QJEEAdv14Phae1e: JEE Advanced 2014 Physics - Class 11 JEEAdv14Phae1e Solution: If the piston is pushed at a speed of 5 mm s 1 , the air comes…

The dynamics of the spray gun follow the principle of conservation of mass or the continuity equation from fluid dynamics. This principle states that the mass flow rate of a fluid entering a system must equal the mass flow rate leaving a system. For incompressible fluids like liquids, or compressed air where changes in density are negligible, this is often simplified to the rate of volume flow or the continuity equation:

$A_1v_1 = A_2v_2$

where

$A_1$ is the cross-sectional area of the piston,

$v_1$ is the speed of the piston,

$A_2$ is the cross-sectional area of the nozzle, and

$v_2$ is the velocity of the fluid (air) leaving the nozzle.

We can use this equation to find the speed of air leaving the nozzle. Given that the piston's speed is 5 mm/s and its radius is 20 mm, we have

$A_1 = \pi (20\, \text{mm})^2 = 400\pi\, \text{mm}^2$

and

$v_1 = 5\, \text{mm/s}$

The nozzle's radius is 1 mm, thus

$A_2 = \pi (1\, \text{mm})^2 = \pi\, \text{mm}^2$

Substituting the values into the continuity equation, we can solve for $v_2$ :

$A_1v_1 = A_2v_2 \rightarrow 400\pi\, \text{mm}^2 \times 5\, \text{mm/s} = \pi\, \text{mm}^2 \times v_2 \rightarrow v_2 = \frac{400\pi\, \text{mm}^2 \times 5\, \text{mm/s}}{\pi\, \text{mm}^2} = 2000\, \text{mm/s}$

Converting this to meters per second, we get

$v_2 = 2000\, \text{mm/s} \times \frac{1\, \text{m}}{1000\, \text{mm}} = 2\, \text{m/s}$

Therefore, the air comes out of the nozzle with a speed of 2 m/s, which corresponds to Option C.