A) \[\left( \frac{3}{4} \right)H\]
B) \[\left( \frac{2}{3} \right)H\]
C) \[\left( \frac{1}{4} \right)H\]
D) \[\left( \frac{1}{2} \right)H\]
E) \[\left( \frac{1}{3} \right)H\]
Correct Answer: D
Solution :
As hole is made in the tank below the free surface of water, so water rushing from this hole follows a parabolic path. The velocity of efflux of liquid, \[v=\sqrt{2gh}\] Time \[t=\sqrt{\frac{2(H-h)}{g}}\] Horizontal range, \[R=vt\] \[R={{\left( 2gh\times \frac{2(H-h)}{g} \right)}^{1/2}}\] ie, \[{{R}^{2}}=4h(H-h)=4(Hh-{{h}^{2}})\] The range is maximum if \[\frac{dR}{dh}=0\] or \[\frac{2RdR}{dh}=4(H-2h)\] or \[0=(H-2h)\] or \[h=\frac{H}{2}\]
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