Two capillaries of same length and radii in the ratio $1 : 2$ are connected in series. A liquid flows through them in streamlined condition. If the pressure across the two extreme ends of the combination is $ 1 m$ of water, the pressure difference across first capillary is...... $m$
Diffcult
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(d)Given, ${l_1} = {l_2} = 1,$and $\frac{{{r_1}}}{{{r_2}}} = \frac{1}{2}$
$V = \frac{{\pi {P_1}r_1^4}}{{8\eta l}} = \frac{{\pi {P_2}r_2^4}}{{8\eta l}}$==> $\frac{{{P_1}}}{{{P_2}}} = {\left( {\frac{{{r_2}}}{{{r_1}}}} \right)^4} = 16$
==> ${P_1} = 16{P_2}$
Since both tubes are connected in series, hence pressure difference across combination,
$P = {P_1} + {P_2}$==> 1 = ${P_1} + \frac{{{P_1}}}{{16}}$ ==> ${P_1} = \frac{{16}}{{17}} = 0.94m$
$V = \frac{{\pi {P_1}r_1^4}}{{8\eta l}} = \frac{{\pi {P_2}r_2^4}}{{8\eta l}}$==> $\frac{{{P_1}}}{{{P_2}}} = {\left( {\frac{{{r_2}}}{{{r_1}}}} \right)^4} = 16$
==> ${P_1} = 16{P_2}$
Since both tubes are connected in series, hence pressure difference across combination,
$P = {P_1} + {P_2}$==> 1 = ${P_1} + \frac{{{P_1}}}{{16}}$ ==> ${P_1} = \frac{{16}}{{17}} = 0.94m$
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