Q 11-09-096JEE MainJEE Main 2023 (8 Apr, Shift 1)Easy
An air bubble of volume $1\ \text{cm}^3$ rises from the bottom of a lake $40$ m deep to the surface at a temperature of $12^\circ\text{C}$. The atmospheric pressure is $1\times10^5$ Pa, the density of water is $1000\ \text{kg m}^{-3}$ and $g=10\ \text{m s}^{-2}$. 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
Answer: (D) $5\ \text{cm}^3$
Pressure at the bottom $=10^5+1000\times10\times40=5\times10^5$ Pa. At constant temperature $P_1V_1=P_2V_2$: $V_2=5\ \text{cm}^3$.
Solution by Sreeraj P, M.Sc Physics