Question

A vessel contains oil (density 0.8g/cc) over mercury (density 13.6g/cc). A homogeneous sphere floats with half its volume immersed in mercury and the other half in oil. The density of the material of the sphere in g/cc is:
A. 3.3
B. 6.4
C. 7.2
D. 12.8

Answer 1

It is given that the sphere floats with half its volume immersed in mercury and the other half in oil. Hence, the weight of the sphere will be equal to the upthrust force on it. So, to solve this problem, find the weight of the sphere. Then, find the upthrust force due to oil and mercury and add them to get the total upthrust force on the sphere. Now, equate the equations for weight of the sphere and upthrust force. Substitute the values in this equation and find the density of the material of the sphere.

Complete solution:
Given:
Density of oil, ${\rho}_{oil}= 0.8 {g}/{cc}$
Density of mercury, ${\rho}_{mercury}= 13.6 {g}/{cc}$
Volume of sphere in oil=Volume of sphere in mercury= $\dfrac {V}{2}$
It is given that the sphere floats in the liquid. Therefore, the weight of the sphere will be equal to the upthrust force on it.
$\Rightarrow W=F$
Weight of the sphere is given by,
$W= V\rho g$ …(1)
Where,
V is the volume of the sphere
$\rho$ is the density
g is the gravity
Upthrust force due to oil will be,
${F}_{1}= \dfrac {V}{2}{\rho}_{oil}g$
Similarly, upthrust force due to mercury will be,
${F}_{2}= \dfrac {V}{2}{\rho}_{mercury}g$
Total upthrust force will be,
$F={F}_{1}+{F}_{2}$
Substituting values in above expression we get,
$F= \dfrac {V}{2}{\rho}_{oil}g +\dfrac {V}{2}{\rho}_{mercury}g$ …(2)
Equating equation. (1) and (2) we get,
$V\rho g= \dfrac {V}{2}{\rho}_{oil}g +\dfrac {V}{2}{\rho}_{mercury}g$
$\Rightarrow Vg\rho =Vg\left( \dfrac { { \rho }_{ oil } }{ 2 } +\dfrac { { \rho }_{ mercury } }{ 2 } \right)$
Cancelling common term on both the sides we get,
$\rho = \dfrac { { \rho }_{ oil } }{ 2 } +\dfrac { { \rho }_{ mercury } }{ 2 }$
$\Rightarrow \rho =\dfrac { { \rho }_{ oil }+{ \rho }_{ mercury } }{ 2 }$
Substituting values in above expression we get,
$\rho =\dfrac { 0.8+13.6 }{ 2 }$
$\Rightarrow \rho= \dfrac {14.4}{2}$
$\Rightarrow \rho= 7.2 {g}/{cc}$
Thus, the density of the material of the sphere in g/cc is $7.2 {g}/{cc}$.

So, the correct answer is option C i.e. 7.2.

Note:
In the given question, the total mass of the sphere, mass of the sphere in oil and mass of the sphere in mercury are not given. Hence, we did not use the direct formula for density. If the masses were given then we would have substituted the value and mass of the sphere in the formula for density. Then, evaluated it and calculated the density of the sphere in the material.