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**Hint:**As a fast stream of air is produced between the balls, the velocity of the air molecules between the balls change. This leads to change in pressure. Try to figure out whether pressure will increase or decrease and accordingly answer.

**Complete step by step answer:**

Important point to understand here is that when pressure between the balls will decrease then the balls will move towards each other. And if the pressure between the balls increases, the balls move away from each other.

Now we have to find an equation which relates velocity, pressure and that equation holds for viscous fluids like air or water. Since we are dealing with air, we have Bernoulli’s equation.

As per Bernoulli’s principle, an increase in a fluid’s speed creates a pressure decrease and a decrease in fluid’s speed creates an increase in pressure.

From the above discussion let’s use Bernoulli’s principles, we have:

$ \Rightarrow {P_2} = {P_1} + \dfrac{1}{2}\rho ({v_1}^2 - {v_2}^2)$ -- equation \[1\]

Where ${P_1},\,{P_2}$ are the initial and final pressures respectively.

$\rho $ is the density of the air which will be same

${v_1},\,{v_2}$ are the initial and final velocities respectively.

It must be noted here that

${v_2} > {v_1}$ as a fast stream of air is produced in the second case.

Using this in equation we have:

$ \Rightarrow {P_2} = {P_1} + \dfrac{1}{2}\rho ({v_1}^2 - {v_2}^2)$

Now as we have achieved earlier that ${v_2} > {v_1}$ hence, the term $({v_1}^2 - {v_2}^2)$ will be negative

$ \Rightarrow {P_2} = {P_1} + (negative\,quantity)$

$ \Rightarrow {P_2} < {P_1}$

Therefore, the pressure after a fast stream of air is produced will be lower.

**As a result, the balls will Come nearer to each other. Option (A) is the correct option.**

**Note:**While dealing with such questions, try to find some definitions or formulas which relate the given quantities. Remember that when velocity of air increases, pressure decreases and when velocity of air decreases then pressure will increase.

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