
v0=v2+2g(2R)v0=n2gR+4gR∴ktop kbottom =v2v02=n2n2+4
JEE Main 2025 — Physics Mechanics
A body of mass ' m ' connected to a massless and unstretchable string goes in verticle circle of radius ' R 'under gravity g. The other end of the string is fixed at the center of circle. If velocity at top of circular path is ngR, where, n⩾1, then ratio of kinetic energy of the body at bottom to that at top of the circle is
Held on 29 Jan 2025 · Verified 6 Jul 2026.
n2+4n2
n2n2+4
nn+4
n+4n
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