Two tubes of radii $$r_1$$ and $$r_2$$ and lengths $$l_1$$ and $$l_2$$, respectively, are connected in series and a liquid flows through each of them in stream line conditions. $$P_1$$ and $$P_2$$ are pressure differences across the two tubes. If $$P_2$$ is $$4P_1$$ and $$l_2$$ is $$\frac{l_1}{4}$$ then the radius $$r_2$$ will be equal to:

We deal with viscous flow through narrow tubes, so we start with the Poiseuille relation. The formula for the pressure difference $$P$$ required to push a steady (stream-line) volume flow rate $$Q$$ through a cylindrical tube of length $$l$$ and radius $$r$$ is stated as

$$P \;=\;\dfrac{8\eta\,l\,Q}{\pi\,r^{4}}$$

where $$\eta$$ is the viscosity of the liquid. We notice that $$\eta$$, $$\pi$$ and the common volume flow rate $$Q$$ are the same for both tubes because the tubes are connected in series; whatever liquid flows through tube 1 per unit time must immediately flow through tube 2 per unit time.

Writing the Poiseuille expression separately for the two tubes, we have

For tube 1: $$P_{1} \;=\;\dfrac{8\eta\,l_{1}\,Q}{\pi\,r_{1}^{4}}$$

For tube 2: $$P_{2} \;=\;\dfrac{8\eta\,l_{2}\,Q}{\pi\,r_{2}^{4}}$$

Now we are given two experimental facts:

$$P_{2}=4P_{1} \quad\text{and}\quad l_{2}=\dfrac{l_{1}}{4}$$

To connect these facts with the radii, we divide the second Poiseuille equation by the first. Doing this cancels the common factors $$8\eta Q/\pi$$ and yields

$$\dfrac{P_{2}}{P_{1}} =\dfrac{\;l_{2}/r_{2}^{4}\;}{\;l_{1}/r_{1}^{4}\;} =\dfrac{l_{2}\,r_{1}^{4}}{l_{1}\,r_{2}^{4}}$$

Substituting the known ratio $$P_{2}/P_{1}=4$$ and the length relation $$l_{2}=l_{1}/4$$, we obtain

$$4=\dfrac{\bigl(l_{1}/4\bigr)\,r_{1}^{4}}{l_{1}\,r_{2}^{4}}$$

The lengths $$l_{1}$$ cancel out, leaving

$$4=\dfrac{r_{1}^{4}}{4\,r_{2}^{4}}$$

Multiplying both sides by $$4\,r_{2}^{4}$$ gives

$$16\,r_{2}^{4}=r_{1}^{4}$$

Now we take the fourth root of both sides to extract the radius ratio:

$$r_{2}=\dfrac{r_{1}}{\,\sqrt[4]{16}\,}=\dfrac{r_{1}}{2}$$

Thus the radius of the second tube is half the radius of the first tube. Among the options provided, this corresponds to Option D.

Hence, the correct answer is Option D.