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Mechanics

Blazing goIITian

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5 Oct 2008 23:36:58 IST

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Fluids2

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Avirup Dasgupta

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6 Oct 2008 09:15:18 IST

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Gourabneo will post the solution.

anchit saini

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6 Oct 2008 10:17:14 IST

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$1)-> \mbox{velocity of efflux } v= \sqrt{2gy} \\ \\ \mbox{height}= h + h - y = 2h -y \\ \\ \mbox{time taken to fall} -> 2h - y = \frac{gt^2}{2} \\ \\ t = \sqrt{\frac{2(2h-y}{g}} \\ \\ x=vt = 2\sqrt{y(2h-y)}\\ \\ \mbox{For max x }\frac{dx}{dt}=0 \\ \\ or \ y = h \\ \\ x= 2h \\ \\ 2)\mbox{Simply use }-> \\ \\ \mbox{volume of water lost }= \mbox{ volume of water fallen down} \\ \\ \pi r^2 \sqrt{2gy}= \pi x^2 \frac{dy}{dt} \\ \\ or \ y=kx^4 \\ \ \mbox{value of k is easy to find}$

Gaurab Mandal

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6 Oct 2008 12:08:24 IST

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PLEASE RATE IF IT HELPS:.....

The radius of the drain hole "r"(which is given as 2mm);

And dy/dx is the rate at which water falls:

Substitute these as the values given at the end:

The answer is given in two posts.......Here it goes:

Gaurab Mandal

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6 Oct 2008 12:09:11 IST

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Re:Fluids2

Gaurab Mandal

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7 Oct 2008 11:43:41 IST

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hey did it help??????

reddevil_2009

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7 Oct 2008 12:04:19 IST

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yup

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