Question

A certain number of spherical drops of a liquid of radius ‘r’ coalesce to form a single drop of radius ‘R’ and volume ‘V’. If ‘T’ is the surface tension of the liquid then:
A. $Energy= 4V T (\dfrac {1}{r}-\dfrac {1}{R})$ is released
B. $Energy= 3V T (\dfrac {1}{r}+\dfrac {1}{R})$ is released
C. $Energy= 3V T (\dfrac {1}{r}-\dfrac {1}{R})$ is released
D. Energy is neither released nor absorbed

Answer 1

Hint: To solve this problem, first find the initial volume of each spherical drop. Using it find the initial volume of n spherical drops. Then, find the final volume of the large single drop. Find the initial area of n spherical drops and then find the final area of a large single drop. Use the formula giving a relationship between energy and surface tension. Substitute the values and find the amount of energy released.

Complete step by step answer:
Let n be the total number of spherical drops
Initial volume of each spherical drop is given by,
$V= \dfrac {4}{3}\pi {r}^{3}$
Initial volume of n number of spherical drops is given by,
$V= n\dfrac {4}{3}\pi {r}^{3}$ …(1)
Final volume of a large single drop is given by,
$V= \dfrac {4}{3}\pi {R}^{3}$ …(2)
Since initial volume is equal to final volume,
$ n\dfrac {4}{3}\pi {r}^{3}= \dfrac {4}{3}\pi {R}^{3}$
$\Rightarrow n= \dfrac {{R}^{3}}{{r}^{3}}$ ….(3)
Initial area of each spherical drop is given by,
$A= 4 \pi {r}^{2}$
Initial area of n number of spherical drops is given by,
$A= n4 \pi {r}^{2}$...(4)
Final area of the large single drop is given by,
${A}_{1}= 4 \pi {R}^{2}$ …(5)
We know, Relation between energy and surface tension is given by,
$E= T \Delta A$
$\Rightarrow E= T ({A}_{1}- A)$
Substituting values in above equation we get,
$E= T(4 \pi {R}^{2}- n4 \pi {r}^{2})$
$\Rightarrow E= 4 \pi T({R}^{2}-n{r}^{2})$
Substituting value of n in above equation we get,
$E= 4\pi T ({R}^{2}- \dfrac {{R}^{3}}{{r}^{3}}{r}^{2})$
$\Rightarrow E= 4 \pi T({R}^{2}- \dfrac {{R}^{3}}{r})$
$\Rightarrow E= 4 \pi T {R}^{3}(\dfrac {1}{R}- \dfrac {1}{r})$
Multiplying and dividing by 3 on right-hand side we get,
$E= 3 (\dfrac {4}{3} \pi {R}^{3})T(\dfrac {1}{R}- \dfrac {1}{r})$
Substituting equation. (2) in above equation we get,
$E= 3VT (\dfrac {1}{R}- \dfrac {1}{r})$ …(6)
But the radius of a large single drop is greater than the radius of a small spherical drop.
$\Rightarrow R > r$
So, equation. (6) can be written as,
$E= 3V T (\dfrac {1}{r}-\dfrac {1}{R})$
Thus, if ‘T’ is the surface tension of the liquid then $3V T (\dfrac {1}{r}-\dfrac {1}{R})$.

So, the correct answer is “Option C”.

Note: When a certain number of spherical drops coalesce to form a large drop only the area changes whereas the density and volume remains the same. To solve these types of problems, students must remember basic mathematical formulas such as area of sphere, volume of sphere, area of cylinder, volume of cylinder etc. Without knowing these formulas most of the problems can never be solved.