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A uniform metal plate shaped like a triangle $\mathrm{ABC}$ has a mass of $540 \mathrm{gm}$. the length of the sides $\mathrm{AB}, \mathrm{BC}$ and CA are $3 \mathrm{~cm}, 5 \mathrm{~cm}$ and $4 \mathrm{~cm}$, respectively. The plate is pivoted freely about the point $\mathrm{A}$. What mass must be added to a vertex, so that the plate can hang with the long edge horizontal ?
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$140 \mathrm{gm}$ at $\mathrm{C}$
$\mathrm{CM}$ of triangular plate is on the median. If we put a mass say $\mathrm{m}_{1}$ on $\mathrm{C}$ it will produce torque about A which balance the torque produce $m g$ about $A$. Thus plate will can be in equilibrium position
$\begin{array}{l}
\mathrm{m}_{1} \mathrm{~g} \times 4 \cos 37=\mathrm{mg} \times \mathrm{y} \\
\mathrm{m}_{1} \mathrm{~g} \times 4 \times \frac{4}{5}=\mathrm{mg} \times \mathrm{y} \\
\mathrm{m}_{1}=\mathrm{m} \times \mathrm{y} \times \frac{5}{16} \\
\frac{\mathrm{m}_{1}}{\mathrm{~m}}=\mathrm{y} \times \frac{5}{16} \\
\mathrm{y} < 3 \quad \therefore \frac{\mathrm{m}_{1}}{\mathrm{~m}} < 1 \\
\mathrm{~m}_{1} < \mathrm{m} \\
\mathrm{m}_{1} < 540 \mathrm{~g}
\end{array}$
from given option Ans. (A)
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