In the diagram shown, the difference in the two tubes of the manometer is $5\, \ cm$, the cross section of the tube at $A$ and $B$ is $6\, mm^2$ and $10\, mm^2$ respectively. The rate at which water flows through the tube is $........ cc/s$
$(g\, = 10\, ms^{-2})$
$(g\, = 10\, ms^{-2})$
JEE MAIN 2014, Diffcult
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According to Bernoulli's theorem $,{p_1} + \frac{1}{2}\rho v_1^2 = {P_2} + \frac{1}{2}\rho v_2^2$
$\therefore v_2^2 - v_1^2 = 2gh$ $...\left( 1 \right)$
According to the equation of continuty
${A_1}{v_1} = {A_2}{v_2}$ $...(2)$
$\frac{{{A_1}}}{{{A_2}}} = \frac{{6\,m{m^2}}}{{10\,m{m^2}}}$
From equation $\left( 2 \right),\frac{{{A_1}}}{{{A_2}}} = \frac{{{v_2}}}{{{v_1}}} = \frac{6}{{16}}$
$or,\,\,{v_2} = \frac{6}{{10}}{v_1}$
Putting this value of $v_2$ in equation $(1)$
${\left( {\frac{6}{{10}}{v_1}} \right)^2} - {\left( {{v_1}} \right)^2} = 2 \times {10^3} \times 5$
$\left[ \begin{array}{l} g = 10m/{s^2} = {10^3}cm/{s^2}\\ and\,h = 5\,cm\end{array} \right]$
Solving we get $,{v_1} = \frac{{10}}{8}$
Therefore the rate at which water flows through the tube
$ = {A_1}{v_1} = {A_2}{v_2} $
$= \frac{{6 \times 10}}{8} = 7.5cc/s$
$\therefore v_2^2 - v_1^2 = 2gh$ $...\left( 1 \right)$
According to the equation of continuty
${A_1}{v_1} = {A_2}{v_2}$ $...(2)$
$\frac{{{A_1}}}{{{A_2}}} = \frac{{6\,m{m^2}}}{{10\,m{m^2}}}$
From equation $\left( 2 \right),\frac{{{A_1}}}{{{A_2}}} = \frac{{{v_2}}}{{{v_1}}} = \frac{6}{{16}}$
$or,\,\,{v_2} = \frac{6}{{10}}{v_1}$
Putting this value of $v_2$ in equation $(1)$
${\left( {\frac{6}{{10}}{v_1}} \right)^2} - {\left( {{v_1}} \right)^2} = 2 \times {10^3} \times 5$
$\left[ \begin{array}{l} g = 10m/{s^2} = {10^3}cm/{s^2}\\ and\,h = 5\,cm\end{array} \right]$
Solving we get $,{v_1} = \frac{{10}}{8}$
Therefore the rate at which water flows through the tube
$ = {A_1}{v_1} = {A_2}{v_2} $
$= \frac{{6 \times 10}}{8} = 7.5cc/s$
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