A block of mass 200 g is kept stationary on a smooth inclined plane by applying a minimum horizontal force F = √(x) N as shown in figure. The value of x = _____________. Numerical answer type. Work it out, then open the answer.

Answer: 12. Smooth incline at 60°, horizontal force F: balance along the incline gives Fcos 60° = mgsin 60°. F = mgtan 60° = 0.2 × 10 × √(3) = 2√(3) N = √(12) N So x = 12.

A block of mass 200 g is kept stationary on a smooth inclined plane… | JEE Main 2022 (25 Jun, Shift 2)

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Connected Bodies & Pulleys

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 = _____________.

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 12

Smooth incline at $60^\circ$ , horizontal force $F$ : balance along the incline gives $F\cos 60^\circ = mg\sin 60^\circ$ .
$F = mg\tan 60^\circ = 0.2 \times 10 \times \sqrt{3} = 2\sqrt{3}\text{ N} = \sqrt{12}\text{ N}$
So $x = 12$ .

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