Question Detail
Question
Two boxes with masses 3kg and 7kg are attached to each end of a massless meter rule. At what distance from the 7kg mass should a pivot be placed in order to keep the rule in equilibrium?
Correct Answer
Option A is the correct answer.
Detailed Explanation
For equilibrium, the net torque about the pivot must be zero. Let the pivot be at distance 'x' from the 7kg mass. Then the distance from the 3kg mass is (100 - x) cm. Clockwise torque = anticlockwise torque: 3g × (100 - x) = 7g × x => 300 - 3x = 7x => 300 = 10x => x = 30 cm from the 7kg mass.
Hint
Apply the principle of moments: Sum of clockwise moments = Sum of anticlockwise moments.