## In This post you will get answer of 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

Solution:

## According to Poiseuille’s formula, rate of flow through a narrow tube $V = frac{pi Pr^4}{8 eta l}$

For given $P$ and $eta , V propto frac{r^4}{l} therefore :: frac{V_1}{V_2} = frac{r^4_1}{r_2^4} times frac{l_2}{l_1}$

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Here, $V_1 = V, r_2 = r_1 /2 , l_2 = 2l_1, V_2 = ?$

So, $frac{V}{V_2} = (2)^4 times 2 = 32 , V_2 = frac{V}{32}$