An air bubble of volume 1 cm³ rises from the bottom of a lake 40 m deep to the surface at a temperature of 12. The atmospheric pressure is 1 × 10⁵ Pa the density of water is 1000 kg/m³ and g = 10 m/s². 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) 4 cm³ (b) 3 cm³ (c) 2 cm³ (d) 5 cm³

Answer: (d). Pressure at the bottom: P₁ = 10⁵ + 1000×10×40 = 5×10⁵ Pa. Same temperature, so Boyle's law: P₁V₁ = P₀V₂. V₂ = (5×10⁵×1)/(10⁵) = 5 cm³ So the answer is (d).

An air bubble of volume 1 cm³ rises from the bottom of a lake 40 m… | JEE Main 2023 (8 Apr, Shift 1)

Q13JEE Main 2023 · 8 APR · SHIFT 1Medium

Pressure in a Fluid

An air bubble of volume $1\text{ cm}^{3}$ rises from the bottom of a lake $40\text{ m}$ deep to the surface at a temperature of $12^\circ\text{C}$ . The atmospheric pressure is $1 \times 10^{5}\text{ Pa}$ 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 $40\text{ m}$ and on the surface. The volume of air bubble when it reaches the surface will be:

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Answer: (d)

Pressure at the bottom: $P_1 = 10^5 + 1000\times10\times40 = 5\times10^5\text{ Pa}$ .
Same temperature, so Boyle's law: $P_1V_1 = P_0V_2$ .
$V_2 = \dfrac{5\times10^5\times1}{10^5} = 5\text{ cm}^3$
So the answer is (d).

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