TS EAMCET 6-Aug-2021 Shift 1 Question Paper

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,
‌

∆V

∆t

=aAvA=40×10−4×1m3∕s
=4000cm3∕s