Given that, pressure difference pA−pB=1500Nm−2 . . . (i) By using Bernoulli's equation, pA+
1
2
ρvA2=pB+
1
2
ρvB2 pA−pB=
1
2
ρ(vB2−vA2) . . . (ii) Density of water, ρ=103kgm−3 Cross-section areas at points A and B, aA=40cm2=40×10−4m2 aB=20cm2=20×10−4m2 By equation of continuity, rate of flow of water through the tube
∆V
∆t
=aAvA=aBvB⇒
vA
vB
=
aB
aA
=
1
2
2vA=vB. . . (iii) Substituting the values from Eqs. (i) and (iii), in Eq. (ii), we get 1500=
1
2
×103[(2vA)2−vA2] 1500=500(4vA2−vA2) vA2=1⇒vA=1m∕s ∴ Rate of flow of water,