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 \[t=\frac{2(H-D)}{g}\] and velocity of water coming out of hole, \[v=\sqrt{2gD}\] \[\therefore \] Horizontal distance covered \[x=\nu \times t\] \[=\,\,\sqrt{2gD}\,\,\times \,\,\sqrt{\frac{2(H-D)}{g}}\,\,=\,\,2\sqrt{D(H-D)}\]=You need to login to perform this action.
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