Question

If momentum [P], area [A] and time [T] are taken as fundamental quantities, then the dimensional formula for the coefficient of viscosity is

• A. [PA^–1T⁰]
• B. [PAT^–1]
• C. [PA^–1T]
• D. [PA^–1T^–1]

Answer 1

The Correct Option is A

 To find the dimensional formula for the coefficient of viscosity using momentum [P], area [A], and time [T] as fundamental quantities, we need to understand the relationship between these quantities and viscosity.

Viscosity (\(\eta\)) is defined as the measure of a fluid's resistance to deform under shear stress. The dimensional formula of viscosity is usually represented as \([ML^{-1}T^{-1}]\), where:

• \(M\) represents mass
• \(L\) represents length
• \(T\) represents time

Given that momentum [P], area [A], and time [T] are taken as fundamental quantities, let's express mass and length in terms of the given quantities:

1. Expressing Mass [M]
Momentum is defined as the product of mass and velocity, i.e., \(P = MV\). The dimensional formula for momentum is \([MLT^{-1}]\). If velocity \((V)\) is M = \frac{P}{V} = [P][L^{-1}T]

2. Expressing Length [L]
Area is given by length squared, i.e., \(A = L^2\). Hence, the dimensional formula for length is:

\(L = A^{\frac{1}{2}}\)

3. Substituting these into Viscosity
We know the dimensional formula of viscosity is \([ML^{-1}T^{-1}]\). Substituting \(M\) and \(L\) into this gives:

\(\eta = \left[\frac{P}{L^{-1}T}\right]\left[A^{-\frac{1}{2}}\right][T^{-1}]\)

Simplifying, this becomes:

\(\eta = [P][A^{-\frac{1}{2}}][T][A^{-\frac{1}{2}}][T^{-1}]\)

\(\eta = [P][A^{-1}][T^0]\)

Thus, the dimensional formula for the coefficient of viscosity is \([PA^{-1}T^0]\).

Therefore, the correct option is: [PA^–1T⁰]