A liquid flows through a horizontal tube. The velocities of the liquid in the two sections, which have areas of cross-section ${A_1}$ and ${A_2}$, are ${v_1}$ and ${v_2}$ respectively. The difference in the levels of the liquid in the two vertical tubes is $ h$
Diffcult
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(d) According to equation of continuity the volume of liquid flowing through the tube in unit time remains constant i.e. ${A_1}{v_1} = {A_2}{v_2},$hence option $ (a)$ is correct
According to Bernoulli's theorem,
${P_1} + \frac{1}{2}\rho v_1^2 = {P_2} + \frac{1}{2}\rho v_2^2$
==> ${P_1} - {P_2} = \frac{1}{2}\rho \left( {v_2^2 - v_1^2} \right)$==> $h\rho g = \frac{1}{2}\rho \left( {v_2^2 - v_1^2} \right)$
$v_2^2 - v_1^2 = 2gh$
Hence option $(c)$ is correct.
Also, according to Bernoulli's theorem option $ (d) $ is correct
According to Bernoulli's theorem,
${P_1} + \frac{1}{2}\rho v_1^2 = {P_2} + \frac{1}{2}\rho v_2^2$
==> ${P_1} - {P_2} = \frac{1}{2}\rho \left( {v_2^2 - v_1^2} \right)$==> $h\rho g = \frac{1}{2}\rho \left( {v_2^2 - v_1^2} \right)$
$v_2^2 - v_1^2 = 2gh$
Hence option $(c)$ is correct.
Also, according to Bernoulli's theorem option $ (d) $ is correct
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