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Which of the following quantities are dimensionless? (symbols have their usual meaning)
(A) Iω2mvr
(B) GρT
(C) ρvrη
(D) τθIω
[Useful relation I=25mr2,F=6πηrv ]

Answer
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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) Iω2mvr
Where I is moment of inertia, ω is angular speed, m is mass, v is velocity and r is perpendicular distance.
We know that I=MR2 , therefore unit of moment of inertia I is kilogram per meter square (kgm2),
Unit of angular speed ω is radian per second but radian is measurement of angle so, unit is s1 ,
Unit of velocity v is meter per second (ms1) and unit of perpendicular distance r is meter (m) .
Putting all the units in the given expression, we have
kgm2×s1kg×ms1×m
After simplify, we get
kgm2s1kgm2s1
Since, numerator and denominator both are equal. Therefore, a given expression has no unit or says it is dimensionless quantity.
Option (B) GρT
Where Gis gravitational constant, we derive unit of gravitational constant from the force formula between two masses F=Gm2r2 G=Fr2m2
Putting unit of force F ; kilogram meter per second square (kgms2) , unit of distance r ; meter (m) , unit of mass m ; kilogram (kg) on above formula we get unit of gravitational constant G .
G=kgms2×m2kg2
After simplify, we get unit of gravitational constant G=m3s2kg
Now, unit of density ρ is kilogram per meter cube (kgm3) and unit of time T is second (s)
Putting units of G, ρ and T on the expression, we get
m3s2×kgm3kg×s
After simplify we get,
s3
Since, it has units, therefore it is not a dimensionless quantity.
Now option (C) ρvrη
We know the unit of density ρ , velocity vand distance r. So we only derive the unit of viscosity coefficient η using the formula F=6πηrv . We get after simplification
η=F6πrv
Since 6π has no unit. So, we put unit of Force F , distance rand velocity von above equation and we have ,
η=kgms2m×ms1
After simplify, unit of viscosity coefficient η=kgs1m
Putting units of η,ρ,v,r on expression and we get
kgm3×m×ms1kgs1m
After simplify we get,
kgs1kgs1
Since, the numerator and denominator are the same. It is unit less or say it is dimensionless quantity.
Now, option (D) τθIω
We know that the unit of I and ω . θ is angle and it has no unit. So, we derive the unit of torque τusing formula τ=Fr .
Put unit of force Fand distance r on above formula and get unit of torque τ=kgms2×m
After simplifying the unit of torque τ=kgm2s2 .
Putting units of τ,ω,I,θ on the given expression we get,
kgm2s2kgm2s1
After simplification we have,
s1
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 [MLT2].
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