Question

One end of a cylindrical pipe has a radius of 1.5cm. Water (density)
= 1.0 x 10 kg/m^j which mass is leaving the pipe is:

(a) 2.5kg/s
(c) 48 kg/s
(b) 4.9kg/s
(d) 7.0 x 10^3kg/s​

Answer 1

We know the density,

ρ = m / V

m = ρV

Then the mass rate is,

m / t = ρV / t

m / t = ρ (V / t)

But by equation of continuity,

m / t = aρv

But, the cross sectional area, a = πr². Then,

m / t = πρr^2v

m / t = 3.14v × 10^3 × (1.5 × 10^(-2))^2

m / t = 0.71v kg s^(-1)

If velocity v is given then we can find the actual answer.