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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 $12 \mathrm{~cm}$ mark, the stick is found to be balanced at $45 \mathrm{~cm}$. What is the mass of the metre stick?
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Verified Answer
Let $\mathrm{m}$ be the mass concentrated at $C$.
For equilibrium about $C$, the moment of the forces should be equal.
$\log (45-12)=\mathrm{mg}(50-45)$
$$
\Rightarrow 10 \times 33=m \times 5 \Rightarrow m=\frac{10 \times 33}{5}=66 \mathrm{~g}
$$
For equilibrium about $C$, the moment of the forces should be equal.
$\log (45-12)=\mathrm{mg}(50-45)$
$$
\Rightarrow 10 \times 33=m \times 5 \Rightarrow m=\frac{10 \times 33}{5}=66 \mathrm{~g}
$$
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