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On heating water, bubbles being formed at the bottom of the vessel detatch and rise. Take the bubbles to be spheres of radius $R$ and making a circular contant fo radius $r$ with the bottom of the vessel. If $r << R$ and the surface tension of water is $T$, value of $r$ just before bubbles detatch is : (density of water is $\rho_w$ )

$\begin{array}{1 1}(A)\;R^2 \sqrt {\large\frac{\rho_wg}{T}} \\ (B)\;R^2 \sqrt {\large\frac{3\rho_wg}{T}} \\(C)\;R^2 \sqrt {\large\frac{\rho_w g}{3T}} \\(D)\;R^2 \sqrt {\large\frac{\rho_w g}{6T}} \end{array} $

1 Answer

$( 2 \pi r T) \sin \theta = \frac{4}{3} \pi R^3 \rho wg$
$T \times \frac{r}{R} \times 2 \pi r = \frac{4}{3} \pi R^3 \rho wg$
$\implies r^2 = \frac{2}{3} \frac{R^4 \rho wg}{T}$
$\implies r =R^2 \sqrt {\large\frac{\rho _w g}{3T}}$ is correct.
Hence C is the correct answer.
answered Apr 7, 2014 by meena.p
edited Nov 25, 2017 by priyanka.c

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