You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Excess Pressure in Bubbles

We have two n...

Question

We have two (narrow) capillary tubes $T_{1}$ and $T_{2}$ . Their length are $l_{1}, l_{2}$ and radii of cross-section are $r_{1}, r_{2}$ , respectively. The rate of flow of water through $T_{1}$ is $8 c m^{3} s^{- 1}$ when the pressure difference across its ends is p. What will be the rate of flow of water through $T_{2}$ under the same pressure difference, given that $l_{1} = l_{2}$ and $r_{1} = 2 r_{2}$ ?

8 c m^{3} s^{- 1}

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

4 c m^{3} s^{- 1}

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

2 c m^{3} s^{- 1}

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

0.5 c m^{3} s^{- 1}

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

Open in App

Solution

The correct option is D $0.5 c m^{3} s^{- 1}$

According to poiseuille's equation, the rate of flow of a liquid a in capillary tube is given by
$Q = \frac{π r^{4} Δ P}{8 μ L}$
$\Rightarrow Q_{1} = \frac{π r_{1}^{4} p}{8 μ l_{1}}$ and $Q_{2} = \frac{π r_{2}^{4} p}{8 μ l_{2}}$
$∵ l_{2} = l_{1}$ and $r_{2} = \frac{r_{1}}{2}$ , we have
$Q_{2} = \frac{π {(\frac{r_{1}}{2})}^{4} p}{8 μ l_{1}} = {(\frac{1}{2})}^{4} Q_{1} = \frac{Q_{1}}{16} = \frac{8}{16} c m^{3} s^{- 1} = 0.5 c m^{3} s^{- 1}$

Suggest Corrections

Similar questions

Q. We have two (narrow) capillary tubes

T_{1}

and

T_{2}

. their lengths are

l_{1}

and

l_{2}

and radii of cross section are

r_{1}

and

r_{2}

respectively. The rate of flow of water under a pressure difference

P

through tube

T_{1}

8 {c m}^{3} / s e c

. If

l_{1} = 2 l_{2}

and

r_{1} = 2 r_{2}

. What will be the rate of flow when the two tuves are connected in series and pressure difference across the combinatin is same as before

[= P]

Q. 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}

4 P_{1}

and

l_{2}

\frac{l_{1}}{4}

, then the radius

r_{2}

will be equal to :

Q. Two capillary tubes of same radius r but of lengths

l_{1}

and

l_{2}

are fitted in parallel to the bottom of a vessel. The pressure head is P. What should be the length of a single tube that can replace the two tubes so that the rate of flow is same as before

Two cylindrical rods of lengths $l_{1}$ and $l_{2}$ , radii $r_{1}$ and $r_{2}$ have thermal conductivities $k_{1}$ and $k_{2}$ respectively. The ends of the rods are maintained at the same temperature difference. If $l_{1} = 2 l_{2}$ and $r_{1} = \frac{r_{2}}{2}$ , the rates of heat flow in them will be the same if $\frac{k_{1}}{k_{2}}$ is

Q. Two capillary tubes A and B are arranged in parallel. A liquid flows through these capillary tubes under the same pressure head. Both the tubes have the same length. The radii of A and B are

r

and

\frac{r}{2}

respectively. If the rate of flow of liquid through A is

8 {c m}^{3} / s

, then the rate of flow through the combination of A and B is:

Join BYJU'S Learning Program

The correct option is D 0.5cm3s−1 According to poiseuille's equation, the rate of flow of a liquid a in capillary tube is given by Q=πr4ΔP8μL⇒Q1=πr41p8μl1 and Q2=πr42p8μl2∵l2=l1 and r2=r12, we have Q2=π(r12)4p8μl1=(12)4Q1=Q116=816cm3s−1=0.5cm3s−1