Let A(h) be the cross-sectional area of the water int he tank at height h and a

Question

Let A(h) be the cross-sectional area of the water int he tank at height h and a the areas of drain hole. the rate at which water if flowing out of the tank at time t can be expressed as the cross-sectional area at height h times the rate at which the height of the water is changing. Alternatively, the rate at which water flows out of the hole can be expressed as the area of the hole times the velocity of the draining water. Set these two equal to each other and insert Torricelli's Law to derive the differential euqation

A(h)*dh/dt= -a*Sqrt(2gh)

please make answer clear and detailed for points

Explanation / Answer

Accrding to Toricelli's law,

speed of water through hole, v = sqrt(2*g*H)

volume flow rate through hole = a*v

= a*sqrt(2*g*h)

but, at height h, speed of water v = -dh/dt

the rate at which voume decreases in the tank = A*v

= -A*dh/dt

Here,the rate at which volume dcreases in the tank = volume flow rate through the hole

-A*dh/dt = sa*sqrt(2*g*h)

==> A*dh/dt = -a*sqrt(2*g*h)

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