Q 11-04-151JEE MainJEE Main 2022 (27 Jun, Shift 2)Medium
A mass of $10$ kg is suspended vertically by a rope of length $5$ m from the roof. A force of $30$ N is applied at the middle point of rope in horizontal direction. The angle made by upper half of the rope with vertical is $\alpha = \tan^{-1}(x\times10^{-1})$. The value of $x$ is ______ .
(Given, $g = 10\ \text{m s}^{-2}$)
Numerical value type. Enter your answer.
Answer: 3
Treat the rope as massless. The lower half holds the 10 kg mass, so it pulls the midpoint down with $T_2 = 100$ N.
At the midpoint, the upper half's tension balances $100$ N downward and $30$ N horizontal:
$$\tan\alpha = \frac{30}{100} = 0.3 = 3\times10^{-1}$$
$x = 3$.
Solution by Sreeraj P, M.Sc Physics