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Physics · PYQ NEET UG 2025
हिन्दीA uniform rod of mass \( 20\text{ kg} \) and length \( 5\text{ m} \) leans against a smooth vertical wall making an angle of \( 60^\circ \) with it. The other end rests on a rough horizontal floor. The friction force that the floor exerts on the rod is (take \( g = 10\text{ m/s}^2 \))
Correct answer · Option B
\( 100\sqrt{3}\text{ N} \)
Explanation
Torque balance about the bottom end: \( N_{wall} \cdot L \cos(60^\circ) = mg \cdot \frac{L}{2} \sin(60^\circ) \). Thus \( N_{wall} = \frac{mg}{2} \tan(60^\circ) \). Friction force \( f = N_{wall} = 100 \times \sqrt{3} = 100\sqrt{3}\text{ N} \).
