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. $V /32$

Answer 1

Answer: (D)

$V /32$

Answer 2

According to Poiseuille's formula, rate of flow through a narrow tube $V = \frac{\pi Pr^4}{8 \eta l}$
For given $P$ and $\eta , V \propto \frac{r^4}{l} \therefore \:\: \frac{V_1}{V_2} = \frac{r^4_1}{r_2^4} \times \frac{l_2}{l_1}$
,
Here, $V_1 = V, r_2 = r_1 /2 , l_2 = 2l_1, V_2 = ?$
So, $\frac{V}{V_2} = (2)^4 \times 2 = 32 , V_2 = \frac{V}{32}$