The water which comes out from the hole at the instant when hole is at height H above the ground, strikes the ground at distance of\[x\]from P. Which of the following is correct for the situation described?
A) The value of\[x\]is\[2\sqrt{\frac{2hH}{3}}\]
B) The value of\[x\]is\[\sqrt{\frac{4hH}{3}}\]
C) The value of\[x\]cant be computed, from information provided
D) The question is irreverent as no water comes out from the hole
Correct Answer: D
Solution :
As vessel is falling freely under gravity, the pressure at all points within the liquid remains the same as the atmospheric pressure. If we apply Bernoullis theorem just inside and outside the hole, then \[{{p}_{inside}}+\frac{\rho v_{inside}^{2}}{2}+\rho gy={{p}_{outside}}\] \[+\frac{\rho v_{outside}^{2}}{2}+\rho gy\] \[{{v}_{inside}}=0,{{p}_{inside}}={{p}_{outside}}=p\](atmospheric pressure) Therefore, \[{{v}_{outside}}=0\]ie, no water comes out.
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