A cord wrapped on a pulley (disc) of mass M and radius R as shown in the 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 is applied and in (B) a block of mass 2M is connected. Let angular acceleration of the disc in A and B are `alpha_(A)` and `alpha_(B)`, respectively, then: [cord is not slipping on pulley]:

Options

A. 2
B. 3
C. `sqrt(2)`
D. `sqrt(3)`

Answer

D

Explanation

Net torque about 0 should be zero, Hence,
`mg(1/2 sin 60^(@)) = Mg(1/2 sin 30^(@)) therefore M/m = (sin 60^(@))/(sin 30^(@)) = sqrt(3)`