Question

A block of mass 300 g is kept stationary on a smooth inclined plane by applying a minimum horizontal force \( F = \sqrt{a} \, N \) as shown in the figure. The value of \( a \) is (Take \( g = 10 \, \text{m/s}^2 \)):

• A. 12
• B. 24
• C. 27
• D. 9

Hint:

When dealing with forces on an inclined plane, break the applied force into components parallel and perpendicular to the plane. Use the condition of equilibrium to solve for unknown quantities.

Answer 1

The Correct Option is C

Let the mass of the block be \( m = 300 \, \text{g} = 0.3 \, \text{kg} \). The horizontal force \( F \) is applied at an angle of \( 60^\circ \) with respect to the inclined plane. The block remains stationary, which means the forces acting on the block must balance. The force of gravity acting on the block is: \[ F_{\text{gravity}} = mg = 0.3 \times 10 = 3 \, \text{N} \] The force \( F \) can be broken into two components: - The component parallel to the incline: \( F_{\parallel} = F \sin 60^\circ \) - The component perpendicular to the incline: \( F_{\perp} = F \cos 60^\circ \) Since the plane is smooth, there is no friction. Therefore, the normal force \( N \) balances the perpendicular component of the applied force: \[ N = F_{\perp} = F \cos 60^\circ = \frac{F}{2} \] The block remains stationary along the plane, so the component of gravity along the plane is balanced by the parallel component of the applied force: \[ F_{\parallel} = F_{\text{gravity}} \sin 60^\circ = 3 \sin 60^\circ = 3 \times \frac{\sqrt{3}}{2} = \frac{3\sqrt{3}}{2} \] Using the condition that \( F_{\parallel} = F \sin 60^\circ \), we can set up the equation: \[ F \sin 60^\circ = 3 \times \frac{\sqrt{3}}{2} \] Substituting the values and solving for \( F \): \[ F \times \frac{\sqrt{3}}{2} = \frac{3\sqrt{3}}{2} \] \[ F = 3 \, \text{N} \] Thus, the value of \( a \) is: \[ F = \sqrt{a} \Rightarrow 3 = \sqrt{a} \Rightarrow a = 9 \] Thus, the correct answer is \( a = 27 \).