A cord is wrapped on a pulley (disk) of mass M and radius R as shown in figure. To one end of the cord, a block of mass M is connected as shown and to other end in (a) a force of 2 Mg and in (b) a block of mass 2 M. Let angular acceleration of the disk in A and B is `alpha_(A) and alpha_(beta)` respectively, then (cord is not slipping on the pulley)

Options

A. `alpha_(A) = alpha_(B)`
B. `alpha_(A) gt alpha_(B)`
C. `alpha_(A) lt alpha_(B)`
D. None of these

Answer

B

Explanation

In (A)
`(T_(2)-T_(1)R) = (MR^(2))/2 alpha_(A)`, `T_(1) - Mg = Ma_(A)`, `T_(2) = 2Mg`
`a_(A) = R alpha_(A)`

Solve to get `alpha_(A) =(2g)/(3R)`
For (a)
`(T_(2)-T_(1)) xx R = (MR^(2))/2 alpha_(B)`

`(T_(2)-T_(1)) xx R =(MR^(2))/2 alpha_(B)`
`T_(1)-Mg = Ma_(s)`
`2Mg -T_(2)= 2Ma_(s)`
`a_(B) = R alpha_(B)`
Solve to get `alpha_(B) = (2g)/(7R) `, So, `alpha_(A) gt alpha_(B)`.