In the diagram shown,the difference in the he | vedclass

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(A) According to Bernoulli's theorem for horizontal flow:$P_A + \frac{1}{2}\rho v_A^2 = P_B + \frac{1}{2}\rho v_B^2$$P_A - P_B = \frac{1}{2}\rho (v_B^

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$7.5$

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(A) According to Bernoulli's theorem for horizontal flow:$P_A + \frac{1}{2}ho v_A^2 = P_B + \frac{1}{2}ho v_B^2$$P_A - P_B = \frac{1}{2}ho (v_B^2 - v_A^2)$Since the pressure difference is measured by the manometer height $h = 5\, cm$,we have $P_A - P_B = ho gh$.Thus,$\frac{1}{2}ho (v_B^2 - v_A^2) = ho gh \implies v_B^2 - v_A^2 = 2gh$.Given $g = 10\, m/s^2 = 1000\, cm/s^2$ and $h = 5\, cm$,we have $v_B^2 - v_A^2 = 2 	imes 1000 	imes 5 = 10000\, cm^2/s^2$.From the equation of continuity,$A_A v_A = A_B v_B$.Given $A_A = 6\, mm^2$ and $A_B = 10\, mm^2$,we have $6 v_A = 10 v_B \implies v_B = 0.6 v_A$.Substituting $v_B$ into the Bernoulli equation:$(0.6 v_A)^2 - v_A^2 = 10000$$0.36 v_A^2 - v_A^2 = 10000$ (Note: The pressure at $A$ is higher due to smaller area,so $v_A > v_B$. Correcting the sign: $v_A^2 - v_B^2 = 2gh$)$v_A^2 - (0.6 v_A)^2 = 10000$$v_A^2 (1 - 0.36) = 10000$$0.64 v_A^2 = 10000 \implies v_A^2 = \frac{10000}{0.64} = 15625$$v_A = \sqrt{15625} = 125\, cm/s$.The flow rate $Q = A_A v_A = 6\, mm^2 	imes 125\, cm/s = 0.06\, cm^2 	imes 125\, cm/s = 7.5\, cc/s$.

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