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

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(A) Let $A_1 = 750 	ext{ cm}^2 = 750 	imes 10^{-4} 	ext{ m}^2$ be the cross-section area of the tank.Let $A_2 = 500 	ext{ mm}^2 = 500 	imes 10^{-

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$2$

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(A) Let $A_1 = 750 	ext{ cm}^2 = 750 	imes 10^{-4} 	ext{ m}^2$ be the cross-section area of the tank.Let $A_2 = 500 	ext{ mm}^2 = 500 	imes 10^{-6} 	ext{ m}^2 = 5 	imes 10^{-4} 	ext{ m}^2$ be the cross-section area of the tap.Let $v_1 = |dh/dt|$ be the speed of the water level falling in the tank.Let $v_2 = 30 	ext{ cm/s} = 0.3 	ext{ m/s}$ be the speed of water coming out of the tap.According to the equation of continuity, $A_1 v_1 = A_2 v_2$.Substituting the values: $(750 	imes 10^{-4}) 	imes v_1 = (5 	imes 10^{-4}) 	imes 0.3$.$750 	imes v_1 = 5 	imes 0.3 = 1.5$.$v_1 = 1.5 / 750 = 15 / 7500 = 1 / 500 = 0.002 	ext{ m/s}$.Since the water level is falling, $dh/dt = -0.002 	ext{ m/s}$.Given $dh/dt = y 	imes 10^{-3} 	ext{ m/s}$, we have $y 	imes 10^{-3} = -2 	imes 10^{-3}$.Taking the magnitude, $y = 2$.

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