A liquid enter at point $A_{1}$ with speed $3.5 m / s$ and leaves at point $A _{2} .$ Then find out the height attained by the liquid above point $A _{2}$ (in $cm$)
A$61.25$
B$51.25$
C$41.25$
D$71.25$
AIIMS 2019, Medium
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A$61.25$
a Consider the following figure
Here
$A _{1}= A _{2} soV _{1}= V _{2}=3.5 m / s$
Use Bernoulli theorem.
$P _{ atm }+\frac{1}{2} \rho(3.5)^{2}+\rho g (0)= P _{ atm }+\frac{1}{2} \rho(0)^{2}+\rho gh$
$h =\frac{(3.5 \times 3.5)^{2}}{2 g }$
$=0.6125 m$
$=61.25 cm$
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