A liquid flows through a horizontal tube. The velocities of the liquid in the two sections, which have areas of cross sections, A1 and A2, are v1 and v2 respectively. The difference in the level of the liquid in the two vertical tubes is h. Then

a) The volume of the liquid flowing through the tube in unit time is A1v1
b) v2 - v1 = √(2gh)
c) v2² - v1² = 2gh
d) The energy per unit mass of the liquid is same in both the tubes

To analyze this problem, we can use the principles of fluid dynamics, particularly the Bernoulli's equation, which relates the speed, pressure, and height of a flowing fluid.

Here's a breakdown of the options provided:

**a) The volume of the liquid flowing through the tube in unit time is \(A_1 v_1\):**

This statement is true. The volumetric flow rate through the tube is given by the product of the cross-sectional area \(A_1\) and the velocity \(v_1\).

**b) \(v_2 - v_1 = \sqrt{2gh}\):**

To derive this, apply Bernoulli's equation between the two sections of the tube:

\[ P_1 + \frac{1}{2} \rho v_1^2 + \rho gh_1 = P_2 + \frac{1}{2} \rho v_2^2 + \rho gh_2 \]

Here, \(h_1 - h_2 = h\) (since \(h\) is the difference in the height of the liquid in the two vertical tubes).

If we assume that the pressure difference between the two sections is negligible (i.e., \(P_1 \approx P_2\)), Bernoulli's equation simplifies to:

\[ \frac{1}{2} \rho v_1^2 + \rho gh = \frac{1}{2} \rho v_2^2 \]

Rearrange this equation to solve for \(v_2 - v_1\):

\[ \frac{1}{2} \rho v_2^2 - \frac{1}{2} \rho v_1^2 = \rho gh \]

\[ v_2^2 - v_1^2 = 2gh \]

So, \(v_2 - v_1\) is not directly \(\sqrt{2gh}\); rather, the correct relation is \(v_2^2 - v_1^2 = 2gh\).

**c) \(v_2^2 - v_1^2 = 2gh\):**

This statement is true based on the derived Bernoulli's equation relation.

**d) The energy per unit mass of the liquid is the same in both the tubes:**

The energy per unit mass of the liquid includes the kinetic energy term \(\frac{1}{2} v^2\) and the potential energy term \(\rho gh\). For Bernoulli's principle, the total energy per unit mass (kinetic + potential) is conserved, so this statement is true if we consider the conservation of mechanical energy.

Therefore, the correct answers are **a, c, and d**.