Question

Calculate the energy spent in spraying a drop of mercury of r=1cm radius into \[N = {10^6}\] droplets all of the same size. If the surface tension of mercury is \[T = 35 \times {10^{ - 3}}N/m\]

Answer 1

Hint: Energy spent is equal to surface tension multiplied by the size of the droplets.


Complete step by step solution:
Let the radius of the small droplets be r and \[\rho \]be the density of mercury
Mass of the drop must be equal to the total mass of the droplets.
Therefore,
After putting it into the formula we get,
\[\rho \times \dfrac{{4\pi }}{3}{R^3} = N \times \dfrac{{4\pi }}{3}{r^3} \times \rho \]
On simplifying the equation, we get,
\[ \Rightarrow {R^3} = N{r^3}\]
We can calculate the change in surface area by,
\[\Delta A = N \times 4\pi {r^2} - 4\pi {R^2}\]
On subtracting we get,
\[\Delta A = 4\pi (N{r^2} - {R^2})\]
Therefore,
From here we can calculate the energy spent,
Hence, energy spent is.
\[E = T\Delta A\]
On putting the values in the equation, we get,
\[E = 4\pi T(N{r^2} - {R^2})\]
Therefore, after simplifying and substituting the values in the equation we the above answer.
Additional Information:
Surface tendency is the referred to as the tendency of the liquid surfaces to shrink into the minimum surface area possible. At liquid-air interfaces, surface tension results from the greater attraction of all liquid molecules to each other due to cohesion than to the molecules in air due to the process of adhesion. Surface tension, therefore, has the dimension of force per unit length, of energy per unit area. Thus, while referring to liquids the terms surface tension is generally used.

Note: Before being able to solve the sum students are required to be able to comprehend the meaning of surface tension in liquids.