Fluid Mechanics

JEE Main level, single correct, medium

37 of 192

Question 37 of 192

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=0.8, specific gravity of ice =0.9=0.9)

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Answer: (A) 1:11: 1

v1= volume immersed in water. v2= volume immersed in oil. v1ρwg+v2ρog=(v1+v2)ρcgv1+v2ρoρw=(v1+v2)ρcρw=v1+0.8v2=0.9v1+0.9v2=0.1v1=0.1v2v1:v2=1:1\begin{aligned} & \mathrm{v}_1=\text { volume immersed in water. } \\ & \mathrm{v}_2=\text { volume immersed in oil. } \\ & \mathrm{v}_1 \rho_{\mathrm{w}} \mathrm{g}+\mathrm{v}_2 \rho_{\mathrm{o}} \mathrm{g}=\left(\mathrm{v}_1+\mathrm{v}_2\right) \rho_{\mathrm{c}} \mathrm{g} \\ & \mathrm{v}_1+\frac{\mathrm{v}_2 \rho_{\mathrm{o}}}{\rho_{\mathrm{w}}}=\left(\mathrm{v}_1+\mathrm{v}_2\right) \frac{\rho_{\mathrm{c}}}{\rho_{\mathrm{w}}} \\ & =\mathrm{v}_1+0.8 \mathrm{v}_2=0.9 \mathrm{v}_1+0.9 \mathrm{v}_2 \\ & =0.1 \mathrm{v}_1=0.1 \mathrm{v}_2 \\ & \mathrm{v}_1: \mathrm{v}_2=1: 1\end{aligned}

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