Physics · 2026
JEE Main · 4 April 2026, Shift 1 · Q47
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…
The surface tension of a soap solution is $\displaystyle 3.5 \times 10^{-2} \mathrm{~N} / \mathrm{m}$. The work required to increase the radius of a soap bubble from $\displaystyle 1$ cm to $\displaystyle 2$ cm is $\displaystyle \alpha \times 10^{-6} \mathrm{~J}$. The value of $\displaystyle \alpha$ is $\displaystyle \_\_\_\_$.
$\displaystyle (\pi=22 / 7)$
Official answer
From NTA’s final answer key for this paper.
264
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JEE Main 2026 Physics question, with the answer from NTA’s final answer key. Where our answers come from.