A metre stick is balanced on a knife edge at its centre. When two coins, each of mass $5 \,g$ are put one on top of the other at the $12 \,cm$ mark, the stick is found to be balanced at $45 \,cm$. The mass of the metre stick is
Updated On: Jul 5, 2022
$56\,g$
$66\,g$
$76\,g$
$86\,g$
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The Correct Option isB
Solution and Explanation
Let $m$ be the mass of the metre stick concentrated at $C$, the $50 \,cm$ mark as shown in the figure.
In equilibrium, taking moments of forces about $C'$, we get
$10 g(45-12) = mg (50-45);$$10g \times 33 = mg \times 5$$ m = \frac{10 \times 33}{5} = 66\,g$
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