The diagram shows a venturimeter through whic | vedclass

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(B) According to Bernoulli's principle for horizontal flow,$P_1 + \frac{1}{2}\rho V_1^2 = P_2 + \frac{1}{2}\rho V_2^2$.Here,$P_1 - P_2 = \rho gh$,wher

Vedclass · Accepted Answer

$32$

Vedclass · Answer

(B) According to Bernoulli's principle for horizontal flow,$P_1 + \frac{1}{2}ho V_1^2 = P_2 + \frac{1}{2}ho V_2^2$.Here,$P_1 - P_2 = ho gh$,where $h = 5.1\,mm = 0.51\,cm$.Substituting this into the equation: $\frac{1}{2}ho V_2^2 - \frac{1}{2}ho V_1^2 = ho gh$.$V_2^2 - V_1^2 = 2gh$.Given $V_1 = 2\,cm/s$,$g = 1000\,cm/s^2$,and $h = 0.51\,cm$.$V_2^2 = V_1^2 + 2gh = (2)^2 + 2 	imes 1000 	imes 0.51 = 4 + 1020 = 1024$.$V_2 = \sqrt{1024} = 32\,cm/s$.

The diagram shows a venturimeter through which water is flowing. The speed of water at $X$ is $2\,cm/s$. The speed of water at $Y$ (taking $g = 1000\,cm/s^2$) is ........ $cm/s$.

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