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

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The acceleration of the system will be $1 m / s^{2}$ forward.

The angular acceleration of the cylinder is 10 rad s^-2.

The cylinder will be slipping but the system does not move forward.

The system cannot move forward for any torque applied to the cylinder.

answer is C, D.

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Detailed Solution

$a = \frac{f_{1} - f_{2}}{m_{1} + m_{2}} = \frac{0 .4 \times 6 g - 0 .5 \times 3 g}{6 + 3}$

$\Rightarrow a = 1 {ms}^{- 2}$

$α = \frac{τ_{applied} - τ_{friction}}{I} = \frac{6 - f_{1} R}{\frac{1}{2} 6 (R)^{2}}$

$= \frac{6 - 0 .4 \times 6 g (0 .2)}{\frac{1}{2} \times 6 \times (0 .2)^{2}} = 10 {rads}^{- 2}$

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