Question

A bent tube is lowered into a water stream as shown in figure. The velocity of the stream relative to the tube is equal to v = 5 m s-1. The closed upper end of the tube has a small orifice. To what height h will the water jet spurt? (g = 10 m s-²)

answer is 1m...kindly explain​

Answer 1

Answer:

Let the velocity of the water jet, near the orifice be v

′

, then applying Bernoulli's theorem,

2

1

ρv

2

=h

0

ρg+

2

1

ρv

2

or, v

′

=

v

2

−2gh

0

(1)

Here the pressure term on both sides is the same and equal to atmospheric pressure.

Now, if it rises upto a height h, then at this height, whole of its kinetic energy will be converted into potential energy. So,

2

1

ρv

′

2

=ρgh or h=

2g

v

′

2

=

2g

v

2

−h

0

=20cm, [using equation (1)]

Answer 2

Explanation:

Let the velocity of the water jet, near the orifice be v

′

, then applying Bernoulli's theorem,

2

1

ρv

2

=h

0

ρg+

2

1

ρv

2

or, v

′

=

v

2

−2gh

0

(1)

Here the pressure term on both sides is the same and equal to atmospheric pressure.

Now, if it rises upto a height h, then at this height, whole of its kinetic energy will be converted into potential energy. So,

2

1

ρv

′

2

=ρgh or h=

2g

v

′

2