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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.**

**Note:-**

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 $\left[ {ML{T^{ - 2}}} \right]$.

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