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?

Options

A. `(2g)/3`
B. `g/2`
C. `(5g)/6`
D. g

Answer

B

Explanation

For the mass m, mg - T = ma
As we know, `a = R alpha`
So, `mg-T = m R alpha`……(i)
As torque `=T xx R = mR^(2) alpha`……..(ii)
From Eqs. (i) and (ii), we get `a= g//2`

Hence, the body will fall with an acceleration `g/2`.