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Which of the following quantities are dimensionless? (symbols have their usual meaning)
(A) $\dfrac{{{\text{I}}{\omega ^2}}}{{mvr}}$
(B) $\dfrac{{G\rho }}{T}$
(C) $\dfrac{{\rho vr}}{\eta }$
(D) $\dfrac{{\tau \theta }}{{I\omega }}$
[Useful relation $I = \dfrac{2}{5}m{r^2}$,$F = 6\pi \eta rv$ ]

Last updated date: 20th Jun 2024
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Hint:-For finding the dimensionless quantity, we will check option by option putting unit of symbols in it. After getting the unit, we convert it into dimension to check dimensionless quantity.

Complete step-by-step solution:
Dimensionless quantity is defined as physical quantity, which has no dimension. They are also called unit less quantity.
First we check option (A) $\dfrac{{I{\omega ^2}}}{{mvr}}$
Where $I$ is moment of inertia, $\omega $ is angular speed, \[m\] is mass, $v$ is velocity and $r$ is perpendicular distance.
We know that $I = M{R^2}$ , therefore unit of moment of inertia \[I\] is kilogram per meter square $\left( {kg{m^2}} \right)$,
Unit of angular speed $\omega $ is radian per second but radian is measurement of angle so, unit is ${s^{ - 1}}$ ,
Unit of velocity $v$ is meter per second $\left( {m{s^{ - 1}}} \right)$ and unit of perpendicular distance $r$ is meter $\left( m \right)$ .
Putting all the units in the given expression, we have
$ \Rightarrow \dfrac{{kg{m^2} \times {s^{ - 1}}}}{{kg \times m{s^{ - 1}} \times m}}$
After simplify, we get
$ \Rightarrow \dfrac{{kg{m^2}{s^{ - 1}}}}{{kg{m^2}{s^{ - 1}}}}$
Since, numerator and denominator both are equal. Therefore, a given expression has no unit or says it is dimensionless quantity.
Option (B) $\dfrac{{G\rho }}{T}$
Where $G$is gravitational constant, we derive unit of gravitational constant from the force formula between two masses $F = G\dfrac{{{m^2}}}{{{r^2}}}$ $ \Rightarrow G = \dfrac{{F{r^2}}}{{{m^2}}}$
Putting unit of force $F$ ; kilogram meter per second square $\left( {kgm{s^{ - 2}}} \right)$ , unit of distance $r$ ; meter $\left( m \right)$ , unit of mass $m$ ; kilogram $\left( {kg} \right)$ on above formula we get unit of gravitational constant $G$ .
$ \Rightarrow G = \dfrac{{kgm{s^{ - 2}} \times {m^2}}}{{k{g^2}}}$
After simplify, we get unit of gravitational constant $G = \dfrac{{{m^3}{s^{ - 2}}}}{{kg}}$
Now, unit of density $\rho $ is kilogram per meter cube $\left( {kg{m^{ - 3}}} \right)$ and unit of time $T$ is second $\left( s \right)$
Putting units of $G$, $\rho $ and $T$ on the expression, we get
$ \Rightarrow \dfrac{{{m^3}{s^{ - 2}} \times kg{m^{ - 3}}}}{{kg \times s}}$
After simplify we get,
$ \Rightarrow {s^{ - 3}}$
Since, it has units, therefore it is not a dimensionless quantity.
Now option (C) $\dfrac{{\rho vr}}{\eta }$
We know the unit of density $\rho $ , velocity $v$and distance $r$. So we only derive the unit of viscosity coefficient $\eta $ using the formula $F = 6\pi \eta rv$ . We get after simplification
$ \Rightarrow \eta = \dfrac{F}{{6\pi rv}}$
Since $6\pi $ has no unit. So, we put unit of Force $F$ , distance $r$and velocity $v$on above equation and we have ,
$ \Rightarrow \eta = \dfrac{{kgm{s^{ - 2}}}}{{m \times m{s^{ - 1}}}}$
After simplify, unit of viscosity coefficient $\eta = \dfrac{{kg{s^{ - 1}}}}{m}$
Putting units of $\eta ,\rho ,v,r$ on expression and we get
$ \Rightarrow \dfrac{{kg{m^{ - 3}} \times m \times m{s^{ - 1}}}}{{\dfrac{{kg{s^{ - 1}}}}{m}}}$
After simplify we get,
$ \Rightarrow \dfrac{{kg{s^{ - 1}}}}{{kg{s^{ - 1}}}}$
Since, the numerator and denominator are the same. It is unit less or say it is dimensionless quantity.
Now, option (D) $\dfrac{{\tau \theta }}{{I\omega }}$
We know that the unit of $I$ and $\omega $ . $\theta $ is angle and it has no unit. So, we derive the unit of torque $\tau $using formula $\tau = Fr$ .
Put unit of force $F$and distance $r$ on above formula and get unit of torque $\tau = kgm{s^{ - 2}} \times m$
After simplifying the unit of torque $\tau = kg{m^2}{s^{ - 2}}$ .
Putting units of $\tau ,\omega ,I,\theta $ on the given expression we get,
$ \Rightarrow \dfrac{{kg{m^2}{s^{ - 2}}}}{{kg{m^2}{s^{ - 1}}}}$
After simplification we have,
$ \Rightarrow {s^{ - 1}}$
Since it has units. Therefore, it is not a dimensionless quantity.

Hence, option (A) and (C) are correct.

Dimensions are represented within the closed brackets and it has no magnitude. But units are represented without any brackets. Dimensions are represented in the form of M, L and T within the closed bracket. For example, the dimension of force is $\left[ {ML{T^{ - 2}}} \right]$.