A steady flow of a liquid of density rho is s | vedclass

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(C) According to the equation of continuity,$A_1 v_1 = A_2 v_2$.Given $A_1 = 2A$,$v_1 = \sqrt{2} \ m \ s^{-1}$,and $A_2 = A$.Substituting these values

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$50$

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(C) According to the equation of continuity,$A_1 v_1 = A_2 v_2$.Given $A_1 = 2A$,$v_1 = \sqrt{2} \ m \ s^{-1}$,and $A_2 = A$.Substituting these values: $(2A)(\sqrt{2}) = (A)v_2 \Rightarrow v_2 = 2\sqrt{2} \ m \ s^{-1}$.For a steady flow,applying Bernoulli's equation at points $1$ and $2$:$P_1 + \rho g h_1 + \frac{1}{2} \rho v_1^2 = P_2 + \rho g h_2 + \frac{1}{2} \rho v_2^2$.Rearranging the terms: $P_1 - P_2 = \rho g (h_2 - h_1) + \frac{1}{2} \rho (v_2^2 - v_1^2)$.Given $P_1 - P_2 = 100 \ N \ m^{-2}$ (since pressure is higher at the wider section $1$ where speed is lower),$h_1 - h_2 = 10 \ cm = 0.1 \ m$,so $h_2 - h_1 = -0.1 \ m$.$100 = \rho [10(-0.1) + \frac{1}{2}((2\sqrt{2})^2 - (\sqrt{2})^2)]$.$100 = \rho [-1 + \frac{1}{2}(8 - 2)]$.$100 = \rho [-1 + 3] = 2\rho$.$\rho = 50 \ kg \ m^{-3}$.

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