The piston and nozzle of a syringe kept horizontal have diameters 5 mm and 1 mm. the piston is pushed with constant velocity of` 0.2ms^(-2)` Find the horizontal distance travelled by water jet before touchig water`[g=10m//s^(2)]`. Height of syringe from ground is 1 m.

Explanation

`d_(1)=5mmthereforer_(1)=2.5mm=2.5xx10^(-3)m`
`d_(2)=1mmthereforer_(2)=0.5mm=0.5xx10^(-3)m`
If velocity of water coming out of a syringe is `v_(2)` , according to continuity equation ,
`A_(1)v_(1)=A_(2)v_(2)`
`thereforev_(2)=((A_(1))/(A_(2)))v_(1)`
`=((pir_(1)^(2))/(pir_(2)^(2)))v_(1)=((r_(1))/(r_(2)))^(2)v_(1)`
`=((2.5)/(0.5))^(2)xx0.2(becausev_(1)=0.2m//s))`
`thereforev_(2)=5m//s`.
Time taken by water jet when it touches the ground using.
`d=(1)/(2)g t^(2)`
`1=(1)/(2)xx10t^(2)(d=1.0m)`
`thereforet=(1)/(sqrt(5))s`
`therefore` Horizontal distance covered by water jet in time t is ,
`x=v_(2)t=5xx(1)/(sqrt(5))`
`thereforex=sqrt(5)m`.
`thereforex=2.23m`.