Physics · 2023
JEE Main · 29 January 2023, Shift 1 · Q13
Surface tension of a soap bubble is 2.0 × 10^-2 Nm^-1. Work done to increase the radius of soap bubble from 3.5 cm to 7 cm will be: Take [π=22/7]
Surface tension of a soap bubble is $\displaystyle 2.0 \times 10^{-2} \mathrm{Nm}^{-1}$. Work done to increase the radius of soap bubble from $\displaystyle 3.5$ cm to $\displaystyle 7$ cm will be:Take $\displaystyle \left[\pi=\frac{22}{7}\right]$
Official answer
From NTA’s final answer key for this paper.
(1)
$\displaystyle 18.48 \times 10^{-4} \mathrm{~J}$
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JEE Main 2023 Physics question, with the answer from NTA’s final answer key. Where our answers come from.