A cube of ice floats partly in water and partly in kerosene oil. The ratio of volume of ice immersed in water to that in kerosene oil (specific gravity of kerosene oil = 0.8, specific gravity of ice = 0.9)

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Solution

Q.
A cube of ice floats partly in water and partly in kerosene oil. The ratio of volume of ice immersed in water to that in kerosene oil (specific gravity of kerosene oil = 0.8, specific gravity of ice = 0.9) [2024]

1 5 : 4

2 9 : 10

3 1 : 1

4 8 : 9

Ans.
(3)

$V_{1} = volume immersed in water$

$V_{2} = volume immersed in oil$

Buoyancy force balance weight at equilibrium

$F_{B_{1}} + F_{B_{2}} = m g$

$V_{1} ρ_{w} g + V_{2} ρ_{o} g = (V_{1} + V_{2}) ρ_{c} g$

$V_{1} + \frac{V_{2} ρ_{o}}{ρ_{w}} = (V_{1} + V_{2}) \frac{ρ_{c}}{ρ_{w}}$

$\Rightarrow V_{1} + 0.8 V_{2} = 0.9 V_{1} + 0.9 V_{2}$

$\Rightarrow 0.1 V_{1} = 0.1 V_{2} \Rightarrow V_{1} : V_{2} = 1 : 1 \Rightarrow V_{1} : V_{2} = 1 : 1$

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