(D) soap bubble has two surfaces (inner and outer). Initially, the surface area of the soap bubble is $A_1 = 2 \times (4 \pi r^2) = 8 \pi r^2$. Under

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(D) soap bubble has two surfaces (inner and outer). Initially, the surface area of the soap bubble is $A_1 = 2 \times (4 \pi r^2) = 8 \pi r^2$. Under isothermal conditions, the radius becomes $2r$. The final surface area is $A_2 = 2 \times (4 \pi (2r)^2) = 2 \times (16 \pi r^2) = 32 \pi r^2$. The increase in surface area is $\Delta A = A_2 - A_1 = 32 \pi r^2 - 8 \pi r^2 = 24 \pi r^2$. The energy spent (work done) is given by $W = T \times \Delta A$. Therefore, $W = T \times 24 \pi r^2 = 24 \pi T r^2$.

Class 11 Physics Fluid Mechanics and Surface Tension Surface Energy

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$A$ soap bubble of radius $r$ is blown up to form a bubble of radius $2r$ under isothermal conditions. If $T$ is the surface tension of the soap solution, the energy spent in the blowing is: (in $\pi T r^2$)

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A
$3$
B
$6$
C
$12$
D
$24$

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$A$ soap bubble of radius $r$ is blown up to form a bubble of radius $2r$ under isothermal conditions. If $T$ is the surface tension of the soap solution, the energy spent in the blowing is: (in $\pi T r^2$)

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