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When a cube is floating in water, 20% of the cube is outside the water. When the same cube is placed in another liquid, 35% of the cube is outside the liquid. The density of the liquid ${gcm}^{- 3}$ is

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$\frac{16}{13}$

$\frac{4}{13}$

$\frac{13}{20}$

$\frac{4}{5}$

answer is A.

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Detailed Solution

when a body is floating in a liquid ,

apparent wt = upthrust
$V_{b} d_{b} g = V_{in} d_{L} g \Rightarrow d_{b} = (\frac{V_{in}}{V_{b}}) d_{L}$
Apply for the liquids,

$d_{b} = \frac{80}{100} d_{W}$ .....(1)
$d_{b} = \frac{65}{100} d_{L}$ ......(2)
From 1 and 2 :

$\frac{80}{100} d_{W} = \frac{65}{100} d_{L} \Rightarrow d_{L} = \frac{80}{65}$
$d_{L} = \frac{16}{13} g / {cm}^{3}$

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