Question

An inverted u−tube has its two limbs in water and kerosene contained in two beakers. If water rises to a height of 10cm, to what height does kerosene (density =0.8gm/cc) rise in the other limb?
A. $10cm$
B. $12.5cm$
C. $15cm$
D. $20cm$

Answer 1

Hint:The water and kerosene are two different liquids. The inverted u tube used as an apparatus in the experiment is also known as inverted U-tube manometer. It is used for the measurement of difference in pressure between two liquids. When two different limbs of a U-tube manometer are dipped in water and kerosene respectively, the water and kerosene will rise in the limbs of the tube to a certain height. The space in the U-tube manometer above the rising level of these liquids will be filled with air.

Formula Used:
We can derive the formula for calculating the height by which kerosene rises in one of the limbs of an inverted U-shaped tube. The formula that can be used is,
${h_2} = \dfrac{{{h_1}{\rho _w}}}{{{\rho _k}}}$
Here, ${h_1}$ is the length of water column in one limb of inverted U tube and ${h_2}$ is the length of column by which kerosene level rises in another limb of inverted U tube. The density of water and kerosene can be denoted by ${\rho _w}$ and ${\rho _k}$ respectively.

Complete step by step answer:
Let us first see the schematic diagram of the experimental setup.

Point A and point B are at the same height from the ground. Therefore, according to Pascal’s law of pressure, the pressure will be the same at both points A and B.In between the water column and kerosene column, there is air. So pressure on the upper surface of the water column and kerosene column will be equal to atmospheric pressure. Let us consider this pressure as ${P_0}$.

Step 1: We have to equate pressure at point A and point to B to get the expressions of height of the kerosene column with respect to height of water column. According to the hydrostatic pressure formula,
Pressure due to h height of fluid (P) = $h \times \rho \times g$
Here, h is the height of liquid in the column, g is the acceleration due to gravity and is equal to ${9.81\,m/{s^2}}$. The density of the liquid is denoted by $\rho $.As we know, the density of water is 1gm/cc.

Let us write the hydrostatic pressure expressions for both ${h_1}$ and ${h_2}$ columns.
Pressure due to ${h_1}$ length of water column = ${h_1}{\rho _w}g$
Pressure due to${h_2}$ length of water column = ${h_2}{\rho _k}g$
The information provided in the question is given below.
${h_1} = 10cm$
${\rho _k} = 0.8gm/cc$
${\rho _w} = 1gm/cc$
Pressure at point A = pressure at point B
On substituting the respective values we get,
${P_0} + {h_1}{\rho _w}g = {P_0} + {h_2}{\rho _k}g$

Step 2: We can simplify the equation (1) further to get,
\[{P_0} + {h_1}{\rho _w}g = {P_0} + {h_2}{\rho _k}g\]
When the two limbs of inverted U-tube are dipped in two liquids of varied densities, the liquid with high density will rise to lesser extent in one of the limbs of inverted U-tube and lighter liquid will rise to the greater extent. Since, the pressure exerted by liquid is the same at all points, the liquid risen in the limbs of the tube to certain heights will exert equal pressure.

Therefore,
${h_1}{\rho _w}g = {h_2}{\rho _k}g$
$\Rightarrow {h_1}{\rho _w} = {h_2}{\rho _k}$………..(2)
Now we need to calculate ${h_2}$. For that we need to rearrange equation (2) and we get,
$ {h_2} = \dfrac{{{h_1}{\rho _w}}}{{{\rho _k}}}\\
\Rightarrow {h_2} = \dfrac{{10 \times 1}}{{0.8}} \\
\therefore {h_2} = 12.5cm$

The kerosene will rise to the height 12.5cm in the other limb of inverted U-tube.So, the correct answer is option B.

Note: In this case, the middle portion in the U-shaped tube between water column and kerosene column, is filled by air. Instead of air, if there is another fluid, then we must consider pressure exerted by that fluid on the water column and kerosene column.Similarly, if we kept the U-shaped tube open then we need to consider the added atmospheric pressure too.