System of Particles & Rotational Motion: Torque and Equilibrium MCQ

Solve this quality-checked NEET multiple-choice question, then review the correct answer and explanation.

Practice and revisionDetailed answer reviewMobile-friendly flow

Torque and EquilibriumHardQuestion 6168

Question

A uniform rod (length 2 m, mass 4 kg) is pivoted 0.5 m from one end. What perpendicular force at the farther end keeps it horizontal (g=10 m/s²)?

A
40/3 N
Correct
B
20 N
C
30 N
D
40 N

Correct answer

40/3 N

Explanation

The rod's weight acts 0.5 m from the pivot, producing $4×10×0.5=20$ N·m of torque. The farther end is 1.5 m from the pivot, so $1.5F=20$ and $F=<strong>40/3\text{ N}</strong>$ .

Review System of Particles & Rotational Motion notes Start timed chapter practice