A cylindrical tank of height 1 m and cross-s | vedclass

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(C) The rate of change of volume of water in the tank is given by the difference between the rate of inflow and the rate of outflow.$A \frac{dh}{dt} =

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(C) The rate of change of volume of water in the tank is given by the difference between the rate of inflow and the rate of outflow.$A \frac{dh}{dt} = a_1 v_1 - a_2 \sqrt{2gh}$Given $A = 4000 \ cm^2 = 0.4 \ m^2$,$a_1 = 1 \ cm^2 = 10^{-4} \ m^2$,$v_1 = 2 \ m/s$,$a_2 = 0.5 \ cm^2 = 0.5 	imes 10^{-4} \ m^2$,and $g = 10 \ m/s^2$.At steady state,the rate of inflow equals the rate of outflow,so $\frac{dh}{dt} = 0$.$a_1 v_1 = a_2 \sqrt{2gh_{steady}}$$1 	imes 10^{-4} 	imes 2 = 0.5 	imes 10^{-4} \sqrt{2 	imes 10 	imes h_{steady}}$$2 = 0.5 \sqrt{20 h_{steady}}$$4 = \sqrt{20 h_{steady}}$$16 = 20 h_{steady}$$h_{steady} = \frac{16}{20} = 0.8 \ m$.As $t$ increases,the water level $h$ increases from $0$ and approaches the steady-state value of $0.8 \ m$. The rate of increase $\frac{dh}{dt}$ decreases as $h$ increases because the efflux velocity increases. Thus,the curve is concave downwards,approaching $0.8 \ m$ asymptotically. This corresponds to the graph in option $C$.

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