Question

If the density of air is $ {\rho _a} $ and that of the liquid is $ {\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) $ \sqrt {\dfrac{{{\rho _a}}}{{{\rho _l}}}} $
(B) $ \sqrt {{\rho _a}{\rho _l}} $
(C) $ \sqrt {\dfrac{{{\rho _l}}}{{{\rho _a}}}} $
(D) $ {\rho _l} $

Answer 1

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,
 $\Rightarrow \dfrac{1}{2}\rho {v^2} = const $
where $ \rho $ is the density and $ v $ is the velocity.

Complete step by step answer
In this question we are given that the density of air is given as $ {\rho _a} $ and the density of the liquid is given as $ {\rho _l} $ .
Now from the equation of continuity, we can write the density and velocity as,
 $\Rightarrow \dfrac{1}{2}\rho {v^2} = const $
Therefore we can take the velocity of air as $ {v_a} $ and that of the liquid at the nozzle as $ {v_l} $ . Therefore, from the equation of continuity we can write,
 $\Rightarrow \dfrac{1}{2}{\rho _a}{v_a}^2 = \dfrac{1}{2}{\rho _l}{v_l}^2 $
Therefore, we can cancel the $ \dfrac{1}{2} $ from both the sides of the equation. So we get,
 $\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,
 $\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,
 $\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,
 $\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 $ \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.