A heavy wheel of radius 20 cm and weight 10kg is to be dragged over a step of height 10cm, by a horizontal force F applied at the centre of the wheel. The minimum value of F is

Given,

radius of wheel r = 20cm

height of step = 10cm

Let the force required to drag the wheel over the step of height = 10cm be F.

Now, balancing torques due to forces F and Mg,

${\mathrm{\tau }}_{\mathrm{F}}={\mathrm{\tau }}_{\mathrm{Mg}}$

$\mathrm{F}\times \mathrm{OB}\text{'}=\mathrm{Mg}\times \mathrm{BB}\text{'}$

$\mathrm{F}\times (20-10)=\mathrm{Mg}\times \sqrt{{20}^2-{10}^2}$

$\mathrm{F}=\frac{10\times 10\sqrt{3}}{10}=10\sqrt{3}\mathrm{kgwt}$

Hence the correct answer is $10\sqrt{3}\mathrm{kgwt}.$