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(B) Given:$A_1 = 0.02 \ m^2$,$V_1 = 2 \ m/s$,$P_1 = 4 	imes 10^4 \ N/m^2$,$A_2 = 0.01 \ m^2$,$ho = 1000 \ kg/m^3$.Using the equation of continuity,

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$3.4 	imes 10^4 \ N/m^2$

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(B) Given:$A_1 = 0.02 \ m^2$,$V_1 = 2 \ m/s$,$P_1 = 4 	imes 10^4 \ N/m^2$,$A_2 = 0.01 \ m^2$,$ho = 1000 \ kg/m^3$.Using the equation of continuity,$A_1 V_1 = A_2 V_2$:$0.02 	imes 2 = 0.01 	imes V_2 \implies V_2 = 4 \ m/s$.Using Bernoulli's equation for a horizontal tube $(h_1 = h_2)$:$P_1 + \frac{1}{2} ho V_1^2 = P_2 + \frac{1}{2} ho V_2^2$$P_2 = P_1 + \frac{1}{2} ho (V_1^2 - V_2^2)$$P_2 = 4 	imes 10^4 + \frac{1}{2} 	imes 1000 	imes (2^2 - 4^2)$$P_2 = 40000 + 500 	imes (4 - 16)$$P_2 = 40000 - 6000 = 34000 \ N/m^2 = 3.4 	imes 10^4 \ N/m^2$.

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