Q 11-04-137JEE MainJEE Main 2022 (25 Jun, Shift 2)Medium
A block of mass $200\ \text{g}$ is kept stationary on a smooth inclined plane by applying a minimum horizontal force $F = \sqrt x\ \text{N}$ as shown in figure. The value of $x$ = ______.
Numerical value type. Enter your answer.
Answer: 12
Resolve along the incline: the component of $F$ up the incline balances the component of weight down it:
$$F\cos60^\circ = mg\sin60^\circ\ \Rightarrow\ F = mg\tan60^\circ = 0.2\times10\times\sqrt3 = 2\sqrt3 = \sqrt{12}\ \text{N}$$
(taking $g = 10\ \text{m s}^{-2}$). So $x = 12$.
Solution by Sreeraj P, M.Sc Physics