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: Understanding the Concept:
This problem requires dimensional analysis. According to the principle of homogeneity of dimensions, all terms in a physical equation must have the same dimensions. We can use this principle to find the dimensions of an unknown quantity.
Step 2: Key Formula or Approach:
1. Write down the dimensions of the known quantities in the equation.
2. Rearrange the equation to solve for the unknown quantity, X.
3. Substitute the dimensions and simplify to find the dimensions of X.
Step 3: Detailed Explanation:
The given equation is \( F = 6\pi\eta X \). The term \( 6\pi \) is a dimensionless constant.
So, dimensionally, the equation is \( [F] = [\eta][X] \).
First, let's establish the dimensions of the known quantities:
- Force (F): From Newton's second law (\( F=ma \)), the dimension is mass times acceleration.
\[ [F] = M \cdot LT^{-2} = MLT^{-2} \] - Coefficient of Viscosity (\( \eta \)): The formula for viscous force is \( F = \eta A \frac{dv}{dx} \), where A is area, dv is change in velocity, and dx is distance.
\[ [\eta] = \frac{[F][dx]}{[A][dv]} = \frac{(MLT^{-2})(L)}{(L^2)(LT^{-1})} = \frac{ML^2T^{-2}}{L^3T^{-1}} = ML^{-1}T^{-1} \] Now, we rearrange the given equation to solve for the dimensions of X:
\[ [X] = \frac{[F]}{[\eta]} \] Substitute the dimensions we found:
\[ [X] = \frac{MLT^{-2}}{ML^{-1}T^{-1}} \] Simplify the expression by applying the rules of exponents:
\[ [X] = M^{1-1} L^{1-(-1)} T^{-2-(-1)} \] \[ [X] = M^0 L^{1+1} T^{-2+1} \] \[ [X] = M^0 L^2 T^{-1} \] Step 4: Final Answer:
The dimension of X is \( M^0 L^2 T^{-1} \).