Properties Of Matter Ques 81
- A drop of liquid of radius $R=10^{-2} \mathrm{~m}$ having surface tension $S=\frac{0.1}{4 \pi} \mathrm{Nm}^{-1}$ divides itself into $K$ identical drops. In this process the total change in the surface energy $\Delta U=10^{-3} \mathrm{~J}$. If $K=10^\alpha$, then the value of $\alpha$ is
(2017 Adv.)
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Answer:
Correct Answer: 81.6
Solution:
Formula:
- From mass conservation,
$$ \begin{aligned} \rho \cdot \frac{4}{3} \pi R^{3}= & \rho \cdot K \cdot \frac{4}{3} \pi r^{3} \Rightarrow R=K^{1 / 3} r \\ \Delta \quad \Delta U & =T \Delta A=T\left(K \cdot 4 \pi r^{2}-4 \pi R^{2}\right) \\ & =T\left(K \cdot 4 \pi R^{2} K^{-2 / 3}-4 \pi R^{2}\right) \\ \Delta U & =4 \pi R^{2} T\left[K^{1 / 3}-1\right] \end{aligned} $$
Putting the values, we get
$$ 10^{-3}=\frac{10^{-1}}{4 \pi} \times 4 \pi \times 10^{-4}\left[K^{1 / 3}-1\right] $$
$$ \begin{array}{rlrl} & & 100 & =K^{1 / 3}-1 \\ \Rightarrow & K^{1 / 3} & \cong 100=10^{2} \\ \text { Given that } & K & =10^{\alpha} \\ \therefore & 10^{\alpha / 3} & =10^{2} \\ \Rightarrow & & \frac{\alpha}{3} & =2 \Rightarrow \alpha=6 \end{array} $$
BETA
