Question

A large number of water drops, each of radius \[r\], combine to have a drop of radius \[R\]. If the surface tension is \[T\] and mechanical equivalent of heat is \[J\], the rise in heat energy per unit volume will be:
A. \[2T/rJ\]
B. \[3T/rJ\]
C. \[\left( {2T/J} \right)\left( {\dfrac{1}{r} - \dfrac{1}{R}} \right)\]
D. \[\left( {3T/J} \right)\left( {\dfrac{1}{r} - \dfrac{1}{R}} \right)\]

Answer 1

Hint: In this question, we need to determine the rise in heat energy per unit volume if a large number of water drops, each of radius \[r\], combine to have a drop of radius \[R\]. If the surface tension is \[T\] and the mechanical equivalent of heat is \[J\]. For this, we need to use the concept that the volume of \[n\] drops is equal to the volume of a bigger drop as a bigger water drop can be formed by combining small water drops.

Formula used::
The volume of a sphere (water drop) is given below.
\[V = \dfrac{4}{3}\pi {r^3}\]
Here, \[V\] is the volume and \[r\] is the radius of a sphere.
And, heat energy is,
\[H = \dfrac{{\vartriangle U}}{J}\]
Here, \[\vartriangle U\] is the decrease in surface energy and \[J\] is heat equivalent.
And, \[\vartriangle U = T \times \] the change in surface area
Here, \[\vartriangle U\] is the decrease in surface energy and \[T\] is surface tension.

Complete step by step solution:
We know that \[r\] is the radius of a smaller drop whereas \[R\] is the radius of a bigger water drop. The small water drops can be combined to form bigger water drops. Thus, we can say that the volume of \[n\] drops is equal to the volume of a bigger drop.
Volume of \[n\] drops \[ = \] volume of a bigger drop
\[n \times \dfrac{4}{3}\pi {r^3} = \dfrac{4}{3}\pi {R^3}\]
Let us simplify this.
\[
  n \times {r^3} = {R^3} \\
   \Rightarrow {n^{1/3}}r = R \\
 \]
Now, the total change in surface energy is given by
Total change in surface area \[ = T\left( {n4\pi {r^2} - 4\pi {R^2}} \right)\]
So, total change in surface area \[ = 4\pi T\left( {n{r^2} - {R^2}} \right)\]
Thus, heat energy per unit volume is given by
\[
  \dfrac{H}{V} = \dfrac{{4\pi T\left( {n{r^2} - {R^2}} \right)}}{{J \times \dfrac{4}{3} \times \pi {R^3}}} \\
   \Rightarrow \dfrac{H}{V} = \dfrac{{3T}}{J} \times \left( {\dfrac{{n{r^2}}}{{{R^3}}} - \dfrac{1}{R}} \right) \\
 \]
Now, put \[n{r^3} = {R^3}\] in the above equation.
Thus, we get
\[\dfrac{H}{V} = \dfrac{{3T}}{J} \times \left( {\dfrac{1}{r} - \dfrac{1}{R}} \right)\]

Therefore, the correct option is (D).

Note: Many students make mistakes in the calculation part of heat energy per unit volume. Also, they may get the wrong result if there is a mistake in the formula of volume of a sphere. Also, it is important to write all the formulas related to heat energy for getting the desired result.