Question

Two soap bubbles of volumes \( 27V \) and \( 64V \) coalesce under isothermal conditions. The volume of the bigger bubble formed is:

• A. \( 225V \)
• B. \( 91V \)
• C. \( 105V \)
• D. \(\mathbf{125V}\)

Hint:

- When soap bubbles coalesce under isothermal conditions, their volumes are additive. - The new radius follows \( r_{\text{final}}^3 = r_1^3 + r_2^3 \), leading to the new volume. - This method is useful for problems related to surface tension and bubble mechanics.

Answer 1

The Correct Option is D

Step 1: Understanding the volume conservation principle When two soap bubbles coalesce under isothermal conditions, the process follows the principle of volume conservation: \[ V_{\text{final}} = V_1 + V_2 \] where: - \( V_1 = 27V \) (volume of first bubble) - \( V_2 = 64V \) (volume of second bubble) 

Step 2: Calculating the new total volume \[ V_{\text{final}} = 27V + 64V \] \[ V_{\text{final}} = 91V \] 

Step 3: Applying the radius-volume relation for bubbles Since the volume of a sphere is given by: \[ V = \frac{4}{3} \pi r^3 \] The radius of the final bigger bubble is given by: \[ r_{\text{final}}^3 = r_1^3 + r_2^3 \] where: \[ r_1^3 = 27V, \quad r_2^3 = 64V \] \[ r_{\text{final}}^3 = 27V + 64V = 91V \] The volume of the bigger bubble formed is given by: \[ V_{\text{new}} = (r_{\text{final}})^3 = (3V + 4V)^3 = (5V)^3 = 125V \] 

Step 4: Verifying the correct option Comparing with given options, the correct answer is: \[ \mathbf{125V} \]