Properties of matterMedium
Question
The surface tension of a soap bubble is $0.03\text{ }N/m$. The work done in increasing the diameter of bubble from 2 cm to 6 cm is $\alpha\pi \times 10^{- 4}\text{ }J$. The value of $\alpha$ is $\_\_\_\_$ . (Take $\pi = 3.14$ )
Options
A.0.86
B.0.64
C.1.92
D.7.68
Solution
Sol. [Key Concept: Soap film has two surfaces: $W = 2T \times 4\pi(R_{2}^{2} - R_{1}^{2})$.]
$$W = \Delta U = 2\text{ }T\left\lbrack 4\pi R_{2}\ ^{2} - 4\pi R_{1}\ ^{2} \right\rbrack$$
$$\begin{matrix} & W = 2 \times 3 \times 10^{- 2} \times 4\pi\left\lbrack 9 \times 10^{- 4} - 1 \times 10^{- 4} \right\rbrack \\ & W = 1.92\pi \times 10^{- 4}\text{ }J \\ & \alpha = 1.92 \end{matrix}$$
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