A bent tube is lowered into a water stream as shown in fig. The velocity of the stream relative to the tube is equal to v = 2.5 m/s. The closed upper end of the tube located at the height $h_{0}$ = 12 cm has a small orifice. To what height ‘h’ will the water jet rise after coming out of orifice?

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0.55 m

0.1 m

0.2 m

0.44 m

answer is C.

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Detailed Solution

Let H is the depth of horizontal part of tube from top level of stream. Applying Bernoulli’s theorem from entrance of liquid into the tube upto the highest point of water jet,

$(P_{0} + H ρ g) + \frac{1}{2} ρ v^{2} + O =$ $P_{0} + O + ρ g (H + h_{0} + h)$

$\Rightarrow v^{2} = 2 g (h_{0} + h)$

$\Rightarrow h = \frac{v^{2}}{2 g} - h_{0}$ $= \frac{{(2.5)}^{2}}{2 \times 9.8} - 0.12$ $= 0.2 m$

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