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JEE Advanced · Physics · 13. Thermodynamics

Two soap bubbles \(A\) and \(B\) are kept in a closed chamber where the air is maintained at pressure \(8 \mathrm{Nm}^{-2}\). The radii of bubbles \(A\) and \(B\) are \(2 \mathrm{~cm}\), respectively. Surface tension of the soap-water used to make bubbles is \(0.04\) \(\mathrm{Nm}^{-1}\). Find the ratio \(\frac{n_B}{n_A}\), where \(n_A\) and \(n_B\) are the number of moles of air in bubbles \(A\) and \(B\), respectively. [Neglect the effect of gravity]

  1. A 2
  2. B 4
  3. C 6
  4. D 8
Verified Solution

Answer & Solution

Correct Answer

(C) 6

Step-by-step Solution

Detailed explanation

Although not given in the question but we will have to assume that temperatures of \(A\) and \(B\) are same.

\(
\begin{aligned}
& \frac{n_B}{n_A}=\frac{p_B V_B / R T}{p_A V_A / R T}=\frac{p_B V_B}{p_A V_A} \\
&= \frac{p+4 s / r_A \times 4 / 3 \pi\left(r_A\right)^3}{\left(p+4 s / r_B\right) \times 4 / 3 \pi\left(r_B\right)^3} \\
&(s=\text { surface tension) }
\end{aligned}
\)
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