A cube of side l and mass M is placed on rough horizontal surface and the friction is sufficient so that it will not move, if a constant force F = Mg is applied horizontally l/4 above the surface. Then the torque due to normal force about center of the cube is equal to
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a
Mgl2
b
Mgl4
c
Mgl8
d
zero
answer is B.
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Detailed Solution
Considering rotational equilibrium , τF-τN=τf ⇒F×l4-τN=f×l2 [ F=Mg , f=F=Mg and N =Mg] ⇒τf= -Mgl4 So magnitude of τf=Mgl4