
mg - T = ma ----- 1

TR=Iα
TR=mR2α
∴T=mRα ----2
For no slippping
a = Rα ---3
putting (3) in (2)
T = ma ---4
Adding 1 and 4
mg = 2ma
∴a=2g
JEE Main 2014 — Physics Mechanics
A mass m is supported by a massless string wound around a uniform hollow cylinder of mass m and radius R. If the string does not slip on the cylinder, with what acceleration will the mass fall on release?

Held on 6 Apr 2014 · Verified 6 Jul 2026.
32g
2g
65g
g
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