Which one of the following represents the correct dimensions of the quantity : `x=(eta)/(rho)`, where `eta`=coefficient of visocosity and `rho` =the density of a liquid?

Options

A. `[ML^(-2)T^(-1)]`
B. `[ML^(-4)T^(-2)]`
C. `[ML^(-5)T^(-2)]`
D. `[M^(0)L^(2)T^(-1)]`

Explanation

To find the correct dimensions of the quantity \( x = \frac{\eta}{\rho} \), where \( \eta \) is the coefficient of viscosity and \( \rho \) is the density of a liquid, we will derive the dimensions of both \( \eta \) and \( \rho \) step by step.

### Step 1: Determine the dimensions of the coefficient of viscosity \( \eta \)

The coefficient of viscosity \( \eta \) can be derived from the formula for viscous force:

\[ F = \eta A \frac{dv}{dx} \]

Where:
- \( F \) is the force,
- \( A \) is the area,
- \( \frac{dv}{dx} \) is the velocity gradient.

Rearranging this formula to express \( \eta \):

\[ \eta = \frac{F \cdot dx}{A \cdot dv} \]

### Step 2: Substitute the dimensions into the equation

Now, we will substitute the dimensions for each of the quantities involved:

- The dimension of force \( F \) is \( [F] = [M L T^{-2}] \).
- The dimension of distance \( dx \) is \( [L] \).
- The dimension of area \( A \) is \( [A] = [L^2] \).
- The dimension of velocity \( dv \) is \( [v] = [L T^{-1}] \).

Substituting these dimensions into the equation for \( \eta \):

\[ \eta = \frac{[M L T^{-2}] \cdot [L]}{[L^2] \cdot [L T^{-1}]} \]

### Step 3: Simplify the dimensions

Now we simplify the right-hand side:

\[ \eta = \frac{[M L^2 T^{-2}]}{[L^2 T^{-1}]} = [M L^2 T^{-2}] \cdot [L^{-2} T] = [M L^{-1} T^{-1}] \]

Thus, the dimensions of the coefficient of viscosity \( \eta \) are:

\[ [\eta] = [M L^{-1} T^{-1}] \]

### Step 4: Determine the dimensions of density \( \rho \)

The density \( \rho \) is defined as mass per unit volume:

\[ \rho = \frac{m}{V} \]

Where:
- The dimension of mass \( m \) is \( [M] \).
- The dimension of volume \( V \) is \( [L^3] \).

Thus, the dimensions of density \( \rho \) are:

\[ [\rho] = \frac{[M]}{[L^3]} = [M L^{-3}] \]

### Step 5: Calculate the dimensions of \( x = \frac{\eta}{\rho} \)

Now we can substitute the dimensions of \( \eta \) and \( \rho \) into the expression for \( x \):

\[ x = \frac{[\eta]}{[\rho]} = \frac{[M L^{-1} T^{-1}]}{[M L^{-3}]} \]

### Step 6: Simplify the dimensions of \( x \)

Simplifying this gives:

\[ x = [M L^{-1} T^{-1}] \cdot [M^{-1} L^{3}] = [L^{2} T^{-1}] \]

### Conclusion

Thus, the dimensions of the quantity \( x \) are:

\[ [x] = [L^{2} T^{-1}] \]

### Final Answer
The correct dimensions of the quantity \( x = \frac{\eta}{\rho} \) are \( [L^{2} T^{-1}] \).

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