Two tubes of radii `r_(1) andr _(2)` and lengths `l_(1) and l_(2)` respectively, are connected in series and a liquid flows through each of them in streamline conditions. `P_(1) and P_(2)` are pressure differences across the two tubes. If` P_(2)` is` 4P_(1)` and `l_(2)` is `(l_(1))/(4)` then the radius r, will be equal to:

Options

A. `r _(1)`
B. `2 r _(1)`
C. `4 r _(1)`
D. `(r _(1))/( 2)`

Answer

To solve the problem, we need to analyze the flow of liquid through two tubes connected in series and apply the principles of fluid mechanics, particularly the equation of continuity and the relationship between pressure, velocity, and radius.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have two tubes with radii \( r_1 \) and \( r_2 \), and lengths \( l_1 \) and \( l_2 \) respectively.
- The pressure difference across the first tube is \( P_1 \) and across the second tube is \( P_2 \).
- Given: \( P_2 = 4P_1 \) and \( l_2 = \frac{l_1}{4} \).

2. **Using the Equation of Continuity**:
- The equation of continuity states that the product of the cross-sectional area \( A \) and the velocity \( v \) is constant along a streamline:
\[ A_1 v_1 = A_2 v_2 \]
- The area \( A \) of a tube is given by \( A = \pi r^2 \). Thus, we can write:
\[ \pi r_1^2 v_1 = \pi r_2^2 v_2 \]
- Simplifying this gives:
\[ r_1^2 v_1 = r_2^2 v_2 \]

3. **Relating Pressure and Velocity**:
- According to Bernoulli's principle, pressure is inversely proportional to the velocity of the fluid:
\[ P \propto \frac{1}{v} \]
- Therefore, we can express the relationship between the pressures and velocities:
\[ \frac{P_1}{P_2} = \frac{v_2}{v_1} \]
- Substituting \( P_2 = 4P_1 \):
\[ \frac{P_1}{4P_1} = \frac{v_2}{v_1} \implies \frac{1}{4} = \frac{v_2}{v_1} \]
- This implies:
\[ v_2 = \frac{1}{4} v_1 \]

4. **Substituting into the Continuity Equation**:
- Now substituting \( v_2 = \frac{1}{4} v_1 \) into the continuity equation:
\[ r_1^2 v_1 = r_2^2 \left(\frac{1}{4} v_1\right) \]
- Canceling \( v_1 \) from both sides (assuming \( v_1 \neq 0 \)):
\[ r_1^2 = \frac{1}{4} r_2^2 \]
- Rearranging gives:
\[ r_2^2 = 4 r_1^2 \]
- Taking the square root:
\[ r_2 = 2 r_1 \]

5. **Conclusion**:
- The radius \( r_2 \) is equal to \( 2 r_1 \). Thus, the correct answer is:
\[ r_2 = 2 r_1 \]

### Final Answer:
The radius \( r_2 \) will be equal to \( 2 r_1 \).