wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Water flows through the tube shown in figure. The areas of cross section of the wide and the narrow portions of the tube are 5 cm2 and 2 cm2 respectively. The rate of flow of water through the tube is 500 cm3 s-1. Find the difference of mercury levels in the U-tube. Assume this tube lying horizontally on a table.


A

1.93 cm

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B

1.93 m

No worries! We‘ve got your back. Try BYJU‘S free classes today!
C

2.58 cm

No worries! We‘ve got your back. Try BYJU‘S free classes today!
D

2.58 m

No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A

1.93 cm


As clear from the sitution,
P1P2=,ρgh
h= difference in height of the mercury column.
Using Bernoulli's equation,
P1+ρgH+12V2=constant
As both the points are at the same horizontal level.
P1+12V21=P2+12V22ρ
P1P2=12[V22V22]ρ
12ρ(V22V21)=,ρgh .....(1)

From equation of continuity,
A1V1=A2V2
5V1=2V2
V2=52V1 ......(2)
also A1V1=A2V2=500cm3s
V1=100cms=1ms......(3)
from (3) &(1)
12ρ{214}=,ρgh
12×1000{214}=13600×10×h
h=0.0193m
h=1.93 cm

flag
Suggest Corrections
thumbs-up
6
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Measurement of Atmospheric Pressure
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon