Torque, Couple & Equilibrium of Rigid Bodies question from JEE Main 2022 (28 Jun, Shift 2)

Answer: (c)

1. Height of the top $= \sqrt{34 - 9} = 5$ m. The smooth wall gives only a horizontal force $F_w$.
2. Moments about the foot: $F_w \times 5 = mg \times 1.5 = 100 \times 1.5 \Rightarrow F_w = 30$ N
3. Floor: normal $= 100$ N, friction $= F_w = 30$ N, so $F_f = \sqrt{100^2 + 30^2} = 10\sqrt{109}$ N
4. $\frac{F_w}{F_f} = \frac{30}{10\sqrt{109}} = \frac{3}{\sqrt{109}}$
5. So the answer is (c) $\frac{3}{\sqrt{109}}$.