A wedgeshaped tank is filled with water at a rate of 500 kgs

A wedge-shaped tank is filled with water at a rate of 5.00 kg/s. The tank is 4.00 m long and 4.0 m tall, and the angle of the wedge is 90

at a certain time the height of water is be h(t)

now area of cross section = 2h(t)x length( which is 4m ) = 8h(t)

rate of flow of water be r

now after a time dt the volume of water increase = r(dt)

height increased d(h) = r(dt)/8h(t)

we have the diif equation dh/dt = r/(8h) ( r = 5 assuming density of water =1)

solution of this diff equation is h^2 = 5/8t

putting time t = 60x60 seconds we get h as 47.43 metres implies tank will be full before 1 hour