An air bubble of volume 1 ~cm^3 rises from the bottom of a lake 40 ~m deep to the surface at a temperature of 12^\circ C . The atmospheric pressure is 1 \times 10^5 ~Pa the density of water is 1000 ~kg / m^3 and g=10 ~m / s^2 . There is no difference of the temperature of water at the depth of 40 ~m and on the surface. The volume of air bubble when it reaches the surface will be:

Properties of Matter PYQ — JEE Main 2023 · 8 Apr · Shift 1 · Q48 with Solution

View written solutionFree

Correct answer: D

Since the temperature remains constant, the air bubble undergoes an isothermal process:

$P_1 V_1 = P_2 V_2$

At depth $h=40\,\text{m}$ , pressure is:

$P_1 = P_{\text{atm}} + \rho g h$

Given:

$P_{\text{atm}} = 1\times 10^5\,\text{Pa}, \quad \rho = 1000\,\text{kg/m}^3, \quad g=10\,\text{m/s}^2, \quad h=40\,\text{m}$

So,

$\rho g h = 1000\times 10\times 40 = 4\times 10^5\,\text{Pa}$

Therefore,

$P_1 = 1\times 10^5 + 4\times 10^5 = 5\times 10^5\,\text{Pa}$

At the surface, pressure is atmospheric:

$P_2 = 1\times 10^5\,\text{Pa}$

$V_1 = 1\,\text{cm}^3$

Using Boyle’s law:

$P_1 V_1 = P_2 V_2$

$V_2 = \frac{P_1 V_1}{P_2} = \frac{5\times 10^5 \times 1}{1\times 10^5} = 5\,\text{cm}^3$

$V_2 = 5\,\text{cm}^3$

So the correct option is D.