The figure shows a liquid of given density fl | vedclass

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(A) Given:Area at $A$,$A_A = 1.5 \, cm^2 = 1.5 	imes 10^{-4} \, m^2$Area at $B$,$A_B = 25 \, mm^2 = 25 	imes 10^{-6} \, m^2$Velocity at $B$,$v_B = 6

Vedclass · Accepted Answer

$175$

Vedclass · Answer

(A) Given:Area at $A$,$A_A = 1.5 \, cm^2 = 1.5 	imes 10^{-4} \, m^2$Area at $B$,$A_B = 25 \, mm^2 = 25 	imes 10^{-6} \, m^2$Velocity at $B$,$v_B = 60 \, cm/s = 0.6 \, m/s$Density,$ho = 1000 \, kg/m^3$Using the equation of continuity,$A_A v_A = A_B v_B$:$1.5 	imes 10^{-4} 	imes v_A = 25 	imes 10^{-6} 	imes 0.6$$v_A = \frac{25 	imes 10^{-6} 	imes 0.6}{1.5 	imes 10^{-4}} = \frac{15 	imes 10^{-6}}{1.5 	imes 10^{-4}} = 10 	imes 10^{-2} = 0.1 \, m/s$Using Bernoulli's equation for a horizontal tube $(h_A = h_B)$:$P_A + \frac{1}{2} ho v_A^2 = P_B + \frac{1}{2} ho v_B^2$$P_A - P_B = \frac{1}{2} ho (v_B^2 - v_A^2)$$P_A - P_B = \frac{1}{2} 	imes 1000 	imes ((0.6)^2 - (0.1)^2)$$P_A - P_B = 500 	imes (0.36 - 0.01)$$P_A - P_B = 500 	imes 0.35 = 175 \, Pa$

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