A clockwise torque of `6N-m` is applied to the circular cylinder as shown in the figure. There is no friction between the cylinder and the block.

Options

A. `F/2`
B. `(3F + 29 Mg sin theta)/57`
C. `(3F + 2Mg sin theta)/19`
D. Coefficient of friction value is not given and hence Cannot be found

Answer

C

Explanation

The free body diagram of the plank and the cylinder is as shown in the figure.
For plank, `F-f_(1)=2 Mg sin theta = 2 Ma_(1)`
For cylinder, `f_(1)-f_(2) = Mg sin theta = Ma_(2)`
`(f_(1) + f_(2))R =(MR^(2))/2 alpha`

For no slipping between plank and cylinder, `a_(2) = Ralpha = a_(1)`
For no slipping between cylinder and incline, `a_(2) = Ralpha`
After, solving the above equation, `f_(1) =(3F + 2Mg sin theta)/15`