TS EAMCET 6-Aug-2021 Shift 2 Question Paper

Radius of each circular opening, r1=1mm
=10−3m
∴ Cross-sectional area of each opening,
A1=πr12=π×10−6m2
Radius of the pipe,
r2=2cm=0.02m
∴ Cross-sectional area of pipe,
A2‌‌=πr22
‌‌=π×(0.02)2m2
The speed of water in the pipe,
v2‌‌=25cm∕s
‌‌=0.25m∕s
Let the speed of water when it exits from the circular openings is v1m∕s.
Number of circular openings, n=25
As, the flow rate is always constant.
∴‌‌25A1v1‌‌=A2v2
⇒‌‌v1‌‌=‌

A2

25A1

v2
‌‌=‌

π×(0.0)2

π×10−6

×‌

0.25

25

=4m∕s