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 ${T_1}$ is $8c{m^3}/\sec $ . If ${l_1} = 2{l_2}$ and${r_1} = {r_2}$, what will be the rate of flow when the two tubes are connected in series and pressure difference across the combination is same as before $\left( { = P} \right)$

$\left( {8/17} \right)c{m^3}/\sec $

NEET UG PHYSICS MCQ: MECHANICAL PROPERTIES OF FLUIDS

1 Medium

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Application +4 | -1

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}

8c{m^3}/\sec

. If

{l_1} = 2{l_2}

and

{r_1} = {r_2}

, what will be the rate of flow when the two tubes are connected in series and pressure difference across the combination is same as before

\left( { = P} \right)

View answer and explanation

Correct option: B

\left( {16/3} \right)c{m^3}/\sec

Option B (

\left( {16/3} \right)c{m^3}/\sec

) is correct because when tubes are connected in series, the flow rate is the same through both, and the pressure drop is additive. By calculating the individual resistances and combining them, the new flow rate can be determined.

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