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 midpoint. 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. When $w$ 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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$\frac{w}{w_{1}}$

$\frac{w l_{1}}{w l - w_{1} l_{2}}$

$\frac{l_{1}}{l_{1} - l_{2}}$

$\frac{l_{1}}{l_{2}}$

answer is C.

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Detailed Solution

In equilibrium condition in air
$w l = w_{1} l_{1} ...... (i)$
Now, in water
$(W - F_{B}) l = W_{1} l_{2}$ (where $F_{B}$ = buoyant force $= \frac{W}{ρ}$ )
Hence, $(W - \frac{W}{ρ}) l = W_{1} l_{2}$
$W (1 - \frac{1}{ρ}) l = W . \frac{l}{l_{1}} . l_{2} {W_{1} = \frac{W}{l_{1}} l} (1 - \frac{1}{ρ}) = \frac{l_{2}}{l_{1}}$