Manipal Entrance Test 2013 Physics Paper

From Bernoulli’s theorem
p1+12ρv12=p2+

1

α

ρv22
∴ p1−p2=

1

2

ρ(v22−v12)
∴ 10=

1

2

×1.25×103(v22−v12)
∴ v22−v12=10×

2

1.25×103

=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