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A uniform rod is suspended horizontally from its mid-point. A piece of metal whose weight is $w$ is suspended at a distance $l$ from the mid-point. Another weight $W_{1}$ is suspended on the other side at a distance $l_{1}$ from the mid-point to bring the rod to a horizontal position. Wben $w$ is completely immersed in water, $w_{1}$ needs to be kept at a distance $l_{2}$ from the mid-point to get the rod back into horizontal position. The specific gravity of the metal piece is
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Verified Answer
The correct answer is:
$\frac{L_{1}}{l_{1}-l_{2}}$
$$
W-F_{B}=W\left(1-\frac{1}{\rho}\right)
$$
where $\rho=$ specific gravity $W\left(1-\frac{1}{\rho}\right)=W_{1} b$
$\left(1-\frac{1}{\rho}\right)=\frac{W_{1} L_{2}}{W}$
$\left(1-\frac{1}{\rho}\right)=\frac{W_{1} b_{2}}{W_{1} L_{1}}$
$1-\frac{1}{\rho}=\frac{L_{2}}{L_{1}} \Rightarrow \frac{1}{\rho}=1-\frac{b}{L_{1}}$
$\frac{1}{\rho}=\frac{l_{1}-L_{2}}{l_{1}}$ or $\rho=\frac{l_{1}}{l_{1}-L_{2}}$
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