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(B) According to the equation of continuity,$A_1 v_1 = A_2 v_2$,where $A_1 = 10^{-4}\,m^2$ and $v_1 = 1.0\,ms^{-1}$.Thus,$A_2 v_2 = A_1 v_1 = 10^{-4}

Vedclass · Accepted Answer

$5	imes 10^{-5}\,m^2$

Vedclass · Answer

(B) According to the equation of continuity,$A_1 v_1 = A_2 v_2$,where $A_1 = 10^{-4}\,m^2$ and $v_1 = 1.0\,ms^{-1}$.Thus,$A_2 v_2 = A_1 v_1 = 10^{-4} 	imes 1 = 10^{-4}\,m^3s^{-1} \dots (1)$.Using Bernoulli's equation for a streamline flow under constant pressure: $P + \frac{1}{2}ho v_1^2 + ho gh_1 = P + \frac{1}{2}ho v_2^2 + ho gh_2$.Since the pressure is constant,we have $\frac{1}{2}ho v_1^2 + ho gh_1 = \frac{1}{2}ho v_2^2 + ho gh_2$.Rearranging gives $v_2^2 - v_1^2 = 2g(h_1 - h_2) = 2gh$,where $h = 0.15\,m$.$v_2 = \sqrt{v_1^2 + 2gh} = \sqrt{1^2 + 2 	imes 10 	imes 0.15} = \sqrt{1 + 3} = \sqrt{4} = 2\,ms^{-1}$.Substituting $v_2$ into equation $(1)$: $A_2 	imes 2 = 10^{-4}$.Therefore,$A_2 = \frac{10^{-4}}{2} = 0.5 	imes 10^{-4} = 5 	imes 10^{-5}\,m^2$.

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