Question

Which of the following has the dimension of $M{L}^{0}{T}^{-2}$?
a).Coefficient of viscosity
b).Surface tension
c).Vapour pressure
d).Kinetic energy

Answer 1

Dimensional analysis is defined as the checking relations between the physical quantities by identifying the dimensions of these physical quantities. Each physical quantity can be expressed using 7 basic dimensions.

Complete answer:
Dimensions of the physical quantity can be defined as the dimensions that are the powers to which the fundamental units are raised to obtain one unit of that quantity. There are 7 basic or fundamental quantities in terms of which every physical quantity is written. These are mass (M), length (L), time (T), current (A), temperature (K), luminous intensity (Cd), and amount of substance (mol).
There are 2 laws of homogeneity for dimensions:
In any correct equation representing the relation between physical quantities, the dimensions of all the terms must be the same on both sides.
A physical quantity Q has dimensions a, b and c in length (L), mass (M) and time (T) respectively, and ${n}_{1}$ is its numerical value in a system in which the fundamental units are ${L}_{1}$, ${M}_{1}$ and ${T}_{1}$ and ${n}_{2}$ is the numerical value in another system in which the fundamental units are ${L}_{2}$, ${M}_{2}$ and ${T}_{2}$ respectively, then
${ n }_{ 2 }\quad =\quad { n }_{ 1 }{ [\cfrac { { L }_{ 1 } }{ { L }_{ 2 } } ] }^{ a }{ [\cfrac { { M }_{ 1 } }{ { M }_{ 2 } } ] }^{ b }{ [\cfrac { { T }_{ 1 } }{ { T }_{ 2 } } ] }^{ c }\quad$
Now, let us look at the options for the dimension $M{L}^{0}{T}^{-2}$
Coefficient of viscosity: The coefficient of viscosity is given as:
$Viscosity\quad =\quad \cfrac { viscous\quad force }{ Area\quad \times \quad velocity\quad gradient }$
$Viscosity\quad =\quad \cfrac { force }{ Area\quad \times \quad \cfrac { dv }{ dx } }$
In terms of dimensions, it can be written as:
$Viscosity\quad =\quad \cfrac { { M }^{ 1 }{ L }^{ 1 }{ T }^{ -2 } }{ { L }^{ 2 }\quad \times \quad \cfrac { { L }^{ 1 }{ T }^{ -1 } }{ { L }^{ 1 } } } \quad =\quad { M }^{ 1 }{ L }^{ -1 }{ T }^{ -1 }$
Surface tension: The surface tension is given as:
$Surface\quad tension\quad =\quad \cfrac { Force }{ Unit\quad length }$
In terms of dimension, it can be written as:
$Surface\quad tension\quad =\quad \cfrac { { M }^{ 1 }{ L }^{ 1 }{ T }^{ -2 } }{ { L }^{ 1 } } \quad =\quad { M }^{ 1 }{ L }^{ 0 }{ T }^{ -2 }$
Vapour pressure: The vapour pressure is given as:
$Vapor\quad pressure_{ solution }\quad =\quad Mole\quad fraction\quad \times \quad vapor\quad pressure_{ solvent }$
In terms of dimensions, it is written as:
$Vapor\quad pressure_{ solution }\quad =\quad { M }^{ 1 }{ L }^{ -1 }{ T }^{ -2 }$
Kinetic energy: The kinetic energy is given as:
$Kinetic\quad energy\quad =\quad \cfrac { 1 }{ 2 } mass\quad \times \quad { velocity }^{ 2 }$
In terms of dimensions, it is written as:
$Kinetic\quad energy\quad =\quad M\quad \times \quad { L }^{ 1 }{ T }^{ -1 }\quad =\quad { M }^{ 1 }{ L }^{ 1 }{ T }^{ -1 }$
Therefore, we can see from the above points that the physical quantity with dimensions $M{L}^{0}{T}^{-2}$ is surface tension.

Hence, the correct answer is option (b).

Note:
While writing or deriving any of the dimensions, you need to keep in mind the laws of homogeneity of the dimensions. And while deriving or writing the dimensions, the numerical part in the formula does not play any role as it is dimensionless.