Question

A U-Tube is partially filled with water. Oil that does not mix with water is poured in next on one side, until water rises by 25cm on the other side. If the density of oil is 0.8, the oil will stand higher than the water level by
A. 6.25cm
B. 12.50cm
C. 31.75cm
D. 62.50cm

Answer 1

Hint: Assume the initial length of water in the tube to be x, and then assume the oil board in the tube to be y cm. Then using these, calculate the total mass of liquid on both the sides of the tube. Since it is balanced on both sides, their masses will be equal.

Complete step by step answer:
There is given a U-Tube which is partially filled with water and then oil is poured from one side. Due to this water level on one side rises by 25 cm, then we must find out the difference between the levels of oil and water on both sides of the tube.
We know that oil does not mix with water, this is because oil and water are immiscible fluids. We know that the density of pure water is 1 gm/cc, therefore the density of oil is lesser than the density of water. Due to this fact, oil will float above the water surface.
Before the oil was poured, there was some water inside the tube. Let the level of water initially be x cm. Then it will be the same on both sides of the tube.
Now when the oil is poured from one side of the tube, the water level on the other side rises by 25 cm. Therefore, the total height of water on one side will be $\left( {x + 25} \right)cm$. Let the length of oil poured be y cm.
If both the sides of the tube are balanced, then the mass of liquids on both the sides will be equal. Assuming the area of cross section of the tube to be a, then
on one side
$ {l_{total}} = {l_{water}} = \left( {x + 25} \right)cm \\
{V_{water}} = a{l_{total}} = a\left( {x + 25} \right)c{m^3} \\
{m_{total}} = {m_{water}} = {\rho _{water}}{V_{water}} = a\left( {x + 25} \right)gm \\ $
on the other side.
$ {l_{water}} = \left( {x - 25} \right)cm,{l_{oil}} = ycm \\
{V_{water}} = a{l_{water}} = a\left( {x - 25} \right)c{m^3},{V_{oil}} = a{l_{oil}} = ayc{m^3} \\
{m_{water}} = {\rho _{water}}{V_{water}} = a\left( {x - 25} \right)gm,{m_{oil}} = {\rho _{oil}}{V_{oil}} = \left( {0.8ay} \right)gm \\
{m_{total}} = {m_{water}} + {m_{oil}} = a\left( {x - 25 + 0.8y} \right)gm \\ $
Therefore
$ a\left( {x + 25} \right) = a\left( {x - 25 + 0.8y} \right) \\
25 = - 25 + 0.8y \\
0.8y = 50 \\
y = \dfrac{{50}}{{0.8}} = 62.5cm \\ $
$ {l_{left}} = \left( {x + 25} \right)cm,{l_{right}} = \left( {x - 25 + y} \right)cm \\
\therefore \Delta l = {l_{right}} - {l_{left}} = \left( {x - 25 + y} \right) - \left( {x + 25} \right) = \left( {y - 50}
\right)cm = 62.5 - 50 = 12.50cm \\ $

So, the correct answer is “Option B”.

Note:
The density of water here is assumed to be one, but it might differ depending upon the quality of the water i.e. density of pure water is 1 gm/cc. While solving it remember the concept of the difference between the levels of oil and water on both sides of the tube.