Under a constant pressure head, the rate of flow of orderly volume flow of liquid through a capillary tube is V. If the length of the capillary is doubled and the diameter of the bore is halved, the rate of flow would become

Question

Under a constant pressure head, the rate of flow of orderly volume flow of liquid through a capillary tube is V. If the length of the capillary is doubled and the diameter of the bore is halved, the rate of flow would become (A) V / 4 (B) V / 8 (C) 16 V (D) V /32

Tardigrade Admin · Accepted Answer

According to Poiseuille's formula, rate of flow through a narrow tube

V = \frac{π P r ^{4}}{8 η l}

For given

P

and

η, V \propto \frac{r ^{4}}{l} ∴ \frac{V _{1}}{V _{2}} = \frac{r _{1}^{4}}{r _{2}^{4}} \times \frac{l _{2}}{l _{1}}

, Here,

V_{1} = V, r_{2} = r_{1} /2, l_{2} = 2 l_{1}, V_{2} = ?

So,

\frac{V}{V _{2}} = (2)^{4} \times 2 = 32, V_{2} = \frac{V}{32}

Q. Under a constant pressure head, the rate of flow of orderly volume flow of liquid through a capillary tube is $V$ . If the length of the capillary is doubled and the diameter of the bore is halved, the rate of flow would become

$V /4$

$V /8$

$16 V$

$V /32$

Q. Under a constant pressure head, the rate of flow of orderly volume flow of liquid through a capillary tube is V. If the length of the capillary is doubled and the diameter of the bore is halved, the rate of flow would become

V/4

V/8

16V

V/32

According to Poiseuille's formula, rate of flow through a narrow tube V=8ηlπPr4​ For given P and η,V∝lr4​∴V2​V1​​=r24​r14​​×l1​l2​​ , Here, V1​=V,r2​=r1​/2,l2​=2l1​,V2​=? So, V2​V​=(2)4×2=32,V2​=32V​

Q. Under a constant pressure head, the rate of flow of orderly volume flow of liquid through a capillary tube is $V$ . If the length of the capillary is doubled and the diameter of the bore is halved, the rate of flow would become

$V /4$

$V /8$

$16 V$

$V /32$