JEE Main 2026 — Jan 23, Shift 2 — Physics solution

JEE Main2026Jan 23, Shift 2Physics

An air bubble of volume $2.9$ cm $^{3}$ rises from the bottom of a swimming pool of $5$ m deep. At the bottom of the pool water temperature is $17^{\circ} C$ . The volume of the bubble when it reaches the surface, where the water temperature is $27^{\circ} C$ , is ______ cm $^{3}$ .

(g = 10 m/s $^{2}$ , density of water $= 1 0^{3}$ kg/m $^{3}$ , and 1 atm pressure is $1 0^{5}$ Pa)

A
4.2
B
2.0
C
3.0
D
4.5

Solution

Air inside the bubble has fixed moles, so $\frac{P _{1} V _{1}}{T _{1}} = \frac{P _{2} V _{2}}{T _{2}}$ .

At the bottom: $P_{1} = P_{a t m} + ρ g h = 1 0^{5} + 1 0^{3} \cdot 10 \cdot 5 = 1.5 \times 1 0^{5}$ Pa, $T_{1} = 290$ K, $V_{1} = 2.9$ cm $^{3}$ .

At the top: $P_{2} = 1 0^{5}$ Pa, $T_{2} = 300$ K.

$V_{2} = V_{1} \cdot \frac{P _{1}}{P _{2}} \cdot \frac{T _{2}}{T _{1}} = 2.9 \cdot \frac{1.5}{1} \cdot \frac{300}{290} = 4.5$ cm $^{3}$ .

← Back to full paper

Practice more Physics

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →