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![[Post New]](/templates/default/images/icon_minipost_new.gif) 1 Apr 2008 11:06:16 IST
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a tank is filled up to a height h with a liquid and is placed on a platform of height h from the ground . to get the maximum range X a small hole is punched at a distance of y from the free surface of the liquid . then a)X=2h b)X=1.5h c)y=h d)y=0.75h
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 1 Apr 2008 11:16:33 IST
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is the answer
c) y=h
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 1 Apr 2008 11:27:02 IST
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is the answer is option A
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 1 Apr 2008 11:32:20 IST
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its y = h
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I DONOT FOLLOW THE RULES I MAKE THEM TO FOLLOW ME. |
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 2 Apr 2008 09:28:49 IST
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Time of fall = sqrt(2*(2h-y)/g) velocity = sqrt(2*g*y) X=2*sqrt(y*(2h-y)) This has a max at y=h
X_max = 2h So (a) is right answer
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Krishna Gopal Singh
B.Tech Chemical Engg
IIT Delhi 2002
Currently doing PhD from IIT Delhi |
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