The figure shows a crude type of perfume atomizer. When bulb at `A` is compressed, air flows swifty through the tiny `BC` with uniform speed `v`, there by causing a reduced pressure at the position of verticall tube `DE`. The liquid of density `500kg//m^(3)`, then rises in the tube, enters tube `BC` and sprayed out. When bulb is in natural position the air in the bulb and tube are at atmospheric pressure `P_(0=15)^(5)N//m^(2)`. When bulb `A` is compressed, it creates an exces pressure `Deltap=0.001P_(0)` inside the bulb `A`. Density of air is `1.3 kg//m^(3)`. If the magnitude of minimum value of speed `v` required to cause the liquid to rise to tube `BC` is `5km//s`. Find the value of `k`. (`g=10m//s^(2)`)

Answer

4

Explanation

Use Bernoulli's equation at point -3 and point -4, we can write
`(P_(0)+DeltaP)/(rho_("liquid"))+1/2 v_("liquid")^(2)=(P_(2))'(rho_("liquid"))`
`implies (P_(BC)+rho_(l"liquid")gh)/(rho_(l"liquid"))+1/2 v_(l)^(2)=(P_(0))/(rho_(l))`
`implies(P_(0)+DeltaP+rho_(l)gh-0.65v^(2))/(rho_(l))+1/2v_(l)^(2)=(P_(0))/l`
`implies1/2rho_(l)v_(l)^(2)=0.65v^(2)-DeltaP-rho_(l)gh ge 0`
`implies v^(2) ge (DeltaP+rho_(l)gh)/0.65`
`=v_("min")=sqrt((DeltaP+rho_(l)gh)/0.65)=100/5=20m//s`