The pressure of water in a pipe when tap is closed is `5.5xx10^5 Nm^(-2)`. When tap gets open, pressure reduces to `5xx10^5Nm^(-2)`. The velocity with

Question

The pressure of water in a pipe when tap is closed is `5.5xx10^5 Nm^(-2)`. When tap gets open, pressure reduces to `5xx10^5Nm^(-2)`. The velocity with

The pressure of water in a pipe when tap is closed is `5.5xx10^5 Nm^(-2)`. When tap gets open, pressure reduces to `5xx10^5Nm^(-2)`. The velocity with which water comes out on opening the tap is
A. `10ms^(-1)`
B. `5ms^(-1)`
C. `20ms^(-1)`
D. `15ms^(-1)`

Monjur Alahi

4 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

Correct Answer - A
Decrease in pressure energy = increase in kinetic energy
or `Deltap=1/2rhoV^2`
`therefore V= sqrt((2(Deltap))/(rho))= sqrt((2xx0.5xx10^5)/(10^3)) =10ms^(-1)`

4 views Answered 2023-02-05 15:29