
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.Find the density of the material of the sphere in g/cc.
(A) 3.3
(B) 6.4
(C) 7.2
(D) 12.8
Answer
507.8k+ views
Hint
According to the question there is oil contained in the vessel whose density is 0.8g/cc,the oil contains over mercury whose density is 13.6g/cc. A homogeneous sphere floats with half its volume immersed in volume and the other half is in oil.
Now, we will find the density of the material in gm/cc.
Complete step by step answer
First of all the weight of the material is in downward direction.
So, let the weight of the material be W.
$\Rightarrow W = {\rho _{Hg}} \times \dfrac{V}{2} \times g + {\rho _{oil}} \times \dfrac{V}{2} \times g$..................[where V is equal to volume of the material]......equation 1
In the equation $\dfrac{V}{2}$ represents the half volume of the material which is immersed in oil as well as mercury.
Now, we can say W of the material is equal to mg, where m is represent the mass of the material
m=density * volume of the material
$ \Rightarrow \rho \times V \times g = \dfrac{{0.8 \times V \times g}}{2} + \dfrac{{13.6 \times V \times g}}{2}$
Canceling $V \times g$from both sides we get,
$\Rightarrow \rho = \dfrac{{0.8}}{2} + \dfrac{{13.6}}{2} $
$\Rightarrow \rho = \dfrac{{14.4}}{2}$
$\Rightarrow \rho = 7.2gm/cc $
Note
The main formula of this type of question is $W = \rho Vg$, rho is the density of the material which we found in the solution. And the other density like oil and mercury will be given in the question.
According to the question there is oil contained in the vessel whose density is 0.8g/cc,the oil contains over mercury whose density is 13.6g/cc. A homogeneous sphere floats with half its volume immersed in volume and the other half is in oil.
Now, we will find the density of the material in gm/cc.
Complete step by step answer
First of all the weight of the material is in downward direction.
So, let the weight of the material be W.
$\Rightarrow W = {\rho _{Hg}} \times \dfrac{V}{2} \times g + {\rho _{oil}} \times \dfrac{V}{2} \times g$..................[where V is equal to volume of the material]......equation 1
In the equation $\dfrac{V}{2}$ represents the half volume of the material which is immersed in oil as well as mercury.
Now, we can say W of the material is equal to mg, where m is represent the mass of the material
m=density * volume of the material
$ \Rightarrow \rho \times V \times g = \dfrac{{0.8 \times V \times g}}{2} + \dfrac{{13.6 \times V \times g}}{2}$
Canceling $V \times g$from both sides we get,
$\Rightarrow \rho = \dfrac{{0.8}}{2} + \dfrac{{13.6}}{2} $
$\Rightarrow \rho = \dfrac{{14.4}}{2}$
$\Rightarrow \rho = 7.2gm/cc $
Note
The main formula of this type of question is $W = \rho Vg$, rho is the density of the material which we found in the solution. And the other density like oil and mercury will be given in the question.
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