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