Question

Calculate the work done in breaking a mercury drop of radius 1 mm into 1000 droplets of the same size. Surface tension of mercury is \[525\times {{10}^{-3}}N{{m}^{-1}}\].

Answer 1

Hint: Here we don’t have the value of the radius of small drops. It can find out from the volume of the big drop of mercury. After finding the radius we can do the calculations to find out the change in surface area during the conversion of mercury big drop into small drops. The change in surface area and surface tension can be used to find the amount of work done.

Formula used:
\[S=4\pi {{R}^{2}}\], where R is the radius of the sphere.
\[W=T\Delta S\], where T is the surface tension and \[\Delta S\] is the change in surface area.

Complete step by step answer:
We can consider a big drop of mercury of radius R. It can split into 1000 droplets of same sized mercury. The radius of the small drop is r.

We can find out the surface area of the big sphere of radius R.

The surface area of a sphere can be written as,

\[S=4\pi {{R}^{2}}\], where R is the radius of the sphere.

Therefore, the surface area of the big sphere will be,

\[{{S}_{b}}=4\pi {{R}^{2}}\], where \[R=1\times {{10}^{-3}}m\]

Since the radius of the big drop is already given, we can assign that value into this.

\[{{S}_{b}}=4\pi {{(1\times {{10}^{-3}})}^{2}}\]
\[{{S}_{b}}=0.1256\times {{10}^{-4}}{{m}^{2}}\]………………………….(1)

Now we have to find out the radius of the small sphere. Since we are dividing one mercury drop into 1000 drops equally, the total volume of small drops and the volume of the big drop will be the same. So we can compare this.

The volume of the big drop can be written as,

\[{{V}_{b}}=\dfrac{4}{3}\pi {{R}^{3}}\]

Therefore, the total volume of the 1000 small drops of mercury will be,

\[{{V}_{b}}=\dfrac{4}{3}\pi {{R}^{3}}\]……………….(2)

If the small drop radius is r, then the volume of each small drop will be,

 \[{{V}_{s}}=\dfrac{4}{3}\pi {{r}^{3}}\]

For 1000 drops, the total volume will be,

\[\Rightarrow {{V}_{s}}\times 1000=1000\times \dfrac{4}{3}\pi {{r}^{3}}\]…………………….(3)

Since equation (2) and equation (3) are the same, we can equate those.

\[\dfrac{4}{3}\pi {{R}^{3}}=1000\times \dfrac{4}{3}\pi {{r}^{3}}\]
\[{{R}^{3}}=1000\times {{r}^{3}}\]
\[r=\dfrac{R}{10}\]……………………………(4)

Now we can find out the surface area of 1000 drops of mercury.

\[{{S}_{s}}=1000\times 4\pi {{r}^{2}}\]

We can assign the equation (4) into this.

\[{{S}_{s}}=1000\times 4\pi {{\left( \dfrac{R}{10} \right)}^{2}}\]

Now we can assign the value of the radius of the big sphere into this.

\[{{S}_{s}}=12.56\times {{10}^{-5}}{{m}^{2}}\]…………………..(5)

We can find the change in surface area by using equation (1) and equation (5).

\[\Delta S=|{{S}_{b}}-{{S}_{s}}|\]
\[\Delta S=|12.56\times {{10}^{-6}}-12.56\times {{10}^{-5}}|\]
\[\Delta S=1.13\times {{10}^{-4}}\]……………………..(6)

As we know, the total work done can be written as,

\[W=T\Delta S\], where T is the surface tension and \[\Delta S\] is the change in surface area.
The surface tension of mercury is already given in the question.

\[T=525\times {{10}^{-3}}N{{m}^{-1}}\]

We can assign this value to find the total work done.

\[W=525\times {{10}^{-3}}\times 1.13\times {{10}^{-4}}\]
\[W=593.25\times {{10}^{-7}}J\]

This much work required to break a mercury drop of 1 mm into 1000 equal-sized smaller drops.

Note: The surface tension is the surface energy per unit area of the surface interface. This surface energy is known as the work done in this case. Since the question is not given any information about the radius of the small drops, we have to find it out from the volume of the small drops. We have used powers in each value appropriately to make the calculations simpler. If we are doing any alterations in the values, we should do the calculation carefully. Since one order can change the entire answer.