A soap bubble of surface tension 0.04 N/m is blown to a diameter of 7 cm. If (15000 − x) μJ of work is done in blowing it further to make its diameter 14 cm, then the value of x is
Q. A soap bubble of surface tension 0.04 N/m is blown to a diameter of 7 cm. If (15000 − x) μJ of work is done in blowing it further to make its diameter 14 cm, then the value of x is ____.

(π = 22/7)

Correct Answer: 11304

Explanation

For a soap bubble, there are two surfaces. Hence, work done in increasing surface area is given by

W = 2T ΔA

Initial radius:

r₁ = 7/2 cm = 3.5 cm = 0.035 m

Final radius:

r₂ = 14/2 cm = 7 cm = 0.07 m

Change in surface area:

ΔA = 4π(r₂² − r₁²)

ΔA = 4 × (22/7) × (0.07² − 0.035²)

ΔA = 4 × (22/7) × (0.0049 − 0.001225)

ΔA = 4 × (22/7) × 0.003675

ΔA = 0.0462 m²

Now, work done:

W = 2 × 0.04 × 0.0462

W = 0.003696 J

Convert into microjoules:

W = 3696 μJ

Given,

15000 − x = 3696

x = 15000 − 3696

x = 11304

Hence, the value of x is 11304.

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