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First, making an FBD of forces acting, we find that :
Force F acts on the topmost side parallel to it.
Force mg acts downwards from the centre of gravity of the hexagon.
Force Fr(friction) acts on the base of the hexagon, opposing the impending motion(opposite in direction to F).
Also, we know that F = μmg.
As the block is to topple, the rotational axis passes through point D and is perpendicular to the plane of the hexagon.
We then have to find the torques due to the various forces.
Let the hexagon have side "a".
Torque Due to F
In triangle BCD, by the cosine rule,
cos120° = a² + a² - BD²2a²
=> BD = √3a
Ï„F = F * √3a, as √3a is the perpendicular distance of the line of force F from the axis of rotation.
Torque due to weight
Clearly, we see that from the line of force mg, the axis of rotation is at half the distance ED, which is a/2.
So Ï„mg = mga2
Torque due to friction
The line of force of friction intersects the axis of rotation, hence there is no torque due to it.
Clearly we see that force F provides a clockwise torque and force mg provides an anticlockwise torque.
For toppling,
Ï„F > Ï„mg
=> √3aF > mga2
=> √3 * μmg > mg2
=> μ > 0.28 > 0.21
Hence (A) is the answer.