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Physics · SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
हिन्दीA uniform meter scale of mass \( 0.2\text{ kg} \) is pivoted at the \( 40\text{ cm} \) mark. A load mass of \( 0.5\text{ kg} \) is suspended at the \( 10\text{ cm} \) mark. To keep the meter scale in rotational equilibrium horizontally, the mass that must be suspended at the \( 90\text{ cm} \) mark is:
Correct answer · Option 2
\( 260\text{ g} \)
Explanation
Taking torque about fulcrum (40 cm): \( (0.5)(40 - 10) = (0.2)(50 - 40) + m_2(90 - 40) \implies 15 = 2 + 50 m_2 \implies m_2 = 13/50 = 0.26\text{ kg} = 260\text{ g} \).
