MCQ
A manometer connected to a closed tap reads $3.5 × 10^5 N/m^2$. When the valve is opened, the reading of manometer falls to $3.0 × 10^5 N/m^2$, then velocity of flow of water is ........ $m/s$
- A$100$
- ✓$10$
- C$1 $
- D$10\sqrt {10} $
✓
Answer
Correct option: B.
$10$
b
(b)Bernoulli's theorem for unit mass of liquid
$\frac{P}{\rho } + \frac{1}{2}{v^2} = $ constant
As the liquid starts flowing, it pressure energy decreases $\frac{1}{2}{v^2} = \frac{{{P_1} - {P_2}}}{\rho }$$ \Rightarrow \frac{1}{2}{v^2} = \frac{{3.5 \times {{10}^5} - 3 \times {{10}^5}}}{{{{10}^3}}} \Rightarrow {v^2}$
$ = \frac{{2 \times 0.5 \times {{10}^5}}}{{{{10}^3}}} \Rightarrow {v^2} = 100 \Rightarrow v = 10\;m/s$
(b)Bernoulli's theorem for unit mass of liquid
$\frac{P}{\rho } + \frac{1}{2}{v^2} = $ constant
As the liquid starts flowing, it pressure energy decreases $\frac{1}{2}{v^2} = \frac{{{P_1} - {P_2}}}{\rho }$$ \Rightarrow \frac{1}{2}{v^2} = \frac{{3.5 \times {{10}^5} - 3 \times {{10}^5}}}{{{{10}^3}}} \Rightarrow {v^2}$
$ = \frac{{2 \times 0.5 \times {{10}^5}}}{{{{10}^3}}} \Rightarrow {v^2} = 100 \Rightarrow v = 10\;m/s$
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