Question

Given the equation → Angular Momentum = \(\left( α β - \frac{γ}{\rm time} \right)\) × velocity which of the following are possible dimensions of α, β and γ?
1. M, L, T
2. M, L, MLT
3. ML, ML, MLT
4. M²L^-2, ML, T

Answer 1

Correct Answer - Option 2 : M, L, MLT

Concept:

• Angular momentum (L): It is a vector quantity that requires both a magnitude and a direction.
• The magnitude of the angular momentum is equal to its linear momentum and perpendicular distance r from the center of rotation to a line.

​L = p × r

Where p is linear momentum and r is the radius. 

• Unit of momentum is kg m /s (mass × velocity). 
• So, the unit of angular momentum is Kg m/s × m = kg m ² s ^-1
• So, the dimension is M L ² T ^-1
• For an equation, it is important that the dimension on the left-hand side is equal to the dimension on the right-hand side. 

Calculation:

Given, Angular Momentum (L) is given as

\(AM = \left( α β - \frac{γ}{\rm time} \right) \times velocity\)

\(\implies AM = \left( α β \times velocity- \frac{γ}{\rm time} \times velocity\right) \)

[AM],  [αβvelocity] and [(γvelocity) / time] has same dimension and that is the dimension M L 2 T -1. 

Dimension of velocity = L T ^-1

So, αβ L T ^{- 1} is dimensionally equal to  M L 2 T -1

[αβ L T ^{- 1}] = [M L 2 T -1]

⇒ αβ L = M L ²

⇒ αβ = M L

So either of αβ can be M or L. 

[(γvelocity) / time]  = [M L 2 T -1]

⇒[ γ L T ^{- 1} / T] = [M L 2 T -1]

⇒ [ γ L T ^{- 2} ] = [M L 2 T -1]

⇒ γ = [ MLT]

So, if we consider α = M, β = L, γ = MLT, equations will become dimensionally correct.