PhysicsQ1–1 of 1 questions
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1
PhysicsMediumMCQAIIMS · 1993
Dimensions of the coefficient of viscosity are
A
$M L^2 T^{-2}$
B
$M L^2 T^{-1}$
C
$M L^{-1} T^{-1}$
D
$M L T$
Solution
(C) The viscous force $F$ acting on a fluid layer is given by Newton's law of viscosity: $F = -\eta A \frac{dv}{dx}$.
Here,$F$ is the force,$\eta$ is the coefficient of viscosity,$A$ is the area,and $\frac{dv}{dx}$ is the velocity gradient.
Rearranging for $\eta$: $\eta = \frac{F}{A (dv/dx)}$.
Substituting the dimensions: $[F] = [M L T^{-2}]$,$[A] = [L^2]$,$[dv] = [L T^{-1}]$,and $[dx] = [L]$.
Thus,$[\eta] = \frac{[M L T^{-2}]}{[L^2] [L T^{-1} / L]} = \frac{[M L T^{-2}]}{[L^2] [T^{-1}]} = [M L^{-1} T^{-1}]$.
Therefore,the correct option is $C$.
Here,$F$ is the force,$\eta$ is the coefficient of viscosity,$A$ is the area,and $\frac{dv}{dx}$ is the velocity gradient.
Rearranging for $\eta$: $\eta = \frac{F}{A (dv/dx)}$.
Substituting the dimensions: $[F] = [M L T^{-2}]$,$[A] = [L^2]$,$[dv] = [L T^{-1}]$,and $[dx] = [L]$.
Thus,$[\eta] = \frac{[M L T^{-2}]}{[L^2] [L T^{-1} / L]} = \frac{[M L T^{-2}]}{[L^2] [T^{-1}]} = [M L^{-1} T^{-1}]$.
Therefore,the correct option is $C$.
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