7×25×3=1415
JEE Main 2014 — Physics Mechanics
A cylinder of mass Mc and sphere of mass Ms are placed at points A and B of two inclines, respectively. (See figure). If they roll on the incline without slipping such that their accelerations are the same, then the ratio sinθssinθc is :

Held on 9 Apr 2014 · Verified 6 Jul 2026.
78
78
1415
1415
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