Question

8 mercury drops coalesce to form one mercury drop, the energy changes by a factor of:
(a). 1
(b). 2
(c). 4
(d). 6

Answer 1

Hint: In this type of question we know that surface energy of a drop is given by, $\text{surface energy}=\text{surface tension}\times \text{area of surface}$. So, we will first find the volume of a single drop and then using it we will find the volume of 8 drops. Then, we will find the radius of the new drop and then using it we will find new energy of the drop and by comparing new and old we will find the answer.

Formula used: $V=\dfrac{4}{3}\pi {{R}^{3}}$, $A=4\pi {{R}^{2}}$

Complete step by step answer:
Now, in question it is given that a new drop is formed from 8 mercury drops, so first of all we will find the total volume of 8 mercury drops.
Now, we know that volume of one mercury drop is given by,
$V=\dfrac{4}{3}\pi {{R}^{3}}$, where R is the radius of the drop.
Now, total volume of 8 droplets is given as,
$V'=8\times \dfrac{4}{3}\pi {{R}^{3}}=\dfrac{4}{3}\pi {{\left( 2R \right)}^{3}}$, thus, the new radius of a big drop combined from 8 drops can be given as,$R'=2R$.
Now, area of drop is given by, $A=4\pi {{R}^{2}}$
Same as volume the total area of 8 droplets is given by,
$A'=4\pi R{{'}^{2}}=4\pi {{\left( 2R \right)}^{2}}$
$\Rightarrow A'=4\pi R{{'}^{2}}=4\times 4\pi {{R}^{2}}$
$\Rightarrow A'=4\times \text{old area}$
Now, we know that $\text{surface energy}=\text{surface tension}\times \text{area of surface}$, so, taking surface tension as constant the relation between surface energy and area can be given by,
$\text{Energy}\ \text{ }\\!\\!\alpha\\!\\!\text{ }\ \text{area}$
From this, we can take energies and areas as,
${{E}_{1}}=4\pi {{R}^{2}}$ …………………..(i)
${{E}_{1}}=4\pi {{R}^{2}}$ ……………….(ii)
Now taking ratios we will get,
$\dfrac{{{E}_{2}}}{{{E}_{1}}}=\dfrac{4\times 4\pi {{R}^{2}}}{4\pi {{R}^{2}}}$
Solving the equation further we will get,
$\dfrac{{{E}_{2}}}{{{E}_{1}}}=\dfrac{4}{1}$
Which can also be written as ${{E}_{2}}=4{{E}_{1}}$.
Hence, we can say that the energy changes by a factor of 4.
Thus, option (c) is correct.

Note: In this type of question students must take care while using formula and considering the constant terms. Here, it was given that the constant remains the same so we are taking it as constant. Now, if tension produced is also given then we have to solve it accordingly. Students must take care while considering the radius as they might confuse in new and old radius and due to that sum may go wrong.