Surface tension of a soap bubble is . Work done to increase the radius of soap bubble from 3.5 cm to 7 cm will be:
Take
Take
- A.J
- B.J
- C.J
- D.J
Show answer and solution
Correct option: D
Solution
A soap bubble has two surfaces, so the total surface area is 2 × 4πr².
Work done = T × ΔA = T × 2 × 4π(r₂² - r₁²)
= 2.0×10⁻² × 8π × [(0.07)² - (0.035)²]
= 2.0×10⁻² × 8 × (22/7) × [0.0049 - 0.001225]
= 2.0×10⁻² × 8 × (22/7) × 0.003675
= 2.0×10⁻² × (8 × 22 × 0.003675)/7
= 2.0×10⁻² × (0.6468)/7
= 2.0×10⁻² × 0.09240
= 18.48×10⁻⁴ J
The key concept is that a soap bubble has two surfaces (inner and outer), so the effective area change is doubled, giving W = 8πT(r₂² - r₁²) = 18.48×10⁻⁴ J.
Work done = T × ΔA = T × 2 × 4π(r₂² - r₁²)
= 2.0×10⁻² × 8π × [(0.07)² - (0.035)²]
= 2.0×10⁻² × 8 × (22/7) × [0.0049 - 0.001225]
= 2.0×10⁻² × 8 × (22/7) × 0.003675
= 2.0×10⁻² × (8 × 22 × 0.003675)/7
= 2.0×10⁻² × (0.6468)/7
= 2.0×10⁻² × 0.09240
= 18.48×10⁻⁴ J
The key concept is that a soap bubble has two surfaces (inner and outer), so the effective area change is doubled, giving W = 8πT(r₂² - r₁²) = 18.48×10⁻⁴ J.
[JEE Main 2023 · 29 Jan · Shift 1]