A vessel contains oil (density =$ 0.8 \;gm/cm^3$) over mercury (density = $13.6\; gm/cm^3$). A homogeneous sphere floats with half of its volume immersed in mercury and the other half in oil. The density of the material of the sphere in $ gm/cm^3$ is
A$3.3$
B$6.4$
C$ 7.2$
D$12.8$
IIT 1998, Diffcult
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C$ 7.2$
c (c)As the sphere floats in the liquid. Therefore its weight will be equal to the upthrust force on it
Weight of sphere
$ = \frac{4}{3}\pi {R^3}\rho g$ …(i) ...... (i)
Upthrust due to oil and mercury
$ = \frac{2}{3}\pi {R^3} \times {\sigma _{oil}}g + \frac{2}{3}\pi {R^3}{\sigma _{Hg}}g$ …(ii)
Equating (i) and (ii)
$\frac{4}{3}\pi {R^3}\rho g = \frac{2}{3}\pi {R^3}0.8g + \frac{2}{3}\pi {R^3} \times 13.6g$$ \Rightarrow 2\rho = 0.8 + 13.6 = 14.4 \Rightarrow \rho = 7.2$
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