A liquid is flowing in a horizontal uniform capillary tube under a constant pressure difference $P$. The value of pressure for which the rate of flow of the liquid is doubled when the radius and length both are doubled is
Medium
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(d)From $V = \frac{{P\pi {r^4}}}{{8\eta l}}$ ==> $P = \frac{{V8\eta l}}{{\pi {r^4}}}$
==>$\frac{{{P_2}}}{{{P_1}}} = \frac{{{V_2}}}{{{V_1}}} \times \frac{{{l_2}}}{{{l_1}}} \times {\left( {\frac{{{r_1}}}{{{r_2}}}} \right)^4}$$ = 2 \times 2 \times {\left( {\frac{1}{2}} \right)^4} = \frac{1}{4}$
==> ${P_2} = \frac{{{P_1}}}{4} = \frac{P}{4}$.
==>$\frac{{{P_2}}}{{{P_1}}} = \frac{{{V_2}}}{{{V_1}}} \times \frac{{{l_2}}}{{{l_1}}} \times {\left( {\frac{{{r_1}}}{{{r_2}}}} \right)^4}$$ = 2 \times 2 \times {\left( {\frac{1}{2}} \right)^4} = \frac{1}{4}$
==> ${P_2} = \frac{{{P_1}}}{4} = \frac{P}{4}$.
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