Determine the physical quantities that would be obtained from the products of each of the following dimensions: (i) [M ¹L ³T ¹] [M ²]L ²]; (ii) [M][L ³] [LT 2][L]; (iii) [M][L ²] [T 2]

We are given the product of dimensions $[M^{-1}L^3T^{-1}][M^2L^{-2}]$.
To simplify, we add the exponents for each base dimension (M, L, T):
For M: $-1 + 2 = 1$
For L: $3 + (-2) = 1$
For T: $-1$
The resulting dimension is $[MLT^{-1}]$. This is the dimension for momentum (mass $\times$ velocity).

We are given the product of dimensions $[M][L^{-3}][T^{-2}][L]$.
To simplify, we add the exponents for each base dimension (M, L, T):
For M: 1
For L: $-3 + 1 = -2$
For T: $-2$
The resulting dimension is $[ML^{-2}T^{-2}]$. This is the dimension for force density (force per unit volume). Force has dimensions $[MLT^{-2}]$ and volume has dimensions $[L^3]$, so force density is $\frac{[MLT^{-2}]}{[L^3]} = [ML^{-2}T^{-2}]$.

We are given the product of dimensions $[M][L^2][T^{-2}]$.
To simplify, we add the exponents for each base dimension (M, L, T):
For M: 1
For L: 2
For T: $-2$
The resulting dimension is $[ML^2T^{-2}]$. This is the dimension for energy or work (e.g., force $\times$ distance). Force has dimensions $[MLT^{-2}]$ and distance has dimensions $[L]$, so work is $[MLT^{-2}] \times [L] = [ML^2T^{-2}]$.

The physical quantities are:

• (i) $\boxed{\text{Momentum}}$
• (ii) $\boxed{\text{Force density}}$
• (iii) $\boxed{\text{Energy or Work}}$