Water flows in a streamlined manner through a capillary tube of radius a the pressure difference being p and the rate of flow Q. If the radius is reduced to a/2 and the pressure increased to 2p. The rate of flow becomes

Question

Water flows in a streamlined manner through a capillary tube of radius a the pressure difference being p and the rate of flow Q. If the radius is reduced to a/2 and the pressure increased to 2p. The rate of flow becomes (A) 4Q (B) Q (C)  (Q/4)  (D)  (Q/8)

Tardigrade Admin · Accepted Answer

V=(π P 4/8 η l) ∴ V propto p r4 ( η and I are contants) ∴ (V2/V1)=((p2/p1))((r2/r1))4 =2 ×((1/2))4=(1/8) V2=(Q/8)

Q. Water flows in a streamlined manner through a capillary tube of radius a the pressure difference being p and the rate of flow $Q$ . If the radius is reduced to $a /2$ and the pressure increased to $2 p$ . The rate of flow becomes

$4 Q$

$Q$

$\frac{Q}{4}$

$\frac{Q}{8}$

$V = \frac{π P ^{4}}{8 η l}$
$∴ V \propto p r^{4}$ ( $η$ and I are contants)
$∴ \frac{V _{2}}{V _{1}} = (\frac{p _{2}}{p _{1}}) (\frac{r _{2}}{r _{1}})^{4}$
$= 2 \times (\frac{1}{2})^{4} = \frac{1}{8}$
$V_{2} = \frac{Q}{8}$

Q. Water flows in a streamlined manner through a capillary tube of radius a the pressure difference being p and the rate of flow Q. If the radius is reduced to a/2 and the pressure increased to 2p. The rate of flow becomes

4Q

Q

4Q​

8Q​

V=8ηlπP4​ ∴V∝pr4 ( η and I are contants) ∴V1​V2​​=(p1​p2​​)(r1​r2​​)4 =2×(21​)4=81​ V2​=8Q​

Q. Water flows in a streamlined manner through a capillary tube of radius a the pressure difference being p and the rate of flow $Q$ . If the radius is reduced to $a /2$ and the pressure increased to $2 p$ . The rate of flow becomes

$4 Q$

$Q$

$\frac{Q}{4}$

$\frac{Q}{8}$

$V = \frac{π P ^{4}}{8 η l}$
$∴ V \propto p r^{4}$ ( $η$ and I are contants)
$∴ \frac{V _{2}}{V _{1}} = (\frac{p _{2}}{p _{1}}) (\frac{r _{2}}{r _{1}})^{4}$
$= 2 \times (\frac{1}{2})^{4} = \frac{1}{8}$
$V_{2} = \frac{Q}{8}$