Water flows in a horizontal pipe of non-uniform area of cross-section at a pressure difference of 1.6 cm of mercury. If the velocity of water at the larger cross-section of pipe is 50 cm/s, find the velocity of water at the other end?

Correct option is C

​$$\text{Bernoulli's equation for two points in the fluid is given by:} \\P + \dfrac{1}{2} \rho v^2 + Z = C \\\text{Where:} \\P = \text{Pressure at a point,} \\\rho = \text{Density of the fluid,} \\v = \text{Velocity at a point,} \\Z = \text{Height at a point,} \\C = \text{Constant for an incompressible, steady fluid.} \\\text{For a horizontal pipe, } Z_1 = Z_2, \text{ and Bernoulli's equation becomes:} \\P_1 + \dfrac{1}{2} \rho v_1^2 = P_2 + \dfrac{1}{2} \rho v_2^2 \\\text{Rearranging for } v_2, \text{ we get:} \\\Delta P = \dfrac{1}{2} \rho \left( v_2^2 - v_1^2 \right)$$

$$\text{Given:} \\v_1 = 50 \, \text{cm/s} = 0.5 \, \text{m/s}, \\\text{Pressure difference } \Delta P = 1.6 \, \text{cm of mercury} = 0.016 \times 13600 \times 9.81 = 2134.65 \, \text{Pa}, \\\rho \text{ for water is approximately } 1000 \, \text{kg/m}^3. \\2134.65 = \dfrac{1}{2} \times 1000 \times \left( v_2^2 - (0.5)^2 \right) \\2134.65 = 500 \times \left( v_2^2 - 0.25 \right) \\\dfrac{2134.65}{500} = v_2^2 - 0.25 \\4.2693 = v_2^2 - 0.25 \\v_2^2 = 4.2693 + 0.25 = 4.5193 \\v_2 = \sqrt{4.5193}$$​​​