Q58 / 75 · JEE Main 2026 (02 April Shift 2) · Physics
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 $\alpha \pi \times 10^{-4}$ J. The value of $\alpha$ is _______. (Take $\pi = 3.14$)
- A$0.86$
- B$0.64$
- Correct answerC$1.92$
- D$7.68$
Solution
Initial radius $r_1 = \dfrac{2}{2} = 1$ cm $= 10^{-2}$ m
Final radius $r_2 = \dfrac{6}{2} = 3$ cm $= 3 \times 10^{-2}$ m
Surface tension $T = 0.03$ N/m
A soap bubble has two free surfaces. The change in total surface area is given by:
$\Delta A = 2 \times 4\pi (r_2^2 - r_1^2)$
$\Delta A = 8\pi ((3 \times 10^{-2})^2 - (10^{-2})^2)$
$\Delta A = 8\pi (9 \times 10^{-4} - 10^{-4}) = 64\pi \times 10^{-4}$ m$^2$
Work done $W = T \Delta A$
$W = 0.03 \times 64\pi \times 10^{-4}$
$W = 1.92\pi \times 10^{-4}$ J
Comparing with $\alpha \pi \times 10^{-4}$ J, we get $\alpha = 1.92$.
Answer: $1.92$
Final radius $r_2 = \dfrac{6}{2} = 3$ cm $= 3 \times 10^{-2}$ m
Surface tension $T = 0.03$ N/m
A soap bubble has two free surfaces. The change in total surface area is given by:
$\Delta A = 2 \times 4\pi (r_2^2 - r_1^2)$
$\Delta A = 8\pi ((3 \times 10^{-2})^2 - (10^{-2})^2)$
$\Delta A = 8\pi (9 \times 10^{-4} - 10^{-4}) = 64\pi \times 10^{-4}$ m$^2$
Work done $W = T \Delta A$
$W = 0.03 \times 64\pi \times 10^{-4}$
$W = 1.92\pi \times 10^{-4}$ J
Comparing with $\alpha \pi \times 10^{-4}$ J, we get $\alpha = 1.92$.
Answer: $1.92$
Hence, the correct answer is option: 'C'.