The surface tension of a soap solution is . The work required to increase the radius of a soap bubble from 1 cm to 2 cm is

Question

The surface tension of a soap solution is $3.5 \times 10^{- 2} N / m$ . The work required to increase the radius of a soap bubble from 1 cm to 2 cm is $α \times 10^{- 6} J$ . The value of $α$ is ______. $(π = \frac{22}{7})$ [2026]

Smart Achievers · Accepted Answer

(264)

$Given: surface tension (T) = 3.5 \times 10^{- 2} N/m$

$initial radius (r_{1}) = 1 cm$

$final radius (r_{2}) = 2 cm$

$work required W = 2 \times T \times Δ A$

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

$W = 2 \times 3.5 \times 10^{- 2} \times 4 \times \frac{22}{7} \times (3 \times 10^{- 4})$

$= 264 \times 10^{- 6} J$

$Now α \times 10^{- 6} = 264 \times 10^{- 6} \Rightarrow α = 264$

Solution

Q.
The surface tension of a soap solution is $3.5 \times 10^{- 2} N / m$ . The work required to increase the radius of a soap bubble from 1 cm to 2 cm is $α \times 10^{- 6} J$ . The value of $α$ is ______. $(π = \frac{22}{7})$ [2026]

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Solution

Q. The surface tension of a soap solution is 3.5×10-2 N/m. The work required to increase the radius of a soap bubble from 1 cm to 2 cm is α×10-6 J. The value of α is ______. (π=227) [2026]

Ans. (264) Given: surface tension (T)=3.5×10-2 N/m initial radius (r1)=1 cm final radius (r2)=2 cm work required W=2×T×ΔA W=2×T×4π(r22-r12) W=2×3.5×10-2×4×227×(3×10-4) =264×10-6 J Now α×10-6=264×10-6⇒α=264

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Q.
The surface tension of a soap solution is $3.5 \times 10^{- 2} N / m$ . The work required to increase the radius of a soap bubble from 1 cm to 2 cm is $α \times 10^{- 6} J$ . The value of $α$ is ______. $(π = \frac{22}{7})$ [2026]