In dimension of critical velocity `v_(c)` liquid following through a take are expressed as `(eta^(x) rho^(y) r^(z))` where `eta, rhoand r `are the coefficient of viscosity of liquid, density of liquid and radius of the tube respectively then the value of `x,y` and `z` are given by

Options

A. `1,-1,-1`
B. `-1,-1,1`
C. `-1,-1,-1`
D. `1,1,1`

Explanation

To determine the values of \( x \), \( y \), and \( z \) in the expression for the critical velocity \( v_c \) in terms of the coefficient of viscosity \( \eta \), density \( \rho \), and radius \( r \), we need to equate the dimensions on both sides of the equation:

\[ v_c = \eta^x \rho^y r^z \]

### Step-by-Step Solution:

1. **Write the dimensional formula for each quantity:**

- Critical velocity \( v_c \): \([v_c] = [L^1 T^{-1}]\)
- Coefficient of viscosity \( \eta \): \([ \eta ] = [M^1 L^{-1} T^{-1}]\)
- Density \( \rho \): \([ \rho ] = [M^1 L^{-3} T^0]\)
- Radius \( r \): \([ r ] = [L^1]\)

2. **Express the dimensional equation:**

\[ [v_c] = [\eta^x \rho^y r^z] \]

Substituting the dimensions:

\[ [L^1 T^{-1}] = [M^1 L^{-1} T^{-1}]^x [M^1 L^{-3} T^0]^y [L^1]^z \]

3. **Expand the dimensions:**

\[ [L^1 T^{-1}] = [M^x L^{-x} T^{-x}] [M^y L^{-3y} T^0] [L^z] \]

4. **Combine the dimensions:**

\[ [L^1 T^{-1}] = [M^{x+y} L^{-x-3y+z} T^{-x}] \]

5. **Equate the dimensions on both sides:**

- For mass \( M \):

\[ x + y = 0 \quad \text{(1)} \]

- For length \( L \):

\[ -x - 3y + z = 1 \quad \text{(2)} \]

- For time \( T \):

\[ -x = -1 \quad \text{(3)} \]

6. **Solve the equations:**

From equation (3):

\[ x = 1 \]

Substitute \( x = 1 \) into equation (1):

\[ 1 + y = 0 \implies y = -1 \]

Substitute \( x = 1 \) and \( y = -1 \) into equation (2):

\[ -1 - 3(-1) + z = 1 \]

Simplify:

\[ -1 + 3 + z = 1 \implies 2 + z = 1 \implies z = -1 \]

### Final Values:

\[ x = 1, \quad y = -1, \quad z = -1 \]