JEE Mains · Physics · STD 11 - 9.2 surface tension
There is an air bubble of radius \(1.0\,mm\) in a liquid of surface tension \(0.075\,Nm ^{-1}\) and density \(1000\,kg\) \(m ^{-3}\) at a depth of \(10\,cm\) below the free surface. The amount by which the pressure inside the bubble is greater than the atmospheric pressure is \(....Pa \left( g =10\,ms ^{-2}\right)\)
- A \(1150\)
- B \(1151\)
- C \(1152\)
- D \(1153\)
Answer & Solution
Correct Answer
(A) \(1150\)
Step-by-step Solution
Detailed explanation
Pressure inside the bubble \(P=P_0+h \rho g+\frac{2 T}{r}\) \(P-P_0=h \rho g+\frac{2 T}{r}\) \(=0.1 \times 1000 \times 10+\frac{2 \times .075}{10^{-3}}\) \(=1000+(0.15)(1000)\) \(=1150\,Pa\)
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