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A tank is filled with water up to a height \(H\). Water is allowed to come out of a hole \(P\) in one of the walls at a depth \(D\) below the surface of water. Express the horizontal distance \(x\) in terms of \(H\) and \(D\)( )
Question: A tank is filled with water up to a height H. Water is allowed to come out of a hole P in one of the walls at a depth D below the surface of water. Express the horizontal distance x in terms of H and D(   )A. x=sqrt {D(H-D)}B. x=sqrt {frac {D(H-D)}{2}}C. x=2sqrt {D(H-D)}D. x=4sqrt {D(H-D)}
A. \(x=\sqrt {D(H-D)}\)
B. \(x=\sqrt {\frac {D(H-D)}{2}}\)
C. \(x=2\sqrt {D(H-D)}\)
D. \(x=4\sqrt {D(H-D)}\)

Answer

C
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Solution

Given : height of tank \(H\)
The velocity of water stream coming out of \(P\) is \(v = \sqrt{2gD}\)
The time taken by the water stream to fall through the height \(H-D\) can be calculated as
\(s = ut + \frac{1}{2}at^2\)
i.e. \(H- D = \frac{1}{2}gt^2\)
\(t = \sqrt{\frac{2(H - D)}{g}}\)
So horizontal distance \(x =v\times t\)
\(x =\sqrt{2gD}\times \sqrt{\frac{2(H - D)}{g}} \)
\(x=2\sqrt {D(H-D)}\)
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