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Question

Physics Question on mechanical properties of fluid

A tank is filled with water upto height HH. When a hole is made at a distance hh below the level of water, what will be the horizontal range of water jet?

A

2h(H−h)2\sqrt{h(H-h)}

B

4h(H+h)4\sqrt{h(H+h)}

C

4h(H−h)4\sqrt{h(H-h)}

D

2h(H+h)2\sqrt{h(H+h)}

Answer

2h(H−h)2\sqrt{h(H-h)}

Explanation

Solution

Applying Bernoulli's theorem, The velocity of water at point AA v=2ghv=\sqrt{2 g h} Time taken to reach point CC is tt So, H−h=12gt2H-h=\frac{1}{2} g t^{2} t=2(H−h)gt=\sqrt{\frac{2(H-h)}{g}} ...(i) Now, horizontal range R=vtR=v t =2gh×2(H−h)g=\sqrt{2 g h} \times \sqrt{\frac{2(H-h)}{g}} =2(H−h)h=2 \sqrt{(H-h) h}