Solveeit Logo

Question

Question: If the density of air is \( {\rho _a} \) and that of the liquid is \( {\rho _l} \) , then for a give...

If the density of air is ρa{\rho _a} and that of the liquid is ρl{\rho _l} , then for a given piston speed the rate (volume per unit time) at which the liquid is sprayed will be proportional to

(A) ρaρl\sqrt {\dfrac{{{\rho _a}}}{{{\rho _l}}}}
(B) ρaρl\sqrt {{\rho _a}{\rho _l}}
(C) ρlρa\sqrt {\dfrac{{{\rho _l}}}{{{\rho _a}}}}
(D) ρl{\rho _l}

Explanation

Solution

Hint We need to use the equation of continuity at the two points. Now we can equate the equation containing the density and the velocity at the two points. So the speed at which the liquid will be sprayed out can be calculated in the terms of the density from there.

Formula Used: In this solution we will be using the following equation,
12ρv2=const\Rightarrow \dfrac{1}{2}\rho {v^2} = const
where ρ\rho is the density and vv is the velocity.

Complete step by step answer
In this question we are given that the density of air is given as ρa{\rho _a} and the density of the liquid is given as ρl{\rho _l} .
Now from the equation of continuity, we can write the density and velocity as,
12ρv2=const\Rightarrow \dfrac{1}{2}\rho {v^2} = const
Therefore we can take the velocity of air as va{v_a} and that of the liquid at the nozzle as vl{v_l} . Therefore, from the equation of continuity we can write,
12ρava2=12ρlvl2\Rightarrow \dfrac{1}{2}{\rho _a}{v_a}^2 = \dfrac{1}{2}{\rho _l}{v_l}^2
Therefore, we can cancel the 12\dfrac{1}{2} from both the sides of the equation. So we get,
ρava2=ρlvl2\Rightarrow {\rho _a}{v_a}^2 = {\rho _l}{v_l}^2
Now we can take the like terms on one side of the equation. So we get,
vl2va2=ρaρl\Rightarrow \dfrac{{{v_l}^2}}{{{v_a}^2}} = \dfrac{{{\rho _a}}}{{{\rho _l}}}
Now we can take the square root on both sides of the equation. So we have,
vlva=ρaρl\Rightarrow \dfrac{{{v_l}}}{{{v_a}}} = \sqrt {\dfrac{{{\rho _a}}}{{{\rho _l}}}}
Now the velocity of the piston is given to be constant. So we can write,
vlρaρl\Rightarrow {v_l} \propto \sqrt {\dfrac{{{\rho _a}}}{{{\rho _l}}}}
Therefore, for a given piston speed the rate (volume per unit time) at which the liquid is sprayed will be proportional to ρaρl\sqrt {\dfrac{{{\rho _a}}}{{{\rho _l}}}}
So the correct answer is option A.

Note
The equation of continuity is physics is an equation which describes the transport of some quantity. In the case of fluid motions, the mass must always be conserved. So in the case the flow is one dimensional, the velocity and the density is conserved over an area.