Question

A boy formed a bubble and a liquid drop from the same soapy water. The pressure difference between inside and outside of the soap bubble is measured to be 100 N/m\(^2\). If the radius of the droplet is half of the radius of the bubble, then the pressure difference between the inside and outside of the droplet is:

• A. 0
• B. 50 N/m\(^2\)
• C. 100 N/m\(^2\)
• D. 200 N/m\(^2\)
• E. 400 N/m\(^2\)

Hint:

Bubble has two surfaces → pressure is double compared to a liquid drop.

Answer 1

The Correct Option is D

Concept:
\[ \Delta P_{\text{bubble}} = \frac{4T}{R}, \quad \Delta P_{\text{drop}} = \frac{2T}{r} \]

Step 1: Use bubble data
\[ 100 = \frac{4T}{R} \Rightarrow T = 25R \]

Step 2: Use radius relation
\[ r = \frac{R}{2} \]

Step 3: Substitute into drop formula
\[ \Delta P = \frac{2T}{R/2} \] \[ = \frac{2(25R)}{R/2} \] \[ = \frac{50R}{R/2} \] \[ = 100 \times 2 = 200 \]

Step 4: Final answer
\[ \boxed{200 \, N/m^2} \]