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(A) According to $Bernoulli's$ theorem for horizontal flow:${P_1} + \frac{1}{2}\rho v_1^2 = {P_2} + \frac{1}{2}\rho v_2^2$${P_1} - {P_2} = \frac{1}{2}

Vedclass · Accepted Answer

$5$

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(A) According to $Bernoulli's$ theorem for horizontal flow:${P_1} + \frac{1}{2}ho v_1^2 = {P_2} + \frac{1}{2}ho v_2^2$${P_1} - {P_2} = \frac{1}{2}ho (v_2^2 - v_1^2) \, ... (i)$Given: ${P_1} - {P_2} = 3 	imes 10^5 \, N \, m^{-2}$,$ho = 1000 \, kg \, m^{-3}$,and $\frac{A}{A'} = 5$.From the equation of continuity,$A v_1 = A' v_2$,so $\frac{v_2}{v_1} = \frac{A}{A'} = 5$,which means $v_2 = 5v_1$.Substituting $v_2 = 5v_1$ into equation $(i)$:$3 	imes 10^5 = \frac{1}{2} 	imes 1000 	imes ((5v_1)^2 - v_1^2)$$3 	imes 10^5 = 500 	imes (25v_1^2 - v_1^2)$$3 	imes 10^5 = 500 	imes 24v_1^2$$3000 = 120v_1^2$$v_1^2 = \frac{3000}{120} = 25$$v_1 = 5 \, m \, s^{-1}$.

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