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UPSEE 2018 Solved Paper

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Question : 3 of 150

Marks: +1, -0

A stream of non-viscous liquid emerges from a very short outlet tube at the base of a large open tank, in which the depth of liquid is

h

. The tube is at a fixed angle

\theta

to the ground as shown in the figure. The maximum height of the stream

y

$h \sin^2 \theta$
$h \sin 2\theta$
$\frac{1}{2} h \sin \theta$
$h \tan^2 \theta$

Solution:

Let the speed of fluid at outlet is

v

.

Applying Bernoulli's theorem,

p_1+\rho g h+\frac{1}{2}\rho v^2=\text{constant}

p_0+\rho g h+0=p_0+0+\frac{1}{2}\rho v^2

∴

v=\sqrt{2gh}

......(i)

At outlet,

v \sin \theta

vertical velocity of the fluid at outlet,

v^2=u^2+2 g h_1

For maximum height,

0=(v \sin \theta)^2-2 g h_1

∴

h_1=\frac{v^2 \sin^2 \theta}{2 g}

=\frac{2 g h \sin^2 \theta}{2 g}

[ { from Eq. (i) }]

=h \sin^2 \theta

h_1 = (h\sin)^2 \theta

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