Question

The rate of flow of liquid in a capillary tube under and constant pressure head is Q. If the diameter of the tube is reduced to half and its length is doubled, then the new rate of flow of liquid will be:
A. $\dfrac{Q}{4}$
B. $\dfrac{Q}{8}$
C. $16Q$
D. $\dfrac{Q}{32}$

Answer 1

Hint: For capillary tubes the rate of flow of liquid can be given as, $Q=\dfrac{\pi {{\Pr }^{4}}}{8\eta l}$. Using this formula, we will find the changes in Q when the diameter of the tube is reduced to half and length is doubled. Then, using those values we will find the new rate of flow of liquid.

Formula used: $Q=\dfrac{\pi {{\Pr }^{4}}}{8\eta l}$

Complete step-by-step answer:
Now, in the question we are given that rate of flow of liquid in capillary tube under pressure is Q. Now, the rate of flow of liquid in capillary can be given mathematically as,
$Q=\dfrac{\pi {{\Pr }^{4}}}{8\eta l}$ ………..(i)
Where, r is radius, Q is rate of flow of liquid, P is pressure, l is length and $\eta$ is coefficient of viscosity.
First of all, we will convert this equation from radius to diameter which can be given as,
$d=\dfrac{r}{2}\Rightarrow r=2d$ ………..(ii)
Replacing r in equation (i) by its value we will get,
$Q=\dfrac{\pi \operatorname{P}{{\left( 2d \right)}^{4}}}{8\eta l}=\dfrac{16\pi \operatorname{P}{{\left( d \right)}^{4}}}{8\eta l}$ ………………..(iii)
Now, in question it is given that the diameter is reduced to half and length is doubled, which can be given as,
$d'=\dfrac{d}{2}$
$l'=\dfrac{l}{2}$
$P'=P$, as pressure is constant.
Now, the new rate of flow of liquid can be given as,
$Q'=\dfrac{16\pi \operatorname{P}'{{\left( d' \right)}^{4}}}{8\eta l'}$ ………………(iv)
Now, substituting the values of $d'$ and $l'$ in equation (iv) we will get,
$Q'=\dfrac{16\pi \operatorname{P}{{\left( \dfrac{d}{2} \right)}^{4}}}{8\eta \left( 2l \right)}$
$\Rightarrow Q'=\dfrac{16\pi \operatorname{P}{{d}^{4}}}{16\times 8\times 2\eta \left( l \right)}=\dfrac{\pi \operatorname{P}{{d}^{4}}}{8\times 2\eta \left( l \right)}$
$\Rightarrow Q'=\dfrac{\pi \operatorname{P}{{d}^{4}}}{16\eta l}$
Now, as we know that value of Q in equation (iv) is, $Q=\dfrac{16\pi \operatorname{P}{{\left( d \right)}^{4}}}{8\eta l}$
So, we will multiply and divide 32 in the equation in we will get the original value, which can be seen mathematically as,
$\Rightarrow Q'=\dfrac{32}{32}\times \dfrac{\pi \operatorname{P}{{d}^{4}}}{16\eta l} = \dfrac{1}{32}\times \dfrac{16\pi \operatorname{P}{{d}^{4}}}{8\eta l}$
$\Rightarrow Q'=\dfrac{Q}{32}$
Hence, the new rate of flow of liquid can be given by $\dfrac{Q}{32}$.
Thus, option (d) is the correct answer.

Note: In such types of questions students might make mistakes in considering the half or double in given quantities and they might get confused in considering the diameter and radius as per the question otherwise they might make mistakes in solving the problem and getting the answer.