Q83 / 90 · JEE Main 2022 (25 Jun Shift 2) · Physics

A block of mass $200 g$ is kept stationary on a smooth inclined plane by applying a minimum horizontal force $F = \sqrt{x} N$ as shown in figure.
The value of $x =_____$ .

Correct answer: 12

Solution

Balancing the forces in horizontal direction $N \sin θ = F$

And balancing the forces in vertical direction $N \cos θ = m g$

$\Rightarrow \tan θ = \frac{F}{m g}$ $\Rightarrow F = m g \tan θ$

$F = 0.2 \times 10 \times \sqrt{3} = \sqrt{12} N$ . Therefore, $x = 12$

Hence, the correct answer is 12.