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

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Question

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

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(D) 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 velocity of water at the tap is $v_2 = 30 \, cm/s = 0.3 \, m/s$.The velocity of the water surface $v_1$ is equal to $|\frac{dh}{dt}|$.Substituting the values: $750 	imes 10^{-4} 	imes |\frac{dh}{dt}| = 500 	imes 10^{-6} 	imes 0.3$.$|\frac{dh}{dt}| = \frac{500 	imes 0.3 	imes 10^{-6}}{750 	imes 10^{-4}} = \frac{150 	imes 10^{-6}}{750 	imes 10^{-4}} = 0.2 	imes 10^{-2} \, m/s = 2 	imes 10^{-3} \, m/s$.Since the height $h$ is decreasing,$\frac{dh}{dt} = -2 	imes 10^{-3} \, m/s$.Comparing with $x 	imes 10^{-3} \, m/s$,we get $x = 2$ (taking the magnitude as per standard convention for such problems).

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