Two solids A and B float in water. It is observed that A floats with 12 of its body immersed in water and B floats with 14 of its volume above the water level. The ratio of the density of A to that of B is

Upthrust = weight of body For A, VA2×ρW×g=VA×ρA×g⇒ρA=ρW2For B, 34VB×ρW×g=VB×ρB×g⇒ρB=34ρW(Since 1/4 of volume of B is above the water surface)∴ρAρB=ρW/23/4 ρW=23

Two solids A and B float in water. It is observed that A floats with 12 of its body immersed in water and B floats with 14 of its volume above the water level. The ratio of the density of A to that of B is

Upthrust = weight of body For A, VA2×ρW×g=VA×ρA×g⇒ρA=ρW2For B, 34VB×ρW×g=VB×ρB×g⇒ρB=34ρW(Since 1/4 of volume of B is above the water surface)∴ρAρB=ρW/23/4 ρW=23

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Two solids A and B float in water. It is observed that A floats with $\frac{1}{2}$ of its body immersed in water and B floats with $\frac{1}{4}$ of its volume above the water level. The ratio of the density of A to that of B is

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4 : 3

2 : 3

3 : 4

1 : 2

answer is B.

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Detailed Solution

Upthrust = weight of body
For A, $\frac{V_{A}}{2} \times ρ_{W} \times g = V_{A} \times ρ_{A} \times g \Rightarrow ρ_{A} = \frac{ρ_{W}}{2}$
For B, $\frac{3}{4} V_{B} \times ρ_{W} \times g = V_{B} \times ρ_{B} \times g \Rightarrow ρ_{B} = \frac{3}{4} ρ_{W}$
(Since 1/4 of volume of B is above the water surface)
$∴ \frac{ρ_{A}}{ρ_{B}} = \frac{ρ_{W} / 2}{3 / 4 ρ_{W}} = \frac{2}{3}$