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 L/2 of A outside the liquid. If the difference in densities of two cylinders is equal to density of the liquid, find $d_{1}$ :

\frac{5}{4} d

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\frac{d}{4}

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\frac{3}{4} d

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\frac{d}{2}

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Solution

The correct option is A $\frac{d}{4}$
Applying the law of flotation, we have weight of the two blocks = weight of water displaced by the blocks
$\Rightarrow V d_{1} g + V d_{2} g = V d g + \frac{V}{2} d g$
$\Rightarrow d_{1} + d_{2} = \frac{3}{2} d$ ......... $(I)$
$d_{2} - d_{1} = d$ ......... $(I I) ∵$ given
$(I) - (I I)$ gives
$2 d_{1} = \frac{d}{2} \Rightarrow d_{1} = \frac{d}{4}$

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