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(C) The work done in blowing a soap bubble of radius $r$ is given by $W = 8 \pi r^2 T$,where $T$ is the surface tension of the soap solution. $A$ soap

Vedclass · Accepted Answer

$W_2 < 4 W_1$

Vedclass · Answer

(C) The work done in blowing a soap bubble of radius $r$ is given by $W = 8 \pi r^2 T$,where $T$ is the surface tension of the soap solution. $A$ soap bubble has two surfaces,so the work done is $W = 2 	imes (4 \pi r^2 T) = 8 \pi r^2 T$.For the first bubble of radius $R$ at room temperature with surface tension $T_1$,the work done is $W_1 = 8 \pi R^2 T_1$.For the second bubble of radius $2R$ at a higher temperature with surface tension $T_2$,the work done is $W_2 = 8 \pi (2R)^2 T_2 = 8 \pi (4R^2) T_2 = 32 \pi R^2 T_2$.Taking the ratio,we get $\frac{W_2}{W_1} = \frac{32 \pi R^2 T_2}{8 \pi R^2 T_1} = 4 \left( \frac{T_2}{T_1} ight)$.Since the soap solution is heated,the surface tension decreases with an increase in temperature,meaning $T_2 Therefore,$\frac{T_2}{T_1} < 1$,which implies $W_2 < 4 W_1$.

The work done in blowing a soap bubble of radius $R$ is $W_1$ at room temperature. Now the soap solution is heated. From the heated solution,another soap bubble of radius $2R$ is blown and the work done is $W_2$. Then:

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