JEE Challenger
More from Mechanical Properties of Fluids
PhyMedMain 2026 • 4th Apr - Shift 1

Work Required to Increase Soap Bubble Radius

The surface tension of a soap solution is 3.5×10−2 N/m3.5 \times 10^{-2}\text{ N/m}. The work required to increase the radius of a soap bubble from 1 cm1\text{ cm} to 2 cm2\text{ cm} is α×10−6 J\alpha \times 10^{-6}\text{ J}. The value of α\alpha is ______. (π=22/7\pi = 22/7)

Official Numerical Answer264

Topics & Concepts

Mechanical Properties of FluidsSurface Tension

Step-by-Step Solution

To find the work required to increase the radius of a soap bubble, we use the concept of surface energy.

A soap bubble has two free surfaces in contact with air (the inner surface and the outer surface). Therefore, the total change in the surface area of the soap bubble when its radius increases from r1r_1 to r2r_2 is given by:

ΔA=2×4π(r22−r12)=8π(r22−r12)\Delta A = 2 \times 4\pi (r_2^2 - r_1^2) = 8\pi (r_2^2 - r_1^2)

The work done (WW) in increasing the radius is equal to the product of surface tension (TT) and the total increase in surface area (ΔA\Delta A):

W=T⋅ΔA=8πT(r22−r12)W = T \cdot \Delta A = 8\pi T (r_2^2 - r_1^2)

Given data:

  • Surface tension, T=3.5×10−2 N/m=72×10−2 N/mT = 3.5 \times 10^{-2} \text{ N/m} = \frac{7}{2} \times 10^{-2} \text{ N/m}
  • Initial radius, r1=1 cm=10−2 mr_1 = 1 \text{ cm} = 10^{-2} \text{ m}
  • Final radius, r2=2 cm=2×10−2 mr_2 = 2 \text{ cm} = 2 \times 10^{-2} \text{ m}
  • Value of π=227\pi = \frac{22}{7}

Now, calculating the change in squared radii:

r22−r12=(2×10−2)2−(10−2)2=(4−1)×10−4 m2=3×10−4 m2r_2^2 - r_1^2 = (2 \times 10^{-2})^2 - (10^{-2})^2 = (4 - 1) \times 10^{-4} \text{ m}^2 = 3 \times 10^{-4} \text{ m}^2

Substituting these values into the formula for work done:

W=8×(227)×(72×10−2 N/m)×(3×10−4 m2)W = 8 \times \left(\frac{22}{7}\right) \times \left(\frac{7}{2} \times 10^{-2} \text{ N/m}\right) \times \left(3 \times 10^{-4} \text{ m}^2\right)

Simplifying the terms:

W=8×11×10−2×3×10−4W = 8 \times 11 \times 10^{-2} \times 3 \times 10^{-4}

W=264×10−6 JW = 264 \times 10^{-6} \text{ J}

Comparing this with W=α×10−6 JW = \alpha \times 10^{-6} \text{ J}, we get:

α=264\alpha = 264

Work Required to Increase Soap Bubble Radius | Physics PYQ Solution - JEE Challenger