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NEET PHYSICSEasy

The force FF on a sphere of radius ‘aa’ moving in a medium with velocity ‘vv’ is given by F=6πηavF = 6\pi\eta av. The dimensions of η\eta are:

A

ML−1T−1ML^{-1}T^{-1}

B

MT−1MT^{-1}

C

MLT−2MLT^{-2}

D

ML−3ML^{-3}

Step-by-Step Solution

To find the dimensions of the coefficient of viscosity (η\eta), we use the provided formula for viscous force (Stokes' Law): F=6πηavF = 6\pi\eta av. Rearranging the formula for η\eta, we get η=F/(6πav)\eta = F / (6\pi av). According to the sources, the dimensional formulae for the relevant quantities are:

  1. Force (FF): [MLT−2][MLT^{-2}] .
  2. Radius (aa): As a measure of length, its dimension is [L][L] .
  3. Velocity (vv): [LT−1][LT^{-1}] .
  4. 6π6\pi: This is a numerical constant and is dimensionless .

Substituting these into the equation for η\eta: [η]=[MLT−2][L][LT−1]=[MLT−2][L2T−1]=[ML−1T−1][\eta] = \frac{[MLT^{-2}]}{[L][LT^{-1}]} = \frac{[MLT^{-2}]}{[L^{2}T^{-1}]} = [ML^{-1}T^{-1}].

This dimensional formula corresponds to Option A .

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