Question

The dimension of specific gravity is:-
A. $\left[ {{M^0}{L^0}{T^0}} \right]$
B. $\left[ {{M^0}L{T^{ - 2}}} \right]$
C. $\left[ {{M^0}L{T^{ - 1}}} \right]$
D. $\left[ {M{L^{ - 3}}{T^0}} \right]$

Answer 1

Hint: The dimensional formula for length and time is $\left[ {{M^0}{L^1}{T^0}{K^0}} \right]$ and $\left[ {{M^0}{L^0}{T^1}{K^0}} \right]$ respectively.
The dimensional formula for velocity $v = \dfrac{d}{t} = \dfrac{{\left[ {{M^0}{L^1}{T^0}{K^0}} \right]}}{{\left[ {{M^0}{L^0}{T^1}{K^0}} \right]}} = \left[ {{M^0}{L^1}{T^{ - 1}}{K^0}} \right]$ . The dimensional formula of specific gravity is the same as acceleration, i.e. $a = \dfrac{v}{t} = \dfrac{{\left[ {{M^0}{L^1}{T^{ - 1}}{K^0}} \right]}}{{\left[ {{M^0}{L^0}{T^1}{K^0}} \right]}} = \left[ {{M^0}{L^1}{T^{ - 2}}{K^0}} \right]$

Complete answer:
The dimension of specific gravity refers to the dimensional formula of specific gravity.
Dimensional formula – Every physical quantity has a unit assigned for it, for example, the force has a unit \[kgm/{s^2}\]. When we express this unit in terms of the fundamental quantities we get the dimensional formula of that physical quantity. The dimensional formula for force is \[[ML{T^{ - 2}}{K^0}]\] .
Here \[M\] represents mass, \[L\] represents the length, \[T\] represents time, and \[K\] represents temperature.
The dimensional formula for specific gravity is the same as the dimensional formula for gravity which is in turn the same as the dimensional formula for acceleration. So, let’s calculate the dimensional formula for acceleration.
We know that the equation for velocity is
$v = \dfrac{d}{t}$
Here, $v = $ Velocity
$d = $ Displacement
$t = $ Time
The Dimensional formula for displacement (same as length) is $\left[ {{M^0}{L^1}{T^0}{K^0}} \right]$ and the dimensional formula for time is $\left[ {{M^0}{L^0}{T^1}{K^0}} \right]$
So, the dimensional formula for velocity $ = \dfrac{{\left[ {{M^0}{L^1}{T^0}{K^0}} \right]}}{{\left[ {{M^0}{L^0}{T^1}{K^0}} \right]}}$
So, the dimensional formula for velocity $ = \left[ {{M^0}{L^1}{T^{ - 1}}{K^0}} \right]$
We know that the equation for acceleration is
$a = \dfrac{v}{t}$
Here, $a = $ Acceleration
$v = $ Velocity
$t = $ Time
Since the dimensional formula for velocity and time is $\left[ {{M^0}{L^1}{T^{ - 1}}{K^0}} \right]$ and $\left[ {{M^0}{L^0}{T^1}{K^0}} \right]$ .
So, the dimensional formula for acceleration $ = \dfrac{{\left[ {{M^0}{L^1}{T^{ - 1}}{K^0}} \right]}}{{\left[ {{M^0}{L^0}{T^1}{K^0}} \right]}}$
So, the dimensional formula for acceleration and hence specific gravity $ = \left[ {{M^0}{L^1}{T^{ - 2}}{K^0}} \right]$

So, the correct answer is “Option B.

Note:
Dimensional formula is unique, whereas the unit can be measured in the SI unit system, metric system, etc. So, for example, the unit for specific gravity can be $m/{s^2}$ in the SI unit whereas the specific gravity is $cm/{s^2}$ in centimeter-scale, but the only dimensional formula for specific gravity is $\left[ {{M^0}{L^1}{T^{ - 2}}{K^0}} \right]$ .