A fixed pulley of radius 20 cm and moment of inertia 0.32 kg.`m^(2)` about its axle has a massless cord wrapped around its rim. A mass M of 2 kg is attached to the end of the cord. The pulley can rotate about its axis without any friction. The acceleration of the mass M is :- (Assuming `g=10m//s^(2)`)

Explanation

For the motion of the block 2g –T = 2a
For the motion of the block ` tau = TR = I alpha`
`because a = alpha R :. T = (Ia)/R^2`
` :." "T = (Ia)/R^2 rArr 2g - (Ia)/R^2 =2a`
`rArr = g/(1+I(2R^2))`