A tap discharges water of constant density uniformly in a jet at a velocity of 2 m/s at the tap outlet. The jet flows down smoothly in a vertical direction as compact smooth stream. The velocity of the jet at 0.6 m below the tap outlet considering acceleration due to gravity as 10 m/s2, is

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A tap discharges water of constant density uniformly in a jet at a velocity of 2 m/s at the tap outlet. The jet flows down smoothly in a vertical direction as compact smooth stream. The velocity of the jet at 0.6 m below the tap outlet considering acceleration due to gravity as 10 m/s2, is

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BPSC AE Paper 4 (General Engineering Science) 10 Nov 2022 Official Paper

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4.5 m/s
4 m/s
5.5 m/s
5 m/s

Answer 1

Concept:

When a jet of water discharges from a tap, it accelerates downward under the influence of gravity. As the jet falls, its velocity increases due to the gravitational pull. We can use the kinematic equations of motion to find the velocity of the jet at a specific point below the tap.

Calculation:

Given:

Initial velocity (\( u \)) at the tap outlet = 2 m/s
Vertical distance below the tap outlet (\( h \)) = 0.6 m
Acceleration due to gravity (\( g \)) = 10 m/s²

Using the kinematic equation:

\( v^2 = u^2 + 2gh \)

Substitute the values:

\( v^2 = (2 \, \text{m/s})^2 + 2 \cdot (10 \, \text{m/s}^2) \cdot (0.6 \, \text{m}) \)

\( v^2 = 4 + 12 \)

\( v^2 = 16 \)

\( v = \sqrt{16} \)

\( v = 4 \, \text{m/s} \)

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Question

A tap discharges water of constant density uniformly in a jet at a velocity of 2 m/s at the tap outlet. The jet flows down smoothly in a vertical direction as compact smooth stream. The velocity of the jet at 0.6 m below the tap outlet considering acceleration due to gravity as 10 m/s2, is

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