A tube of length L and radius $R$ is joined to another tube of length $\frac{L}{3}$ and radius $\frac{R}{2}$ . A fluid is flowing through this tube. If the pressure difference across the first tube is $P$ , then the pressure difference across the second tube is

\frac{16 P}{3}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{4 P}{3}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

P

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{3 P}{16}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $\frac{16 P}{3}$
Since we know that flow rate remains constant, then
flow rate in 1 = flow rate in 2.
$\frac{π P_{1} r_{1}^{4}}{8 g l_{1}} = \frac{π P_{1} r_{2}^{4}}{8 g l_{2}}$
$\frac{P_{1}}{P_{2}} = \frac{l_{1}}{l_{2}} \times {(\frac{r_{2}}{r_{1}})}^{4} [n o w s i n c e r_{2} = \frac{r_{1}}{2} & l_{2} = \frac{L_{1}}{3}]$
So,
$\frac{P}{P_{2}} = \frac{3 L_{1}}{L_{1}} = {(\frac{r_{1}}{{2 r}_{1}})}^{4} = \frac{3}{16}$
$P_{2} = \frac{16}{3} P$

Suggest Corrections

Similar questions

Q. A tube of radius R and length L is connected in series with another tube of radius

\frac{R}{2}

and length

\frac{L}{8}

. If the pressure across the tubes taken together is P, the pressure across the two tubes separately are:

The rate of steady volume flow of water through a capillary tube of length 'l' and radius 'r' under a pressure difference of P is V. This tube is connected with another tube of the same length but half the radius in series. Then the rate of steady volume flow through them is (The pressure difference across the combination is P)

Q. A capillary tube of length