A block having equilateral triangular cross-section of side a and mass `m` is placed on a rough inclined surface, so that it remains in equilibrium as shown in figure. The torque of normal force acting on the block about its centre of mass is

Options

A. `(sqrt(3))/2 mga sin theta`
B. `1/(2sqrt(3))mga sin theta`
C. `1/(2sqrt(3)) mga cos theta`
D. zero

Answer

B

Explanation

Here torque due to normal force balances the torque due to frictional force which is equal to `1/(2sqrt(3)) mga sin theta`.