A uniform solid cylinder is placed on a horizontal board. The coefficient of friction between them is $0.3$ . The board is imparted a constant acceleration of $a = 6 m / s^{2}$ in a horizontal direction at right angles of the cylindrical axis. Then (take $g = 10 m / s^{2}$ )

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The acceleration of the cylinder axis in the absence of slipping will be $1.5 m / s^{2}$

The acceleration of the cylinder axis in the absence of slipping will be $2 m / s^{2}$

The limiting value of ‘ $a$ ’ for which there will be no slipping is $12 m / s^{2}$

The limiting value of ‘ $a$ ’ for which there will be no slipping is $9 m / s^{2}$

answer is B, D.

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Detailed Solution

$m a_{c} = F_{f r} I \propto = r F_{f r} = r m a_{c} a - a_{c} = α r = \frac{r m a_{c}}{I} r o r a - a_{c} = \frac{m r^{2} a_{c}}{\frac{m r^{2}}{2}} = 2 a_{c}$

or $3 a_{c} = a, a_{c} = \frac{a}{3} = \frac{6}{3} = 2 m / s^{2}$

So, $f_{F R} = \frac{m a}{3}$ so maximum value of this force is equal $μ m g$
Hence $μ m g = \frac{m a_{l i m}}{3}$ or, $a_{\lim} = 3 μ g = 3 \times 0.3 \times 10 = 9 m / s^{2}$