An air bubble of volume $1 \mathrm{~cm}^{3}$ rises from the bottom of a lake 40 m deep to the surface at a temperature of $12^{\circ} \mathrm{C}$. The atmospheric pressure is $1 \times 10^{5} \mathrm{~Pa}$, the density of water is $1000 \mathrm{~kg} / \mathrm{m}^{3}$ and $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}$. 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

JEE Main · 2023 · 08 Apr 2023 · Shift 1 · Heat & Thermodynamics

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

1 2^{\circ} C

. The atmospheric pressure is

1 \times 1 0^{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

(A) $5 cm^{3}$ ✓ Correct
(B) $2 cm^{3}$
(C) $4 cm^{3}$
(D) $3 cm^{3}$

Answer: (A)

Solution

At depth 40 m, pressure

P = P_{0} + ρ g h = 1 0^{5} + 1 0^{3} \times 10 \times 40 = 5 \times 1 0^{5}

Pa. At the surface, pressure is

P_{0} = 1 0^{5}

Pa. Since temperature is constant,

P V = P_{0} V_{0}

. Therefore

5 \times 1 0^{5} \times 1 = 1 0^{5} \times V_{0}

, giving

V_{0} = 5

cm³.

JEE Main · 2023 · 08 Apr 2023 · Shift 1 · Single correct option

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JEE Main · 2023 · 08 Apr 2023 · Shift 1 · Heat & Thermodynamics

Solution

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