Q 11-04-182JEE MainJEE Main 2020 (7 Jan, Shift 2)Easy
A mass of $10$ kg is suspended by a rope of length $4$ m from the ceiling. A force $F$ is applied horizontally at the mid-point of the rope such that the top half of the rope makes an angle of $45^\circ$ with the vertical. Then $F$ equals: (Take $g = 10\ \text{m s}^{-2}$ and the rope to be massless)
Answer: (A) $100$ N
The lower half of the rope hangs vertically with tension $mg = 100$ N.
At the mid-point, with $T$ the tension in the upper half:
Vertical: $T\cos45^\circ = 100$ N
Horizontal: $T\sin45^\circ = F$
Dividing: $F = 100\tan45^\circ = 100$ N.
Solution by Sreeraj P, M.Sc Physics