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A liquid flows through a horizontal tube. The velocities of the liquid in the two sections, which have areas of crosssection $A_1$ and $A_2$, are ${ }^{v_1}$ and ${ }^{v_2}$ respectively. The difference in the levels of the liquid in the two vertical tubes is $h$
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The volume of the liquid flowing through the tube in unit time is $A_1 v_1$, $v_2^2-v_1^2=2 g h$, The energy per unit mass of the liquid is the same in both sections of the tube
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 (1) is correct
According to Bernoulli's theorem,
$P_1+\frac{1}{2} \rho v_1^2=P_2+\frac{1}{2} \rho v_2^2$
$\Rightarrow P_1-P_2=\frac{1}{2} \rho\left(v_2^2-v_1^2\right) \Rightarrow h \rho g=\frac{1}{2} \rho\left(v_2^2-v_1^2\right)$
$\therefore v_2^2-v_1^2=2 g h$
Hence option (3) is correct.
Also, according to Bernoulli's theorem option (4) is correct
According to Bernoulli's theorem,
$P_1+\frac{1}{2} \rho v_1^2=P_2+\frac{1}{2} \rho v_2^2$
$\Rightarrow P_1-P_2=\frac{1}{2} \rho\left(v_2^2-v_1^2\right) \Rightarrow h \rho g=\frac{1}{2} \rho\left(v_2^2-v_1^2\right)$
$\therefore v_2^2-v_1^2=2 g h$
Hence option (3) is correct.
Also, according to Bernoulli's theorem option (4) is correct
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