MyQuestionIcon

2

You visited us 2 times! Enjoying our articles? Unlock Full Access!

Density

10.0 cm from ...

Question

$10.0$ cm from the centre of the pipe (half way between the centre and the walls)

Open in App

Solution

$v = \frac{P_{1} - P_{2}}{4 η L} (R^{2} - r^{2})$
or $v \propto R^{2} - r^{2}$
$\frac{v_{1}}{v_{2}} = \frac{R^{2} - r_{2}^{2}}{R^{2} - r_{2}^{2}}$
or $\frac{3}{v_{2}} = \frac{400 - 0}{400 - 100}$
$∴$ $v_{1} = 2.25$ m/s

Suggest Corrections

thumbs-up

0

Similar questions

Q. Hoop tension in a pressure pipe (where 'p' is the pressure at the pipe centre, 'd' is the diameter of pipe and 't' is the thickness of pipe wall) is given by

15 m

ladder weighing

50 k g

rest against a smooth wall at a point

12 m

above the ground. The centre of gravity of the ladder is one third the way up. A

80

kg man climbs half way up the ladder. Find the force exerted by the system on the ground and the wall.

Q. In fully developed laminar flow in a circular pipe, if the velocity at midway between the wall surface and the centerline is measured to be 6 m/s, then the velocity at the centre of pipe is

Q. A horizontal plane mirror and a vertical wall are placed in such a way that the centre of the horizontal plane mirror is at a distance

d

from the vertical wall. The mirror is rotated with

ω

in clockwise direction.

Incident light falls at centre

C

as shown in the figure, then linear velocity of spot

(S)

formed on wall is

Q. From a uniform circular disc of radius

20 cm

, two small circular discs of radius

10 cm

are removed in such a way that both have common tangent as shown in the figure given below. The distance of centre of mass of the remaining part from the centre of original disc is:

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Density

CHEMISTRY

Watch in App

explore_more_icon

Explore more

Standard XII Chemistry

Join BYJU'S Learning Program

CrossIcon