1. Physical idea

To just lift the cylinder over the step, the cylinder will start rotating about the corner of the step.

Let the corner of the step be point $P$. At the instant the cylinder is about to climb:

• the ground reaction becomes zero,
• the cylinder pivots about $P$,
• we balance torques about $P$.

2. Geometry of the situation

The cylinder has radius $R$, and the step has height $a$.

At the instant of just lifting, the center $O$ of the cylinder is still at height $R$ from the ground. The pivot point $P$ is at height $a$.

So the vertical distance between $O$ and $P$ is $O_y-P_y = R-a.$

Since $OP=R$ (because the corner touches the cylinder), the horizontal distance between the center and the corner is $x=\sqrt{R^2-(R-a)^2}.$

Thus, $x=R\sqrt{1-\left(\frac{R-a}{R}\right)^2}.$

3. Torques about the corner $P$

Suppose the cylinder is pulled horizontally by force $F$ through its center.

• The weight $Mg$ acts vertically downward through the center.
• The perpendicular distance of $Mg$ from $P$ is the horizontal distance $x$.

So torque due to weight is $\tau_g = Mg\,x.$

• The force $F$ acts horizontally through the center.
• Its perpendicular distance from $P$ is the vertical distance between the center and the corner, i.e. $R-a.$

So torque due to pulling force is $\tau_F = F(R-a).$

At the threshold of climbing, $F(R-a)=Mg\,x.$

Substitute $x$: $F(R-a)=Mg\sqrt{R^2-(R-a)^2}.$

Therefore, $F=Mg\frac{\sqrt{R^2-(R-a)^2}}{R-a}.$

Simplify: $F=Mg\sqrt{\frac{R^2-(R-a)^2}{(R-a)^2}}$ $F=Mg\sqrt{\frac{R^2}{(R-a)^2}-1}$ $F=Mg\sqrt{\left(\frac{R}{R-a}\right)^2-1}.$

4. Match with the options

This is exactly Option D: $Mg\sqrt{\left(\frac{R}{R-a}\right)^2-1}.$

5. Check Option A

Option A is $Mg\sqrt{1-\left(\frac{R-a}{R}\right)^2}.$

This equals $Mg\frac{\sqrt{R^2-(R-a)^2}}{R},$ which is not the same as the required expression, because the correct denominator is $(R-a)$, not $R$.

So Option A is incorrect for a horizontal pull through the center.

6. Final answer

The correct option is D.