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A body floats with $\frac{1^{t h}}{4}$ volume outside water and half of the volume outside a liquid, then the density of liquid is

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1.5 kg m^-3

1500 kg m^-3

100 kg m^-3

2000 kg m^-3

answer is B.

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Detailed Solution

Vdg= veg

$V d g = \frac{3 V}{4} \times 1 \times g$
$d = \frac{3}{4}$
$V d g = V^{1} e^{1} g$ $V \times \frac{3}{4} = \frac{V}{2} e^{1}$
$e^{1} = \frac{3}{2} g m / c c = 1500 k g / m^{3}$

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