Water is flowing through a horizontal tube having cross-sectional areas of its two ends being $A$ and $A'$ such that the ratio $A/A'$ is $5$ છે.જો If the pressure difference of water between the two ends is $3 \times 10^5\, N\, m^{-2}$, the velocity of water with which it enters the tube will be ......... $m s^{-1}$ (neglect gravity effects)
AIEEE 2012Diffcult
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According to $Bernoulli's$ theorem
${P_1} + \frac{1}{2}\rho v_1^2 = {P_2} + \frac{1}{2}\rho v_2^2\,\,\,...\left( i \right)$
From question,
${P_1} - {P_2} = 3 \times {10^5},\frac{{{A_1}}}{{{A_2}}} = 5$
According to equation of constinuity
${A_1}{v_1} = {A_2}{v_2}$
$or,\frac{{{A_1}}}{{{A_2}}} = \frac{{{v_2}}}{{{v_1}}} = 5$
$ \Rightarrow \,\,{v_2} = 5{v_1}$
From equation $(i)$
${P_1} - {P_2} = \frac{1}{2}\rho \left( {v_2^2 - v_1^2} \right)$
$or\,\,3 \times {10^5} = \frac{1}{2} \times 1000\left( {5v_1^2 - v_1^2} \right.$
$ \Rightarrow 600 = 6{v_1} \times 4{v_1}$
$ \Rightarrow v_1^2 = 25$
$\therefore \,{v_1} = 5\,m/s$
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Figure: $Image$
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