A spray gun is shown in the figure where a piston pushes air out of a nozzle. A thin tube of uniform cross section is connected to the nozzle. The other end of the tube is in a small liquid container. As the piston pushes air through the nozzle, the liquid from the container rises into the nozzle and is sprayed out. For the spray gun shown, the radii of the piston and the nozzle are 20mm and 1mm respectively. The upper end of the container is open to the atmosphere.

If the density of air is `rho_a`, and that of the liquid `rho_l`, then for a given piston speed the rate (volume per unit time) at which the liquid is sprayed will be proportional to

Options

A. ` sqrt((rho _a)/(rho_1))`
B. `sqrt(rho_a rho _1) `
C. `sqrt((rho _ 1 )/(rho_a)) `
D. `rho _1`

Answer

A

Explanation

Pressure at the point A in the nozzle is P.
Now, ` P = P _ 0 - (1)/(2) rho _ 0 v _ a ^ 2 " " `…(i)
and als ` P = P _ 0 - (1)/(2) rho _l v _ l ^2 - rho _l gh" " `…(ii)
From equation (i) and (ii)
` (1)/(2) rho _a v_a ^ 2 = (1)/(2) rho _lv _ l ^ 2 + rho _ l gh rArr v _1 = sqrt((rho_a)/(p_l) v _ a ^ 2 - 2 gh ) ` So , For the given ` v _ a and h, v _ 1 prop sqrt((rho_a)/(rho _ l )) `
If a is area of cross-section of thin tube, then rate of flow of liquid will be equal to ` av _ 1 `.
Hence the rate (volume per unit time) at which the liquid is sprayed ` (av_1 ) prop sqrt((rho _a)/(rho_1)) `