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Physics · SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
हिन्दीA uniform meter scale of mass \( 100\,\text{g} \) is balanced horizontally at the \( 40\,\text{cm} \) mark when an unknown mass \( m \) is suspended at the \( 10\,\text{cm} \) mark. The value of the unknown mass \( m \) is:
Correct answer · Option A
33.3 g
Explanation
The weight of the uniform meter scale acts at its centre of gravity, i.e., at \( 50\,\text{cm} \). Taking torque about fulcrum at \( 40\,\text{cm} \): \( m \times (40 - 10) = 100 \times (50 - 40) \implies 30 m = 1000 \implies m = \frac{100}{3} \approx 33.3\,\text{g} \).
