Question

The surface tension of soap solution is 3.5 × 10–2 Nm–1. The amount of work done required to increase the radius of soap bubble from 10 cm to 20 cm is________× 10–4 J. (take π = 22/7)

Answer 1

Correct Answer: 264

To find the work done when increasing the radius of a soap bubble, we use the concept of surface tension. The work done (W) to increase a bubble's surface area is given by the formula: W = 2 × T × ΔA, where T is the surface tension, and ΔA is the change in surface area. A soap bubble has two surfaces (inner and outer), hence the factor of 2.

First, calculate the initial and final surface areas. The formula for the surface area of a sphere is A = 4πr².

Initial radius (r₁) = 10 cm = 0.1 m. Final radius (r₂) = 20 cm = 0.2 m.

Initial surface area (A₁) = 4πr₁² = 4 × (22/7) × (0.1)².

Initial surface area (A₁) = 4 × (22/7) × 0.01 = 0.1257 m² (approximately).

Final surface area (A₂) = 4πr₂² = 4 × (22/7) × (0.2)².

Final surface area (A₂) = 4 × (22/7) × 0.04 = 0.5028 m² (approximately).

Change in surface area (ΔA) = A₂ - A₁ = 0.5028 - 0.1257 = 0.3771 m².

Given, T = 3.5 × 10⁻² Nm⁻¹.

Work done (W) = 2 × T × ΔA.

W = 2 × 3.5 × 10⁻² × 0.3771 = 0.026397 J.

Convert the work done to the required units: 0.026397 J = 263.97 × 10⁻⁴ J.

This value is approximately 264 × 10⁻⁴ J, which is within the given range.