A uniform cube of side b and mass M rests on a rough horizontal table. A horizontal force F is applied normal to one of the face to a point, at a height `(3*b)/4` above the base.

About the bottom right edge, there is a torque created due to two forces, one of them is the weight of the the cube. This is equal to `F*r*sin theta` = `(m*g)*(b/2)*sqrt 2*(1/sqrt 2)` = `(m*g*b)/2`

The other is due to the force F being applied at a height `(3*b)/4` . The torque in this case is `F*(3*b)/4` .

The net torque is `F*(3*b)/4 - m*g*b/2` . If the box topples over before it starts to slide, `F*(3*b)/4 - M*g*b/2 > 0` and `F < mu*M*g`

`F*3/2 > M*g`

=>` F > (2*M*g)/3`

As `mu*M*g > F`

`mu*M*g > (2*M*g)/3`

=> `mu > 2/3`

The coefficient of friction should be greater than `2/3` .

**The correct answer is option A**

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