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Discussion Response Post to: Fluids2

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anchit saini (4396)

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

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