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9-1.Fluid Mechanics
hard
A cube of ice floats partly in water and partly in kerosene oil. The radio of volume of ice immersed in water to that in kerosene oil (specific gravity of Kerosene oil $=0.8$, specific gravity of ice $=0.9$ )
A
$8: 9$
B
$5: 4$
C
$9: 10$
D
$1: 1$
(JEE MAIN-2024)
Solution
$\mathrm{v}_1=\text { volume immersed in water. }$
$\mathrm{v}_2=\text { volume immersed in oil. }$
$\mathrm{v}_1 \rho_{\mathrm{w}} \mathrm{g}+\mathrm{v}_2 \rho_{\mathrm{o}} \mathrm{g}=\left(\mathrm{v}_1+\mathrm{v}_2\right) \rho_{\mathrm{c}} \mathrm{g}$
$\mathrm{v}_1+\frac{\mathrm{v}_2 \rho_0}{\rho_w}=\left(\mathrm{v}_1+\mathrm{v}_2\right) \frac{\rho_{\mathrm{c}}}{\rho_w}$
$=\mathrm{v}_1+0.8 \mathrm{v}_2=0.9 \mathrm{v}_1+0.9 \mathrm{v}_2$
$=0.1 \mathrm{v}_1=0.1 \mathrm{v}_2$
$\mathrm{v}_1: \mathrm{v}_2=1: 1$
Standard 11
Physics
