In the arrangement as shown, A and B are two cylinders each of length L and density $d_{1}$ and $d_{2}$ $(d_{1} < d_{2})$ respectively. They are joined together and submerged in the liquid of density d with length $\frac{L}{2}$ of A outside the liquid. If the difference in densities of two cylinders is equal to density of the liquid, find $d_{1}$

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$\frac{5}{4} d$

$\frac{d}{4}$

$\frac{3}{4} d$

$\frac{d}{2}$

answer is B.

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Detailed Solution

Upthrust = Net weight $\frac{3}{2} LAdg = (d_{1} + d_{2}) LAg$

$\Rightarrow d_{1} + d_{2} = \frac{3}{2} d and d_{2} - d_{1} = d$

$\Rightarrow d_{2} = \frac{5}{4} d and d_{1} = \frac{d}{4}$

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