Question

The radius $R$ of the soap bubble is doubled under iso-thermal condition. If $T$ be the surface tension of soap bubble. The work done in doing so is given by:

• A. $32\pi R^{2}T$
• B. $24\pi R^{2}T$
• C. $8\pi R^{2}T$
• D. $4\pi R^{2}T$

Answer 1

Answer: (B)

$24\pi R^{2}T$

Initial radius of the bubble $=R$

Initial outer surface area of the bubble $A_o=4\pi R^2$

As the radius of the bubble gets doubled i.e $R^{\prime }=2R$

$\therefore$ Final outer surface area of the bubble $A_o^{\prime }=4\pi (2R{)}^2=16\pi R^2$

Change in outer surface area $\Delta A=16\pi R^2-4\pi R^2=12\pi R^2$

As the bubble has two surface (inner and outer), thus total change in surface area gets doubled.

$⟹$ $\Delta A_T=24\pi R^2$

Work done $W=\Delta A_T\times T=24\pi R^{2}T$