Two capillary tubes of same radius r but of lengths l1 and l2 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?

Question

Two capillary tubes of same radius r but of lengths l1 and l2 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? (A)  l1+l2  (B)  (l1l2/l1+l2)  (C)  (1/l1+l2)  (D)  (1/l1)+(1/l2)

Tardigrade Admin · Accepted Answer

For parallel combination (1/Reff)=(1/R1)+(1/R2) ⇒ (π r4/8η l)=(π r4/8η l1)+(π r4/8η l2) ⇒ (1/l)=(1/l1)+(1/l2) ∴ l=(l1l2/l1+l2)

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?

$l_{1} + l_{2}$

$\frac{l _{1} l _{2}}{l _{1} + l _{2}}$

$\frac{1}{l _{1} + l _{2}}$

$\frac{1}{l _{1}} + \frac{1}{l _{2}}$

Q. Two capillary tubes of same radius r but of lengths l1​ and l2​ 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?

l1​+l2​

l1​+l2​l1​l2​​

l1​+l2​1​

l1​1​+l2​1​

For parallel combination Reff​1​=R1​1​+R2​1​ ⇒ 8ηlπr4​=8ηl1​πr4​+8ηl2​πr4​ ⇒ l1​=l1​1​+l2​1​ ∴ l=l1​+l2​l1​l2​​

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?

$l_{1} + l_{2}$

$\frac{l _{1} l _{2}}{l _{1} + l _{2}}$

$\frac{1}{l _{1} + l _{2}}$

$\frac{1}{l _{1}} + \frac{1}{l _{2}}$