Q 11-04-187JEE MainJEE Main 2019 (9 Jan, Shift 2)Easy
A mass of $10\ \text{kg}$ is suspended vertically by a rope from the roof. When a horizontal force is applied on the rope at some point, the rope deviated at an angle of $45^\circ$ at the roof point. If the suspended mass is at equilibrium, the magnitude of the force applied is $(g = 10\ \text{m s}^{-2})$
Answer: (A) $100\ \text{N}$
Take the point where the force acts (with the rope below it and the mass) as the system. The upper part of the rope pulls along itself with tension $T$ at $45^\circ$ to the vertical.
Vertical: $T\cos45^\circ = mg = 100\ \text{N}$
Horizontal: $T\sin45^\circ = F$
So $F = mg\tan45^\circ = 100\ \text{N}$.
Solution by Sreeraj P, M.Sc Physics