
Which of the following are dimensionless quantities, (symbols have their usual meaning)
[ $ \eta = $ viscosity, $ \rho = $ density, r=radius, k=thermal conductivity, c=heat capacity.]
(A) $ \dfrac{k}{{\rho \times c}} $
(B) $ \dfrac{{\rho \times v \times r}}{\eta } $
(C) Specific gravity
(D) Rate of change of angle (in radians) of rotation.
Answer
584.4k+ views
Hint
We will first write the dimensional formulae of all the quantities given in the question. Now, substitute the dimensional formulae in the given relations. If on solving, all the values get cancelled, then the quantity is dimensionless.
Complete step by step answer
The dimensional formulae of the given quantities are:
$\Rightarrow k = {M^1}{L^{ - 1}}{T^{ - 3}}{\theta ^{ - 1}} $
$\Rightarrow \eta = M{L^{ - 1}}{T^{ - 1}} $
$\Rightarrow c = {L^2}M{T^{ - 2}}{\theta ^{ - 1}} $
$\Rightarrow \rho = M{L^{ - 3}} $
$\Rightarrow r = L $
$\Rightarrow v = L{T^{ - 1}} $
Now, let’s check each option by putting the dimensional values in the given relations.
- $\Rightarrow \dfrac{k}{{\rho \times c}} = \dfrac{{M{L^{ - 3}}T{\theta ^{ - 1}}}}{{M{L^{ - 3}} \times {L^2}M{T^{ - 2}}{\theta ^{ - 1}}}} = M{L^{ - 2}}T $ (So, this relation has a dimension)
- $\Rightarrow \dfrac{{\rho \times v \times r}}{\eta } = \dfrac{{M{L^{ - 3}} \times L{T^{ - 1}} \times L}}{{M{L^{ - 1}}{T^{ - 1}}}} = {M^0}{L^0}{T^0} $ (Hence, it is a dimensionless quantity)
- Specific gravity is the ratio of the density of a substance to the density of a given reference material. Since, it is a ratio of two quantities, it is dimensionless.
- The rate of change of angles while a body is rotating is known as its angular velocity. It has a dimensional formula.
Therefore, option (B) and (C) are correct.
Note
It is feasible to write the dimensional formulae of all the quantities given in the question. This will reduce the chances of a mistake to occur. Also, it’s better to memorize the common dimensional formulae.
There are many applications of dimensional analysis, such as:
- To find the unit of a physical quantity in a given system of units.
- To convert a physical quantity from one system to another.
- To check the dimensional correctness of a given physical relation.
We will first write the dimensional formulae of all the quantities given in the question. Now, substitute the dimensional formulae in the given relations. If on solving, all the values get cancelled, then the quantity is dimensionless.
Complete step by step answer
The dimensional formulae of the given quantities are:
$\Rightarrow k = {M^1}{L^{ - 1}}{T^{ - 3}}{\theta ^{ - 1}} $
$\Rightarrow \eta = M{L^{ - 1}}{T^{ - 1}} $
$\Rightarrow c = {L^2}M{T^{ - 2}}{\theta ^{ - 1}} $
$\Rightarrow \rho = M{L^{ - 3}} $
$\Rightarrow r = L $
$\Rightarrow v = L{T^{ - 1}} $
Now, let’s check each option by putting the dimensional values in the given relations.
- $\Rightarrow \dfrac{k}{{\rho \times c}} = \dfrac{{M{L^{ - 3}}T{\theta ^{ - 1}}}}{{M{L^{ - 3}} \times {L^2}M{T^{ - 2}}{\theta ^{ - 1}}}} = M{L^{ - 2}}T $ (So, this relation has a dimension)
- $\Rightarrow \dfrac{{\rho \times v \times r}}{\eta } = \dfrac{{M{L^{ - 3}} \times L{T^{ - 1}} \times L}}{{M{L^{ - 1}}{T^{ - 1}}}} = {M^0}{L^0}{T^0} $ (Hence, it is a dimensionless quantity)
- Specific gravity is the ratio of the density of a substance to the density of a given reference material. Since, it is a ratio of two quantities, it is dimensionless.
- The rate of change of angles while a body is rotating is known as its angular velocity. It has a dimensional formula.
Therefore, option (B) and (C) are correct.
Note
It is feasible to write the dimensional formulae of all the quantities given in the question. This will reduce the chances of a mistake to occur. Also, it’s better to memorize the common dimensional formulae.
There are many applications of dimensional analysis, such as:
- To find the unit of a physical quantity in a given system of units.
- To convert a physical quantity from one system to another.
- To check the dimensional correctness of a given physical relation.
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