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JEE Mains · Physics · STD 11 - 9.2 surface tension

An air bubble of radius 0.1 cm lies at a depth of 20 cm below the free surface of a liquid of density \(1000 \mathrm{~kg} / \mathrm{m}^3\). If the pressure inside the bubble is \(2100 \mathrm{~N} / \mathrm{m}^2\) greater than the atmospheric pressure, then the surface tension of the liquid in SI unit is (use \(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2\) )

  1. A \(0.1\)
  2. B \(0.05\)
  3. C \(0.02\)
  4. D \(0.25\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(0.05\)

Step-by-step Solution

Detailed explanation

\(\begin{aligned} & P_1=P_0+\rho g h=P_0+1000 \times 10 \times \frac{20}{100} \\ & \Rightarrow P_1=P_0+2000 \end{aligned}\) So, \(P_2-P_1=\frac{2 S}{R}=\left(\frac{2 S}{1 \times 10^{-3}}\right)\) \(\Rightarrow P_2=P_0+2100\) (given)…
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