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Torque about a Point

A uniform rod...

Question

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. 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

\frac{w}{w_{1}}

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

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

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

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Solution

The correct option is C $\frac{l_{1}}{l_{1} - l_{2}}$
where $ρ =$ specific gravity
$W l (1 - \frac{1}{ρ}) = W_{1} l_{2}$
$(1 - \frac{1}{ρ}) = \frac{W_{1} l_{2}}{W l}$ [from equation (i)]
$(1 - \frac{1}{ρ}) = \frac{W_{1} l_{2}}{W_{1} l_{1}}$
$1 - \frac{1}{ρ} = \frac{l_{2}}{l_{1}} \Rightarrow \frac{1}{ρ} = 1 - \frac{l_{2}}{l_{1}}$
$\frac{1}{ρ} = \frac{l_{1} - l_{2}}{l_{1}}$ or $ρ = \frac{l_{1}}{l_{1} - l_{2}}$

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