The surface of water in a water tank of cross | vedclass

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Question

(A) According to the equation of continuity, the volume flow rate at the tank surface must equal the volume flow rate at the tap: $A_1 v_1 = A_2 v_2$.

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(A) According to the equation of continuity, the volume flow rate at the tank surface must equal the volume flow rate at the tap: $A_1 v_1 = A_2 v_2$. Here, $A_1 = 750 \,cm^2 = 750 	imes 10^{-4} \,m^2$ and $A_2 = 500 \,mm^2 = 500 	imes 10^{-6} \,m^2$. The speed of water at the tap is $v_2 = 30 \,cm/s = 0.3 \,m/s$. The speed of the water surface falling is $v_1 = \frac{dh}{dt}$. Substituting the values: $(750 	imes 10^{-4}) \cdot \frac{dh}{dt} = (500 	imes 10^{-6}) 	imes (0.3)$. $\frac{dh}{dt} = \frac{500 	imes 10^{-6} 	imes 0.3}{750 	imes 10^{-4}} = \frac{150 	imes 10^{-6}}{750 	imes 10^{-4}} = 0.2 	imes 10^{-2} \,m/s = 2 	imes 10^{-3} \,m/s$. Given $\frac{dh}{dt} = x 	imes 10^{-3} \,m/s$, we find $x = 2$.

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