JEE Mains · Physics · STD 11 - 9.1 fluid mechanics
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 \(12^{\circ}\,C\). The atmospheric pressure is \(1 \times 10^5 Pa\), 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 \(..........\,cm^{3}\)
- A \(5\mathrm{~cm}^3\)
- B \(2\mathrm{~cm}^3\)
- C \(4\mathrm{~cm}^3\)
- D \(3\mathrm{~cm}^3\)
Answer & Solution
Correct Answer
(A) \(5\mathrm{~cm}^3\)
Step-by-step Solution
Detailed explanation
\(P = P _0+\rho gh =10^5\,Pa +10^3 \times 10 \times 40=5 \times 10^5\,Pa\) At \(T\) is constant \(PV = P _0 V _0\) \(\Rightarrow 5 \times 10^5\,Pa \times 1\,cm ^3=10^5 Pa \times V _0 \Rightarrow V _0=5\,cm ^3\)
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