Here, cross-section of the tube, a1=8.0 cm2=8.0×10−4 m2
The speed of liquid in the tube, v1=1.5m min−1=

1.5

60

m s−1
=0.025 m s−1
diameter of a hole, D=1.0 mm=10−3 m
Therefore, cross-section of a hole,

πD2

4

=

π

4

×(10−3 )2
=

π

4

×10−6 m2
Therefore, total cross-section of 40 holes, a2=

π

4

×10−6×40 m2
If v2 is the speed of ejection of the liquid through the holes, then a1v1=a2v2
or v2=

a1v1

a2

=

8.0×10−4×0.025

( π 4 )×10−6×40

=0.637 ms−1.