Consider a horizontally oriented syringe containing water locate at a height of `1.25 m` above the ground. The diameter of the plunger is `8 mm` and the diameter of the nozzle is `2 mm`. The plunger is pushed with a constant speed of `0.25m//s.` Find the horizontal range of water stream on the grond. Take `g=10 m//s^(2)`

Answer

2

Explanation

From equation of continuity (Av = constant )
` (pi )/(4 ) (8 ) ^ 2 (0.25) = (pi )/(4) (2 ) ^2 (v) " " `…(i)
Here, v is the velocity of water with which water comes out of the syringe (Horizontally) Solving equation (i), we get ` v = 4m//s `
The path of water after leaving the syringe will be a parabola.
Range , ` R = ut = u sqrt ((2h )/(g)) rArr R = 4 sqrt ((2 xx 1.25 )/(10 )) rArr R = (4)/(10 ) xx 5 = 2 m `