SSC JE Mechanical Engineering 25 Sep 2019 Shift 1 Paper

Question : 141 of 200

Marks: +1, -0

A simple pitot tube is placed 1 m below the water surface in an open channel. The water level in the open end of the tube is 1 m above the water surface. The water flow velocity is:

3 km/hr
$(4g)^{0.5 }$ m/s
2.05 m/s
$(2g)^{0.5 }$ m/s

Validate

Solution:

{v}_{1}=\√{ {2 {gh}}}=\√{{2 g}}

{v}_{1}=(2 {g})^{0.5} {m} \/ {s}

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