Physics · 2026
JEE Main · 2 April 2026, Shift 2 · Q33
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 × 10^-4 J. The value…
The surface tension of a soap bubble is $\displaystyle 0.03 \mathrm{~N} / \mathrm{m}$. The work done in increasing the diameter of bubble from $\displaystyle 2$ cm to $\displaystyle 6$ cm is $\displaystyle \alpha \pi \times 10^{-4} \mathrm{~J}$. The value of $\displaystyle \alpha$ is $\displaystyle \_\_\_\_$. (Take $\displaystyle \pi=3.14$ )
Official answer
From NTA’s final answer key for this paper.
(3)
$\displaystyle 1.92$
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JEE Main 2026 Physics question, with the answer from NTA’s final answer key. Where our answers come from.