=16×10−3 .....(i) Also from equation of continuity =A1v1=A2v2 πr12v1=πr22v2 ∴
v1
v2
=[
r2
r1
]2=
0.04
0.1
=0.4 v1=0.4v2 ....(ii) Substituting this value in Eq. (i) v22−(0.4v2)2=16×10−3 v2=1.38×10−1=0.138ms−1 Rate of flow of glycrine v=A2v2 =πr22v2 =6.93×10−4m3s−1