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Physics · SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
हिन्दीA uniform metre stick is balanced on a knife edge at its centre (\( 50.0 \text{ cm} \) mark). When two identical coins, each of mass \( 5.0 \text{ g} \), are placed one on top of the other at the \( 12.0 \text{ cm} \) mark, the stick is found to be balanced about the knife edge at the \( 45.0 \text{ cm} \) mark. The mass of the metre stick is:
Correct answer · Option 1
\( 66.0 \text{ g} \)
Explanation
Total mass of two coins \( m = 10.0 \text{ g} \). Lever arm of coins from new fulcrum \( = 45.0 - 12.0 = 33.0 \text{ cm} \). Lever arm of CM of stick (at \( 50.0 \text{ cm} \)) from fulcrum \( = 50.0 - 45.0 = 5.0 \text{ cm} \). By torque balance: \( 10.0 \times 33.0 = M \times 5.0 \implies M = \frac{330}{5} = 66.0 \text{ g} \).
