Water is flowing with a velocity of $2\,m/s$ in a horizontal pipe where cross-sectional area is $2 \times 10^{-2}\, m^2$ at pressure $4 \times 10^4\, pascal$. The pressure at cross-section of area $0.01\, m^2$ in pascal will be
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$A_{1} V_{1}=A_{2} V_{2}$

$(2)$ $\left(2 \times 10^{-2}\right)=\mathrm{V}_{2}(0.01)$

$\mathrm{V}_{2}=4 \mathrm{m} / \mathrm{s}$

$4 \times 10^{4}+\frac{1}{2}\left(10^{3}\right)(2)^{2}=P+\frac{1}{2}\left(10^{3}\right)(4)^{2}$

$P=4 \times 10^{4}-6 \times 10^{3}$

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