(A) soap bubble has two surfaces (inner and outer). The initial surface area $A_1 = 2 \times (4 \pi r^2) = 8 \pi r^2$.When the radius is doubled to $2

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(A) soap bubble has two surfaces (inner and outer). The initial surface area $A_1 = 2 \times (4 \pi r^2) = 8 \pi r^2$.When the radius is doubled to $2r$,the final surface area $A_2 = 2 \times (4 \pi (2r)^2) = 2 \times (16 \pi r^2) = 32 \pi r^2$.The change in surface area is $\Delta A = A_2 - A_1 = 32 \pi r^2 - 8 \pi r^2 = 24 \pi r^2$.The work done (energy required) is $W = T \times \Delta A$.Therefore,$W = T \times 24 \pi r^2 = 24 \pi r^2 T$.

Class 11 Physics Mechanical Properties of Fluids Surface Energy

Difficult

The radius of a soap bubble is $r$. The surface tension of the soap solution is $T$. Keeping the temperature constant,the radius of the soap bubble is doubled. The energy necessary for this process will be: (in $\pi r^2 T$)

A
$24$
B
$8$
C
$12$
D
$16$

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The radius of a soap bubble is $r$. The surface tension of the soap solution is $T$. Keeping the temperature constant,the radius of the soap bubble is doubled. The energy necessary for this process will be: (in $\pi r^2 T$)

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