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9-1.Fluid Mechanics
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Water is flowing through two horizontal pipes of different diameters which are connected together. The diameters of the two pipes are $3\, cm$ and $6\, cm$ respectively. If the speed of water in narrower pipe is $4\, m/sec$ and the pressure is $2.0\times10^4$ pascal, then the speed of water in the wider pipe is ........ $m/sec$
A
$4$
B
$2$
C
$1$
D
$16$
Solution
$A_{1} V_{1}=A_{2} V_{2}$
$=\pi\left(\frac{3}{2}\right)^{2} \times 4=\pi\left(\frac{6}{2}\right)^{2} \times v$
$=\mathrm{v}=1 \mathrm{m} / \mathrm{s}$
Standard 11
Physics
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