Torque, Couple & Equilibrium of Rigid Bodies question from JEE Main 2020 (7 Jan, Shift 2)

Answer: 50

1. To just slide, the horizontal force is $F = \mu mg$. It acts at height $\frac{a}{2} + b$ above the floor.
2. No toppling about the front edge: $F\left(\frac{a}{2} + b\right) \le mg\frac{a}{2} \Rightarrow b \le \frac{a(1-\mu)}{2\mu} = 0.75a$
3. But the force must act on the face, so $b \le \frac{a}{2}$. This is less than $0.75a$, so the box never topples.
4. Maximum possible $b = \frac{a}{2}$, so $100 \times \frac{b}{a} = 50$.