Angular acceleration of the cylinder `C` shows in the figure is (all strings and the pulley are ideal)

Options

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

Answer

B

Explanation

`|vec(a)_(A)|=a+Ralpha`
`|vec(a)_(B)|=a-Ralpha`
`mg-T_(1)=m(a+Ralpha)......(1)`
`T_(1)+mg-T_(2)=ma......(2)`
`T_(2)=m(a-Ralpha)......(3)`
`T_(1)R+T_(2)R=(mR^(2))/2 alpha`
On solving these eqn. we get
`a=(2g)/3, alpha=(2g)/(5R)`