Question

Consider a water jar of radius $$R$$ that has water filled up to height $$H$$ and is kept on a stand of height $$h$$ (see figure). Through a hole of radius $$r$$ $$\left( r\ll R \right)$$ at its bottom, the water leaks out and the stream of water coming down towards the ground has a shape like a funnel as shown in the figure. If the radius of the cross-section of water stream when it hits the ground is $$x$$. Then:

A
x=r(HH+h)14
B
x=r(HH+h)2
C
x=r(HH+h)12
D
x=r(HH+h)

Solution

The correct option is A $$x=r{ \left( \dfrac { H }{ H+h } \right) }^{ \dfrac { 1 }{ 4 } }$$By energy conservation, $$\dfrac { \dfrac { 1 }{ 2 } \rho { v }_{ 1 }^{ 2 } }{ \dfrac { 1 }{ 2 } \rho { v }_{ 2 }^{ 2 } } =\dfrac { \rho gH }{ \rho g(H+h) }$$Hence, $$\dfrac { { v }_{ 2 } }{ { v }_{ 1 } } =\sqrt { \dfrac { H+h }{ H } }$$,Also, since by mass conservation, $${ A }_{ 1 }{ v }_{ 1 }={ A }_{ 2 }{ v }_{ 2 }$$, $${ r }^{ 2 }{ v }_{ 1 }={ x }^{ 2 }{ v }_{ 2 }$$From the two relations, we get, $$\dfrac { x }{ r } ={ (\dfrac { H }{ H+h } ) }^{ \dfrac { 1 }{ 4 } }$$.                                                      $${ x } =r{ (\dfrac { H }{ H+h } ) }^{ \dfrac { 1 }{ 4 } }$$.Physics

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