Turpentine oil is flowing through a capillary tube of length l and radius r- The pressure difference between the two ends of the tube is p- The viscosity of oil is given by- -pr2-x24vl- Here v is velocity of oil at a distance x from the axis of the tube- From this relation- the dimensional formula of - is

The correct option is D [ML^ 1T^ 1] [η ]= [p] [ r ^ 2 x ^ 2 ] [v] ^ 1 [l] ^ 1 #160; #160; #160; = M L ^ 1 T ^ 2 L ^ 2 L ^ ...

You visited us 3 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Dimensional Analysis

Turpentine oi...

Question

Turpentine oil is flowing through a capillary tube of length $l$ and radius $r$ . The pressure difference between the two ends of the tube is $p$ . The viscosity of oil is given by: $η = \frac{p (r^{2} - x^{2})}{4 v l}$ . Here v is velocity of oil at a distance $x$ from the axis of the tube. From this relation, the dimensional formula of $η$ is

[M L^{- 1} T^{- 1}]

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

[M L T^{- 1}]

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

[M L^{2} T^{- 2}]

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

[M^{0} L^{0} T^{0}]

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

Open in App

Solution

The correct option is D $[M L^{- 1} T^{- 1}]$
$[η] = [p] [r^{2} - x^{2}] {[v]}^{- 1} {[l]}^{- 1}$
$= M L^{- 1} T^{- 2} L^{2} L^{- 1} T L^{- 1}$
$= M L^{- 1} T^{- 1}$

Suggest Corrections

Similar questions

Q. Turpentine oil is flowing through a tube of length

^{'} l^{'}

and radius

^{'} r^{'}

. The pressure difference between the two ends of tube is

P

. The viscosity of oil is given by

η = \frac{P (r^{2} - x^{2})}{4 v l}

, where

v

is velocity of oil at a radial distance '

x

' from the axis of tube. The dimensions of

η

are

Q. Turpentine oil is flowing through a tube of length l and radius r. The pressure difference between the two ends of the tube is P, the viscosity of the oil is given by

η = \frac{p (r^{2} + x^{2})}{4 v l}

, where

v

is the velocity of oil at a distance x from the axis of the tube. From this relation,the dimensions of viscosity

η

is:

Q. Turpentine oil is flowing through a tube of length l and radius r. The pressure difference between the two ends of the tube is p. The viscosity of oil is given

η = \frac{p (r^{2} - x^{2})}{4 v l}

, where v is velocity of oil at a distance x from the axis of the tube. The dimensional formula of viscosity is

Q. Turpentine oil is flowing through a tube of length

^{'} l^{'}

and radius

^{'} r^{'}

. The pressure difference between the two ends of the tube is

P

. The viscosity of oil is given by

η = \frac{P (r^{2} - x^{2})}{4 v l}

, where

v

is the velocity of oil at a radial distance '

x

' from the axis of the tube. The dimensions of

η

are

Q. A dimensionally consistent relation for the volume

V

of a liquid of coefficient of viscosity

^{'} η^{'}

flowing per second, through a tube of radius

r

and length

l

and having a pressure difference

P

across its ends, is :

Join BYJU'S Learning Program

The correct option is D [ML−1T−1][η]=[p][r2−x2][v]−1[l]−1 =ML−1T−2L2L−1TL−1 =ML−1T−1

The correct option is D $[M L^{- 1} T^{- 1}]$
$[η] = [p] [r^{2} - x^{2}] {[v]}^{- 1} {[l]}^{- 1}$
$= M L^{- 1} T^{- 2} L^{2} L^{- 1} T L^{- 1}$
$= M L^{- 1} T^{- 1}$