Question
How much work is required to form a bubble of 2 cm radius from the soap solution having surface tension 0.07 N/m.

Answer

Data $: r =4 cm =4 \times 10^{-2} m , T =25 \times 10^{-3} N / m$ Initial surface area of soap bubble $=0$
Final surface area of soap bubble $=2 \times 4 \pi r^2$
$\therefore$ Increase in surface area $=2 \times 4 \pi r^2$
The work done
$=$ surface tension $\times$ increase in surface area
$=T \times 2 \times 4 \pi r^2$
$=25 \times 10^{-3} \times 2 \times 4 \times 3.142 \times\left(4 \times 10^{-2}\right)^2$
$=1.005 \times 10^{-3} J$
The work done $=0.07 \times 8 \times 3.142 \times\left(2 \times 10^{-2}\right)^2$ $=7.038 \times 10^{-4} J$

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