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  • An air bubble of volume  rises from the bottom of a lake <img alt="40 m" src="https://entrancec

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°C. The atmospheric pressure is 1 = 105 Pa , the density of water is 1000 kg/m^{3} 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:

Option: 1

3 cm ^{3}


Option: 2

4cm^{3}


Option: 3

2cm^{3}


Option: 4

5cm ^{3}


Answers (1)

best_answer

Pressure at surface = P _{atm}= 1\times 10^{5 }Pa
V _{surface}= ?
Pressure at h = 40 m depth 
P= P_{atm}+pgh
\begin{aligned} & \mathrm{P}=\mathrm{P}_{\text {atm }}+\rho g h \\ & \mathrm{P}=10^5+10^3 \times 10 \times 40 \\ & \mathrm{P}=5 \times 10^5 \mathrm{~Pa} \\ & \mathrm{v}=1 \mathrm{~cm}^3 \end{aligned}
Temp. is constant
\begin{aligned} & \mathrm{P}_1 \mathrm{~V}_1=\mathrm{P}_2 \mathrm{~V}_2 \\ & 10^5 \times \mathrm{v}=5 \times 10^5 \times 1 \\ & \mathrm{v}=5 \mathrm{~cm}^3 \end{aligned}

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Nehul

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