Water is moving with a speed of $5.0\,m/s$ through a pipe of cross sectional area $4.0\,cm^2$ . The water gradually descends $10\,m$ as the pipe increase in area to $8.0\,cm^2$ . If the pressure at the upper level is $1.5 \times 10^5\,Pa$ , the pressure at lower level will be
, Diffcult
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According to equation of continuity
${\mathrm{A}_{1} \mathrm{v}_{1}=\mathrm{A}_{2} \mathrm{v}_{2}}$
or ${\mathrm{v}_{2}=\frac{\mathrm{A}_{1} \mathrm{v}_{1}}{\mathrm{A}_{2}}=\frac{4 \times 10^{-4} \times 5}{8 \times 10^{-4}}=\frac{5}{2} \mathrm{m} / \mathrm{s}}$
According to Bernoulli's theorem
$\mathrm{P}_{1}+\rho \mathrm{gh}_{1}+\frac{1}{2} \rho \mathrm{v}_{1}^{2}=\mathrm{P}_{2}+\rho \mathrm{gh}_{2}+\frac{1}{2} \rho \mathrm{v}_{2}^{2}$
$\mathrm{P}_{2}=\mathrm{P}_{1}+\rho \mathrm{g}\left(\mathrm{h}_{1}-\mathrm{h}_{2}\right)+\frac{1}{2} \rho\left(\mathrm{v}_{1}^{2}-\mathrm{v}_{2}^{2}\right)$
$=1.5 \times 10^{5}+10^{3} \times 10 \times 10+\frac{1}{2} \times 10^{3} \times\left(5^{2}-\left(\frac{5}{2}\right)^{2}\right)$
$=(1.5+1+0.21) \times 10^{5}=2.7 \times 10^{5} \mathrm{Pa}$
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