(D) soap bubble has two surfaces (inner and outer). Therefore,the total surface area is $A = 2 \times (4\pi R^2) = 8\pi R^2$.The work done $W$ in blow

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(D) soap bubble has two surfaces (inner and outer). Therefore,the total surface area is $A = 2 \times (4\pi R^2) = 8\pi R^2$.The work done $W$ in blowing the bubble is given by the change in surface energy: $W = T \times \Delta A$.Given: Surface tension $T = 2 \times 10^{-2} \ N/m$ and radius $R = 1 \ cm = 10^{-2} \ m$.Substituting the values:$W = T \times 8\pi R^2$$W = (2 \times 10^{-2}) \times 8\pi \times (10^{-2})^2$$W = 16\pi \times 10^{-2} \times 10^{-4}$$W = 16\pi \times 10^{-6} \ J$.

Class 11 Physics Fluid Mechanics and Surface Tension Excess Pressure and coalesce of Bubble and drop

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The surface tension of a soap solution is $2 \times 10^{-2} \ N/m$. To blow a bubble of radius $1 \ cm$,the work done is:

A
$4\pi \times 10^{-6} \ J$
B
$8\pi \times 10^{-6} \ J$
C
$12\pi \times 10^{-6} \ J$
D
$16\pi \times 10^{-6} \ J$

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Fluid Mechanics and Surface Tension

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Excess Pressure and coalesce of Bubble and drop

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The surface tension of a soap solution is $2 \times 10^{-2} \ N/m$. To blow a bubble of radius $1 \ cm$,the work done is:

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$Assertion :$ $A$ bubble comes from the bottom of a lake to the top.
$Reason :$ Its radius increases.