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Two cubical blocks identical in dimension float in water in such a way that 1^st block floats with half part immersed in water and second block floats with $\frac{3}{4} t h$ of its volume inside the water the ratio of densities of blocks is

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$2 : 3$

$3 : 4$

$1 : 3$

$1 : 4$

answer is A.

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Detailed Solution

Given

Weight of block in water =Weight of block in air

$v_{2} ρ_{2} g = v_{1} ρ_{1} g (\begin{matrix} ∵ w = m g \\ ∵ m = ρ v \end{matrix})$

Given $v_{2} = \frac{3}{4} v, v_{1} = \frac{1}{2} v$

$\Rightarrow \frac{3}{4} v \times ρ_{2} \times g = \frac{1}{2} v \times ρ_{1} \times g$

$\Rightarrow \frac{ρ_{2}}{ρ_{1}} = \frac{2}{3}$

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