Question

If $ F = 6\pi \eta x $, find the dimension of $ x $.

• A. \( [M^1 L^1 T^{-1}] \)
• B. \( [M^0 L^0 T^1] \)
• C. \( [M^1 L^2 T^{-2}] \)
• D. \( [M^0 L^1 T^0] \)

Hint:

In dimensional analysis, ensure that the powers of \( M \), \( L \), and \( T \) balance correctly on both sides of the equation.

Answer 1

The Correct Option is A

We are given the equation: \[ F = 6\pi \eta x \] where \( F \) is force, \( \eta \) is viscosity, and \( x \) is the quantity we need to find the dimensions for. We know the dimensional formula for force \( F \) is: \[ [F] = [M L T^{-2}] \] The dimensional formula for viscosity \( \eta \) is: \[ [\eta] = [M L^{-1} T^{-1}] \] Now, let the dimensional formula of \( x \) be \( [x] = [M^a L^b T^c] \). 
Using the given equation: \[ [M L T^{-2}] = [M L^{-1} T^{-1}] [M^a L^b T^c] \] Equating the powers of \( M \), \( L \), and \( T \) on both sides, we get the following system of equations: \[ M: 1 = 1 + a \] \[ L: 1 = -1 + b \] \[ T: -2 = -1 + c \] Solving these equations: - From \( M: 1 = 1 + a \), we get \( a = 0 \) 
- From \( L: 1 = -1 + b \), we get \( b = 2 \) 
- From \( T: -2 = -1 + c \), we get \( c = -1 \) 
Therefore, the dimensional formula for \( x \) is: \[ [x] = [M^1 L^1 T^{-1}] \] 
Thus, the correct answer is \( [M^1 L^1 T^{-1]} \).