Question

The excess pressure inside a soap bubble A in air is half the excess pressure inside another soap bubble B in air. If the volume of the bubble A is \( n \) times the volume of the bubble B, then, the value of \( n \) is \underline{\hspace{1cm}} .

Hint:

The excess pressure inside a soap bubble is inversely proportional to its radius. The volume of a soap bubble is proportional to the cube of its radius. Use the relationship between the excess pressures to find the relationship between the radii, and then use the relationship between the radii to find the relationship between the volumes.

Answer 1

Correct Answer: 8

The excess pressure \( \Delta P \) inside a soap bubble of radius \( R \) is given by: \[ \Delta P = \frac{4T}{R} \] where \( T \) is the surface tension of the soap solution. Let the excess pressure inside soap bubble A be \( \Delta P_A \) and its radius be \( R_A \). Let the excess pressure inside soap bubble B be \( \Delta P_B \) and its radius be \( R_B \). According to the problem, the excess pressure inside bubble A is half the excess pressure inside bubble B: \[ \Delta P_A = \frac{1}{2} \Delta P_B \] Using the formula for excess pressure: \[ \frac{4T}{R_A} = \frac{1}{2} \left( \frac{4T}{R_B} \right) \] \[ \frac{1}{R_A} = \frac{1}{2 R_B} \] \[ R_A = 2 R_B \] The volume of a spherical bubble is given by \( V = \frac{4}{3} \pi R^3 \). Let the volume of bubble A be \( V_A \) and the volume of bubble B be \( V_B \). \[ V_A = \frac{4}{3} \pi R_A^3 \] \[ V_B = \frac{4}{3} \pi R_B^3 \] We are given that the volume of bubble A is \( n \) times the volume of bubble B: \[ V_A = n V_B \] \[ \frac{\frac{4}{3} \pi R_A^3}{\frac{4}{3} \pi R_B^3} = n \] \[ \left( \frac{R_A}{R_B} \right)^3 = n \] We found that \( R_A = 2 R_B \), so \( \frac{R_A}{R_B} = 2 \). Substituting this into the equation for \( n \): \[ n = (2)^3 = 8 \] The value of \( n \) is 8.