Q 11-04-091JEE MainJEE Main 2024 (9 Apr, Shift 2)Easy
A $1\ \text{kg}$ mass is suspended from the ceiling by a rope of length $4\ \text{m}$. A horizontal force $F$ is applied at the mid point of the rope so that the rope makes an angle of $45^\circ$ with respect to the vertical axis as shown in the figure. The magnitude of $F$ is: (Assume that the system is in equilibrium and $g = 10\ \text{m/s}^2$)
Answer: (A) $10\ \text{N}$
Lower half: $T_2 = mg = 10\ \text{N}$ (vertical).
At the midpoint, balance of forces:
$$T_1\cos45^\circ = T_2 = 10\ \text{N},\qquad T_1\sin45^\circ = F$$
So $F = T_2\tan45^\circ = 10\ \text{N}$.
Solution by Sreeraj P, M.Sc Physics