A liquid flows through two capillary tubes connected in series. Their lengths are L and 2 L and radii r and 2 r respectively. Then the pressure differences across the first and the second tube are in the ratio

Question

A liquid flows through two capillary tubes connected in series. Their lengths are L and 2 L and radii r and 2 r respectively. Then the pressure differences across the first and the second tube are in the ratio (A) 8 (B) 4 (C) 2 (D) 1 / 8

Tardigrade Admin · Accepted Answer

V=(π P r4/8 η l) As tubes are in series V1=V2 ⇒ (P1 r14/l1)=(P2 r24/l2) ⇒ (P1/P2)=((l1/l2))((r2/r1))4 =((L/2 L))((2 r/r))4=8

Q. A liquid flows through two capillary tubes connected in series. Their lengths are $L$ and $2 L$ and radii $r$ and $2 r$ respectively. Then the pressure differences across the first and the second tube are in the ratio

8

4

2

$1/8$

$V = \frac{π P r ^{4}}{8 η l}$
As tubes are in series
$V_{1} = V_{2} \Rightarrow \frac{P _{1} r _{1}^{4}}{l _{1}} = \frac{P _{2} r _{2}^{4}}{l _{2}} \Rightarrow \frac{P _{1}}{P _{2}} = (\frac{l _{1}}{l _{2}}) (\frac{r _{2}}{r _{1}})^{4}$
$= (\frac{L}{2 L}) (\frac{2 r}{r})^{4} = 8$

Q. A liquid flows through two capillary tubes connected in series. Their lengths are L and 2L and radii r and 2r respectively. Then the pressure differences across the first and the second tube are in the ratio

8

4

2

1/8

V=8ηlπPr4​ As tubes are in series V1​=V2​⇒l1​P1​r14​​=l2​P2​r24​​⇒P2​P1​​=(l2​l1​​)(r1​r2​​)4 =(2LL​)(r2r​)4=8

Q. A liquid flows through two capillary tubes connected in series. Their lengths are $L$ and $2 L$ and radii $r$ and $2 r$ respectively. Then the pressure differences across the first and the second tube are in the ratio

$1/8$

$V = \frac{π P r ^{4}}{8 η l}$
As tubes are in series
$V_{1} = V_{2} \Rightarrow \frac{P _{1} r _{1}^{4}}{l _{1}} = \frac{P _{2} r _{2}^{4}}{l _{2}} \Rightarrow \frac{P _{1}}{P _{2}} = (\frac{l _{1}}{l _{2}}) (\frac{r _{2}}{r _{1}})^{4}$
$= (\frac{L}{2 L}) (\frac{2 r}{r})^{4} = 8$