MHTCET 10 May 2024 Shift 1 Paper

Concept:The problem uses Bernoulli’s principle for the flow of liquid from a pipe when pressure changes.Explanation:Initially, the liquid is at rest inside the closed pipe, so the initial velocity is $v_1 = 0$.The initial gauge pressure is $P_1$.After the valve is opened, the pressure becomes $P_2$ and the liquid flows out with speed $v_2$.Applying Bernoulli’s equation between the two points:$$P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2$$Since $v_1 = 0$, this simplifies to:$$P_1 = P_2 + \frac{1}{2}\rho v_2^2$$Rearranging gives:$$\frac{1}{2}\rho v_2^2 = P_1 - P_2$$Therefore, the speed of water flowing out is:$$v_2 = \sqrt{\frac{2(P_1 - P_2)}{\rho}}$$Answer:The correct option is C: $\left[\frac{2\left(\mathrm{P}_{1}-\mathrm{P}_{2}\right)}{\rho}\right]^{1 / 2}$.