Directions : In the following question more than one of the answers given may be correct. Select the correct answer and mark it according to the code:
The tank is filled upto a height h with a liquid and is placed on a platform of height h from the ground. To get maximum range\[{{x}_{m}}a\]small hole is punched at a distance of y from the free surface of the liquid. Then
(1) \[{{x}_{m}}=2h\] (2) \[{{x}_{m}}=1.5h\] (3) \[y=h\] (4)\[y=0.75\,h\]
A) 1, 2 and 3 are correct.
B) 1 and 2 are correct.
C) 2 and 4 are correct.
D) 1 and 3 are correct.
Correct Answer: D
Solution :
Velocity of liquid through orifice,\[v=\sqrt{2gy}\]and time taken by liquid to reach the ground \[t=\sqrt{\frac{2(h+h-y)}{g}}=\sqrt{\frac{2(2h-y)}{g}}\] \[\therefore \]Horizontal distance covered by liquid \[x=v.t=\sqrt{2gy}\times \sqrt{\frac{2(2h-y)}{g}}=\sqrt{4y(2h-y)}\] \[\Rightarrow \] \[{{x}^{2}}=4y(2h-y)\] \[\Rightarrow \] \[\frac{d({{x}^{2}})}{dy}=8h-8y\] For\[x\]to be maximum, \[\frac{d({{x}^{2}})}{dy}=0\] \[\therefore \] \[8h-8y=0\] or \[h=y\] So, \[{{x}_{m}}=\sqrt{4h(2h-h)}=2h\]
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