JEE Physics 2023 Question with Solution

Given: radius of bubble $r = 1.0\times10^{-3}\,m$, surface tension $T = 0.075\,N m^{-1}$, density $\rho = 1000\,kg m^{-3}$, depth $h = 0.10\,m$, and $g = 10\,m s^{-2}$.

Find: the excess pressure inside the bubble over atmospheric pressure.

The excess pressure has two contributions:

1. hydrostatic pressure due to depth
2. excess pressure due to surface tension

Using

$$P_{hyd} = \rho g h$$

we get

$$P_{hyd} = 1000 \times 10 \times 0.10 = 1000\,Pa$$

For an air bubble in a liquid,

$$P_{st} = \frac{2T}{r}$$

So,

$$P_{st} = \frac{2 \times 0.075}{1.0 \times 10^{-3}} = 150\,Pa$$

Hence, total excess pressure is

$$P_{excess} = P_{hyd} + P_{st} = 1000 + 150 = 1150\,Pa$$

Therefore, the pressure inside the bubble exceeds atmospheric pressure by $1150\,Pa$.

Given: the bubble is at depth $h$ inside a liquid and has radius $r$.

Find: pressure inside the bubble above atmospheric pressure.

For an air bubble in a liquid, add hydrostatic pressure and surface tension pressure directly:

$$P_{excess} = \rho g h + \frac{2T}{r}$$

Substituting the values,

$$P_{excess} = 1000\times10\times0.10 + \frac{2\times0.075}{1.0\times10^{-3}} = 1000 + 150 = 1150\,Pa$$

Therefore, the required answer is $1150\,Pa$.