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Question

A manometer connected to a closed tap reads $$3.5 \times 10^5 \, N/m^2$$. When the value is opened,the reading of manometer fall to $$3.0 \times 10^5\, N/m^2$$, then velocity of flow of water is 


A
100m/s
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B
1m/s
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C
10m/s
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D
1010m/s
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Solution

The correct option is C $$10\, m/s$$

Bernoulli's theorem for unit mass of liquid $$\dfrac{P}{\rho } + \dfrac{1}{2}{v^2} = cons\tan t\,\,$$ 

As the liquid starts flowing . it  pressure energy decreases .

$$\begin{array}{l} \, \,  \\ \dfrac { 1 }{ 2 } { v^{ 2 } }=\dfrac { { { P_{ 1 } }-{ P_{ 2 } } } }{ \rho  }  \\ \Rightarrow \dfrac { 1 }{ 2 } { v^{ 2 } }=\dfrac { { 3.5\times { { 10 }^{ 5 } }-3\times { { 10 }^{ 5 } } } }{ { { { 10 }^{ 3 } } } }  \\ \Rightarrow { v^{ 2 } }=\dfrac { { 2\times 0.5\times { { 10 }^{ 5 } } } }{ { { { 10 }^{ 3 } } } }  \\ \therefore { v^{ 2 } }=100 \\ \therefore v=10\, \, m/s \end{array}$$


Physics

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