Question

Work done in increasing the size of a soap bubble from a radius of 3 cm to 5cm is nearly (Surface tension of soap solution = 0.03 Nm–1) :-
(1) 2 π mJ
(2) 0.4 π mJ
(3) 4π mJ
(4) 0.2 π mJ

Answer 1

Answer on Question#41096 – Physics - Mechanics | Kinematics | Dynamics

Work done in increasing the size of a soap bubble from a radius of 3 cm to 5 cm is nearly (Surface tension of soap solution = 0.03 Nm⁻¹):

(1) $2\pi \mathrm{mJ}$

(2) $0.4\pi \mathrm{mJ}$

(3) $4\pi \mathrm{mJ}$

(4) $0.2\pi \mathrm{mJ}$

**Solution:**

$$T = 0.03 \frac{N}{m} - \text{surface tension of soap};$$ $$r_1 = 0.03 \, \mathrm{m} - \text{initial radius of the bubble};$$ $$r_2 = 0.05 \, \mathrm{m} - \text{final radius of the bubble};$$

Work done = (surface tension) × (increase in area):

$$W = T(S_2 - S_1)$$

Increase in area of the soap bubble:

$$S_2 - S_1 = 4\pi r_1^2 - 4\pi r_2^2 = 4\pi (r_1^2 - r_2^2)$$

(2) in (1):

$$W = T \cdot 4\pi (r_1^2 - r_2^2) = 0.03 \frac{N}{m} \cdot 4 \cdot \pi \cdot ((0.05 \, \mathrm{m})^2 - (0.03 \, \mathrm{m})^2) = 0.2\pi \cdot 10^{-3} \, \mathrm{J}$$ $$= 0.2 \, \mathrm{mJ}$$

Answer: (4) $0.2\pi \mathrm{mJ}$.