An air bubble of radius 1.0 mm is observed at a depth of 20 cm below the free surface of a liquid having…
Solution

$\Delta \mathrm{P}=\mathrm{P}_{\mathrm{in}}-\mathrm{P}_0$
$\begin{aligned} & =\rho \mathrm{gh}+\frac{2 \mathrm{~T}}{\mathrm{R}}=\frac{1000 \times 10 \times 20}{100}+\frac{2 \times 0.095}{10^{-3}} \\ & =2000+190 \\ & =2190\end{aligned}$
Asked in: JEE Main 2025 (23 Jan Shift 2)
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