Question

The dimension of $X$ in the equation $F=6\pi\eta X$ is: (F - Force; $\eta$ - Coefficient of viscosity)

• A. $M^0 L^2 T^{-1}$
• B. $ML^2 T^{-2}$
• C. $M^0 L^2 T^{-2}$
• D. $M^0 L^3 T^{-2}$
• E. $ML^2 T^{-1}$

Hint:

$6\pi$ is a dimensionless constant and does not affect the dimensional analysis.

Answer 1

The Correct Option is A

Step 1: Concept
Use Stokes' Law $F = 6\pi\eta r v$. Here $X$ represents $r \times v$.

Step 2: Analysis
Dimensions of $F = [MLT^{-2}]$. Dimensions of $\eta = [ML^{-1}T^{-1}]$. $[X] = \frac{[F]}{[\eta]} = \frac{[MLT^{-2}]}{[ML^{-1}T^{-1}]} = [L^2 T^{-1}]$.

Step 3: Conclusion
In $M^0$ notation, the dimensions are $M^0 L^2 T^{-1}$. Final Answer: (A)