# 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 center' 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 : Option: 1 $Mg\sqrt{1-\left ( \frac{R-a}{R} \right )^{2}}$   Option: 2 $Mg\sqrt{\left ( \frac{R}{R-a} \right )^{2}-1}$ Option: 3 $Mg\frac{a}{R}$   Option: 4 $Mg\sqrt{1-\frac{a^{2}}{R^{2}}}$

For minimum force to topple

Let the net torque about point P tends to zero

i.e

$\begin{array}{l} (\tau)_{P}=0 \\ F_{min}.R. -\mathrm{mgx}=0 \end{array}$

And $x=\sqrt{R^{2}-(R-a)^{2}}$

So

$\begin{array}{l} \mathrm{F_{min}}=\mathrm{mg} \frac{\mathrm{x}}{\mathrm{R}} \\ \\ \mathrm{F_{min}}=\mathrm{mg} \sqrt{1-\left(\frac{\mathrm{R}-\mathrm{a}}{\mathrm{R}}\right)^{2}} \end{array}$

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