
When one drop of a liquid is broken into a number of droplets, which of the statements is correct?
A. Surface area decreases
B. Surface energy decreases
C. Temperature of liquid increases
D. Surface area increases.
Answer
443.1k+ views
Hint :In order to this question, to know the correct option, first we will equate or compare the volumes of both the drop and new droplets. And, then we will compare the surface areas of both the cases. That’s how we will know the exact answer.
Complete Step By Step Answer:
When the drop is broken into many droplets, the total volume of the droplets is constant.
Thus,
$ \begin{gathered}
\dfrac{4}{3}\pi {R^3} = n\dfrac{4}{3}\pi {r^3} \\
\Rightarrow r = \dfrac{R}{{\sqrt[3]{n}}} \\
\end{gathered} $
Here, $ \dfrac{4}{3}\pi {R^3} $ is the volume of the first drop.
$ n\dfrac{4}{3}\pi {r^3} $ is the volume of the number of new droplets.
Now, The ratio of surface area of both the first drop and the other new droplets:
$ \dfrac{{{S_1}}}{{{S_2}}} = \dfrac{{{R^2}}}{{n{r^2}}} = \dfrac{{{R^2}}}{{n \times \dfrac{{{R^2}}}{{{n^{\dfrac{2}{3}}}}}}} = \dfrac{1}{{{n^{\dfrac{1}{3}}}}} < 1 $
Now, according to the upper conclusion, the surface area of the new droplets is increasing as compared to the drop.
Hence, the correct option is (D.) Surface area increases.
Note :
If a number of little droplets of water, each of radius $ r $ , coalesce to form a single drop of radius $ R $ , and the energy released is converted into kinetic energy then find out the velocity acquired by the bigger drop.
Complete Step By Step Answer:
When the drop is broken into many droplets, the total volume of the droplets is constant.
Thus,
$ \begin{gathered}
\dfrac{4}{3}\pi {R^3} = n\dfrac{4}{3}\pi {r^3} \\
\Rightarrow r = \dfrac{R}{{\sqrt[3]{n}}} \\
\end{gathered} $
Here, $ \dfrac{4}{3}\pi {R^3} $ is the volume of the first drop.
$ n\dfrac{4}{3}\pi {r^3} $ is the volume of the number of new droplets.
Now, The ratio of surface area of both the first drop and the other new droplets:
$ \dfrac{{{S_1}}}{{{S_2}}} = \dfrac{{{R^2}}}{{n{r^2}}} = \dfrac{{{R^2}}}{{n \times \dfrac{{{R^2}}}{{{n^{\dfrac{2}{3}}}}}}} = \dfrac{1}{{{n^{\dfrac{1}{3}}}}} < 1 $
Now, according to the upper conclusion, the surface area of the new droplets is increasing as compared to the drop.
Hence, the correct option is (D.) Surface area increases.
Note :
If a number of little droplets of water, each of radius $ r $ , coalesce to form a single drop of radius $ R $ , and the energy released is converted into kinetic energy then find out the velocity acquired by the bigger drop.
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