Two liquids of densities `d_(1)` and `d_(2)` are flowing in identical capillaries under same pressure difference. If `t_(1)` and `t_(2)` are the time taken for the flow of equal quantities of liquid, then the ratio of coefficients of viscosities of liquids must be

Options

A. `(d_(1)d_(2))/(t_(1)t_(2))`
B. `(d_(1)t_(1))/(d_(2)t_(2))`
C. `(d_(1)t_(2))/(d_(2)t_(1))`
D. `sqrt(((d_(1)t_(1))/(d_(2)t_(2))))`

Answer

To solve the problem, we need to find the ratio of the coefficients of viscosities (\(\eta_1\) and \(\eta_2\)) of two liquids flowing through identical capillaries under the same pressure difference. The time taken for the flow of equal quantities of liquid is \(t_1\) and \(t_2\).

### Step-by-Step Solution:

1. **Understanding the Flow of Liquids**:
We know that the quantity of liquid flowing through a capillary can be described by Poiseuille's law. For a liquid flowing through a capillary, the volume flow rate \(Q\) is given by:
\[ Q = \frac{\pi \Delta P r^4}{8 \eta L} \]
where \(\Delta P\) is the pressure difference, \(r\) is the radius of the capillary, \(\eta\) is the coefficient of viscosity, and \(L\) is the length of the capillary.

2. **Setting Up the Equations**:
For the first liquid, we have:
\[ Q_1 = \frac{\pi \Delta P r^4}{8 \eta_1 L} \]
For the second liquid, we have:
\[ Q_2 = \frac{\pi \Delta P r^4}{8 \eta_2 L} \]

3. **Relating the Quantities**:
Since the liquids flow equal quantities, we can relate the two quantities:
\[ Q_1 \eta_1 = Q_2 \eta_2 \]

4. **Expressing Flow Rates in Terms of Time**:
The flow rates can also be expressed in terms of the volume \(V\) and time \(t\):
\[ Q_1 = \frac{V_1}{t_1} \quad \text{and} \quad Q_2 = \frac{V_2}{t_2} \]
Since the question states that equal quantities of liquids are flowing, we have:
\[ V_1 = V_2 \]

5. **Using Densities**:
The mass of each liquid can be expressed as:
\[ m_1 = d_1 V_1 \quad \text{and} \quad m_2 = d_2 V_2 \]
Since the masses are equal:
\[ d_1 V_1 = d_2 V_2 \]
Substituting \(V_1\) and \(V_2\) in terms of \(Q_1\) and \(Q_2\):
\[ d_1 \left(Q_1 t_1\right) = d_2 \left(Q_2 t_2\right) \]

6. **Substituting for \(Q_1\) and \(Q_2\)**:
From the earlier equations, we can substitute \(Q_1\) and \(Q_2\):
\[ d_1 \left(\frac{\pi \Delta P r^4}{8 \eta_1 L} t_1\right) = d_2 \left(\frac{\pi \Delta P r^4}{8 \eta_2 L} t_2\right) \]
The common terms cancel out, leading to:
\[ d_1 t_1 \eta_2 = d_2 t_2 \eta_1 \]

7. **Finding the Ratio of Viscosities**:
Rearranging gives:
\[ \frac{\eta_1}{\eta_2} = \frac{d_1 t_1}{d_2 t_2} \]

### Final Result:
Thus, the ratio of the coefficients of viscosities of the two liquids is:
\[ \frac{\eta_1}{\eta_2} = \frac{d_1 t_1}{d_2 t_2} \]