In the figure shown acceleration with which the mass m falls down when released is ( consider the string to be massless g-acceleration due to gravity )

Options

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

Answer

B

Explanation

Let tension in string is T and tension T is rotating the hollow cylinder. Torque produced in hollow cylinder, `tua = I alpha ` Moment of inerita of cylinder `I= MR^(2)` where M = mass of cylinder and R = radius of cylinder.

Angular acceleration `alpha = (a)/(R ), ` where a= acceleration.
`implies tau= MR^(2). (a)/(R ) = MR a " " cdots (i)`
Also torque `tau = F.R = TR " " cdots (ii)`
Equating Eqs . (i) and (ii) we get TR = MRa
` implies T = Ma `
Now `" "` Mg -T = Ma
Mg - Ma = Ma `implies 2 Ma = Mg `
`implies a= g//2`