JEE MAIN - Physics (2014 (Offline) - No. 21)
On heating water, bubbles being formed at the bottom of the vessel detach and rise. Take the bubbles to be spheres of radius $$R$$ and making a circular contact of radius $$r$$ with the bottom $$R$$ and making a circular contact of 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 detach is: (density of water is $${\rho _w}$$)
_en_21_1.png)
$${R^2}\sqrt {{{{\rho _w}g} \over {3T}}} $$
$${R^2}\sqrt {{{{\rho _w}g} \over {6T}}} $$
$${R^2}\sqrt {{{{\rho _w}g} \over {T}}} $$
$${R^2}\sqrt {{{{2\rho _w}g} \over {3T}}} $$
Explanation
When the bubble gets detached, Buoyant force $$=$$ force due to surface tension
_en_21_2.png)
Force due to excess pressure $$=$$ upthrust
Access pressure in air bubble $$ = {{2T} \over R}$$
$${{2T} \over R}\left( {\pi {r^2}} \right) = {{4\pi {R^3}} \over {3T}}{\rho _w}g$$
$$ \Rightarrow {r^2} = {{2{R^4}{\rho _w}g} \over {3T}}$$
$$ \Rightarrow r = {R^2}\sqrt {{{2{\rho _w}g} \over {3T}}} $$
Force due to excess pressure $$=$$ upthrust
Access pressure in air bubble $$ = {{2T} \over R}$$
$${{2T} \over R}\left( {\pi {r^2}} \right) = {{4\pi {R^3}} \over {3T}}{\rho _w}g$$
$$ \Rightarrow {r^2} = {{2{R^4}{\rho _w}g} \over {3T}}$$
$$ \Rightarrow r = {R^2}\sqrt {{{2{\rho _w}g} \over {3T}}} $$
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