Question

The excess of pressure inside a soap bubble is twice the excess pressure inside a soap bubble. The volume of the first bubble is n times the volume of the second where n is:
A. 0.125
B. 0.250
C. 1
D. 2

Answer 1

Hint: The excess pressure inside a soap bubble is twice the soap bubble, if the volume of first bubble is given n times and we have to find out the volume of the second bubble for that we are assuming the radius ${r_1}$and ${r_2}$, for this we are taking the Laplace’s equation.

Complete step-by-step solution:
From laplace’s equation
$\Delta {P_1} = 2\Delta {P_2}$
Here $\Delta {P_1}$is the excess pressure inside a soap bubble and when it doubles the exces pressure inside a soap bubble then it is $2\Delta {P_2}$
Here $
  \Delta {P_1} = \dfrac{{4T}}{{{r_1}}} \\
  \Delta {P_2} = \dfrac{{4T}}{{{r_2}}} \\
 $
Then by substituting the above values we get,
$
  \dfrac{{4T}}{{{r_1}}} = 2\dfrac{{4T}}{{{r_2}}} \\
  {r_2} = 2{r_1} \\
 $
Here we get the radius,
The volume of the first bubble is n times the second bubble then,
Volume of first bubble is, ${V_1} = \dfrac{4}{3}\pi {r_1}^3$
Volume of second bubble is, ${V_2} = \dfrac{4}{3}\pi {r_2}^3$
Already we have found the radius, ${r_2} = 2{r_1}$
By substituting the radius in volume of second then we get,
$
  {V_2} = \dfrac{4}{3}\pi {\left( {2{r_1}} \right)^3} \\
  {V_2} = {2^3} \times \left( {\dfrac{4}{3}\pi {r_1}^3} \right) \\
 $
Therefore ${V_2} = 8{V_1}$
From the question the first bubble is n times to second that means, ${V_1} = n{V_2}$
From the above we get, ${V_1} = \dfrac{1}{8}{V_2}$
Thus the volume of first bubble inside is n times to second bubble is, 0.125 times
Thus the correct option is 1.

Note: From the question we have calculated the volume of the second bubble is 0.125 times the volume of the first bubble. For this type of questions we can use the formulas of Laplace’s equation and volume formula. Here we are using Laplace’s and volume equation for calculation.