A uniform cylinder of mass M and radius R is to be pulled over a step of height a (a<R) by applying a force F at its centre ‘O’ perpendicular to the plane through the axes of the cylinder on the edge of the step (see figure). The minimum value of F required is:

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$M g \sqrt{1 - \frac{a^{2}}{R^{2}}}$

$M g \sqrt{1 - {(\frac{R - a}{R})}^{2}}$

$M g \sqrt{{(\frac{R}{R - a})}^{2} - 1}$

$M g \frac{a}{R}$

answer is B.

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Detailed Solution

Assume no slipping about P

Torque of F about P> torque of gravity

$\begin{array}{l} \Rightarrow F R > M g x \\ \Rightarrow F R > M g R \sin θ \\ \Rightarrow F_{\min} = M g \sin θ = M g \frac{x}{R} \\ = M g \frac{\sqrt{R^{2} - {(R - a)}^{2}}}{R} (f r o m f i g u r e) \\ = M g \sqrt{\frac{R^{2} - {(R - a)}^{2}}{R^{2}}} \\ = M g \sqrt{1 - {(\frac{R - a}{R})}^{2}} \end{array}$

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