A) \[x=\sqrt{D(H-D)}\]
B) \[x=\sqrt{\frac{D(H-D)}{2}}\]
C) \[x=2\sqrt{D(H-D)}\]
D) \[x=4\sqrt{D(H-D)}\]
Correct Answer: C
Solution :
Time taken by water to reach the bottom = \[t=\sqrt{\frac{2(H-D)}{g}}\] and velocity of water coming out of hole, \[v=\sqrt{2gD}\] \ Horizontal distance covered \[x=v\times t\] = \[\sqrt{2gD}\times \sqrt{\frac{2(H-D)}{g}}\]= \[2\sqrt{D(H-D)}\]
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