Question

The lower end of a capillary tube of diameter $2mm$ is dipped in water $8cm$ below the surface in a beaker so that an air bubble is formed at the end of the tube inside water. The excess pressure inside the bubble will be (in Pascal) (\[T - 73{\text{ }}dyne/cm\])
(a) $146$
(b) $1.078 \times {10^5}$
(c) $1.0193 \times {10^5}$
(d) $1.02 \times {10^4}$

Answer 1

Hint:Surface tension is defined as the quantity of force that is exerted on the water surfaces to contract or diminish in its least area. Surface tension is calculated by taking the ratio of force to the unit length or the ratio of energy to the area.

Complete step by step solution:
Now we have to take some constant which is useful to solve the problem

Let The surface tension of the water at temperature in the experiments is given as
 \[7.30{\text{ }} \times {10^{ - 2}}N{m^{ - 1}}\]

We know that $1atm = 1.01 \times {10^5}Pa$

We know that the density of water can be given by $\rho = 1000Kg/{m^3}$ and the acceleration due to gravity $g = 9.8m/s{}^2$

First we have to calculate the outside pressure (${p_0}$) in the bubble in order to calculate the excess pressure

So to calculate the outside pressure we have the formulae

${p_0} = {p_a} + hdg$

In the above equation ${p_a}$represents the atmospheric pressure and $hdg$be the height, density, acceleration due to gravity

Now substitute the values in the above equation

${p_0} = 1.01 \times {10^5}Pa + (0.08 \times 1000 \times 9.8)$

After calculating the values in the above equation we get

${p_0} = 1.01784 \times {10^5}Pa$

Now we have to calculate the inside pressure
So to calculate the pressure inside the bubble we have formulae

${p_i} = {p_0} + \dfrac{{2s}}{r}$

Here ${p_0}$ represent the outside pressure and $s$be the surface tension
Now substitute the values in the above equation

${p_i} = 1.01784 \times {10^5} + \dfrac{{2 \times 7.3 \times {{10}^{ - 2}}}}{{{{10}^{ - 3}}}}$

Now after calculating the values in the above equation

${p_i} = (1.01784 + 0.00146) \times {10^5}Pa$

Now, ${p_i} = 1.02 \times {10^5}Pa$

Where the radius of bubble is equal to the radius of the capillary tube hence the excess pressure in the bubble is $146Pa$

Note:The pressure is defined as the force applied perpendicular to the surface of the object per unit area over which the force is distributed.