JEE Main · 2026 · 02 Apr 2026 · Shift 2 · Fluid Mechanics

The surface tension of a soap bubble is

0.03 N / m

. The work done in increasing the diameter of bubble from 2 cm to 6 cm is

α π \times 1 0^{- 4} J

. The value of

α

is ____. (Take

π = 3.14

)

(A)0.86
(B)0.64
(C)1.92✓ Correct
(D)7.68

Answer: (C)

Solution

A soap bubble has two surfaces, so the work done equals the surface tension times the total change in area,

W = T \times 2 \times 4 π (R_{2}^{2} - R_{1}^{2})

. With

R_{1} = 1 cm

and

R_{2} = 3 cm

W = 0.03 \times 8 π [(3 \times 1 0^{- 2})^{2} - (1 \times 1 0^{- 2})^{2}] = 0.24 π (9 - 1) \times 1 0^{- 4} = 1.92 π \times 1 0^{- 4} J

. Comparing with

α π \times 1 0^{- 4}

, the value of

α

1.92

JEE Main · 2026 · 02 Apr 2026 · Shift 2 · Single correct option

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