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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 $2 R$ is blown and the work done is $W_2$. Then
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The correct answer is:
$W_2<4 W_1$
The work done in forming a soap in forming a soap bubble of radius $r$ is $W_1=8 \pi r^2 T_1$, where $T_1$ is the surface tension at room temperature and the work done to form a soap bubble of radius $2 r$ is
$\therefore \frac{W_2}{W_1}=\frac{328 \pi r^2 T_2}{8 \pi r^2 T_1}=\frac{4 T_2}{T_1}$
If $T_1=T_2$, then $W_2=4 W_1$
But it is given that the solution is heated. As the temperature is increased, the surface tension of the soap solution is decreases
$\therefore T_2
$\therefore \frac{W_2}{W_1}=\frac{328 \pi r^2 T_2}{8 \pi r^2 T_1}=\frac{4 T_2}{T_1}$
If $T_1=T_2$, then $W_2=4 W_1$
But it is given that the solution is heated. As the temperature is increased, the surface tension of the soap solution is decreases
$\therefore T_2
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