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Dimensional Analysis

The dimension...

Question

The dimensional formula for coefficient of kinematic viscosity is :

A

M^{0} L^{- 1} T^{- 1}

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B

M^{0} L^{2} T^{- 1}

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C

{M L}^{2} T^{- 1}

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D

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

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Solution

The correct option is A $M^{0} L^{2} T^{- 1}$
Kinematic viscocity $= \frac{c o e f f o f v i s c o c i t y}{d e n s i t y}$
$= \frac{M L^{- 1} T^{- 1}}{M L^{- 3}}$

$= M^{0} L^{2} T^{- 1}$

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Similar questions

Q. Write the SI unit of the physical quantity having following dimensional formula

[M^{0} L^{2} T^{- 2} K^{- 1}]

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Q. The dimensional formula for viscosity is

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

. One poise, i.e, C.G.S. unit of viscosity is:

Q. Match the following physical quantities with their respective dimensional formula:

(a) Angular Momentum	(e) $[{M L}^{2} T^{- 3}]$
(b) Impulse	(f) $[{M L}^{2} T^{- 1}]$
(c) Pressure	(g) $[M L T^{- 1}]$
(d) Power	(h) $[{M L}^{- 1} T^{- 2}]$

Q. Match List I with II and select the correct answer:

List-I	List-II
A.Spring constant	1. $M^{1} L^{2} T^{- 2}$
B.pascal	2. $M^{0} L^{0} T^{- 1}$
C.hertz	3. $M^{1} L^{0} T^{- 2}$
D.joule	4. $M^{1} L^{- 1} T^{- 2}$

Q. Names of units of some physical quantities are given in List - I and their dimensional formulae are given in List - II. Match the correct pair of the lists.

List - I	List - II
a) Pa-s	e) $[L^{2} T^{- 2} K^{- 1}]$
b) ${N m K}^{- 1}$	f) ${M L T}^{- 3} K^{- 1}$
c) ${J k g}^{- 1} k^{- 1}$	g) ${M L}^{- 1} T^{- 1}$
d) ${W m}^{- 1} k^{- 1}$	h) $[{M L}^{2} T^{- 2} K^{- 1}]$

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