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Step-by-Step Solution
Step 1: Understand the Problem
Two tubes with different radii and lengths are connected in series. A liquid flows through both tubes under the same flow rate, and the pressures across each tube are given. We need to find the radius of the second tube, given relationships between pressures and lengths.
Step 2: Recall the Formula for Flow Rate
The volume flow rate $ \frac{dv}{dt} $ for a liquid flowing through a narrow tube under streamlined (Poiseuille) conditions is given by:
$ \frac{dv}{dt} = \frac{\pi \, P \, r^{4}}{8 \, \eta \, l} $
where:
$P$ is the pressure difference across the tube.
$r$ is the radius of the tube.
$\eta$ is the coefficient of viscosity of the liquid.
$l$ is the length of the tube.
Step 3: Apply the Series Connection Condition
Since the same liquid flows in series through both tubes, their flow rates must be equal. Hence:
$ \frac{\pi P_{1} \, r_{1}^{4}}{8 \, \eta \, l_{1}}
= \frac{\pi P_{2} \, r_{2}^{4}}{8 \, \eta \, l_{2}} $
The common constants cancel out, leaving:
$ \frac{P_{1} \, r_{1}^{4}}{l_{1}}
= \frac{P_{2} \, r_{2}^{4}}{l_{2}}. $
Step 4: Substitute the Given Conditions
$P_{2} = 4 \, P_{1}$
$l_{2} = \frac{l_{1}}{4}$
Using these in the flow-rate equality, we get:
$ \frac{P_{1} \, r_{1}^{4}}{l_{1}}
= \frac{4 \, P_{1} \, r_{2}^{4}}{\frac{l_{1}}{4}}. $
Simplify this equation:
$ \frac{P_{1} \, r_{1}^{4}}{l_{1}}
= \frac{4 \, P_{1} \, r_{2}^{4}}{\frac{l_{1}}{4}}
\quad \Longrightarrow \quad
\frac{r_{1}^{4}}{l_{1}}
= \frac{4 \, r_{2}^{4}}{\frac{l_{1}}{4}}.
$
Step 5: Solve for $r_{2}$
Further simplifying:
$ \frac{r_{1}^{4}}{l_{1}}
= \frac{4 \, r_{2}^{4}}{\frac{l_{1}}{4}}
\quad \Longrightarrow \quad
4 \, r_{2}^{4} \times \frac{4}{l_{1}}
= \frac{r_{1}^{4}}{l_{1}}
$
Or equivalently:
$ r_{2}^{4}
= \frac{r_{1}^{4}}{16}.
$
Taking the fourth root of both sides gives:
$ r_{2} = \frac{r_{1}}{2}.
$
Therefore
The radius of the second tube $r_{2}$ is
$ \displaystyle \frac{r_{1}}{2}.
$
