1. Use Poiseuille’s law for streamline flow through a tube:

$Q = \frac{\pi r^4 \Delta P}{8\eta l}$

where:

• $Q$ = volume flow rate
• $r$ = radius of tube
• $l$ = length of tube
• $\Delta P$ = pressure difference across the tube

2. Since the two tubes are connected in series, the same liquid flows through both, so the flow rate is the same in both tubes.

Thus,

$\frac{\pi r_1^4 P_1}{8\eta l_1} = \frac{\pi r_2^4 P_2}{8\eta l_2}$

Cancel common factors:

$\frac{r_1^4 P_1}{l_1} = \frac{r_2^4 P_2}{l_2}$

3. Given:

$P_2 = 4P_1, \qquad l_2 = \frac{l_1}{4}$

Substitute into the equation:

$\frac{r_1^4 P_1}{l_1} = \frac{r_2^4 (4P_1)}{l_1/4}$

Now simplify the right-hand side:

$\frac{r_2^4 (4P_1)}{l_1/4} = \frac{16 r_2^4 P_1}{l_1}$

So,

$\frac{r_1^4 P_1}{l_1} = \frac{16 r_2^4 P_1}{l_1}$

Cancel $\dfrac{P_1}{l_1}$:

$r_1^4 = 16 r_2^4$

$r_2^4 = \frac{r_1^4}{16}$

Taking fourth root:

$r_2 = \frac{r_1}{2}$

4. Therefore, the correct option is:

$\boxed{\text{D: } \frac{r_1}{2}}$

5. Comparison with stored answer:

Stored correct answer = D

Our derived answer = D

So they agree.