MHTCET 16 Apr 2026 Shift 1 Paper

Concept:When the valve is opened, the pressure energy of the stationary liquid is converted into kinetic energy of the flowing water.This conversion is described by Bernoulli’s principle for flow at the same height.Explanation:Initially, the pipe is closed, so the liquid is at rest and the gauge reads the static pressure $P_1$.After the valve is opened, the water flows out, and part of the pressure energy is transformed into kinetic energy, reducing the pressure to $P_2$.Applying Bernoulli’s equation for the same horizontal level:$$P_1 = P_2 + \frac{1}{2}\varrho v^2$$Here, $\varrho$ is the density of water and $v$ is the speed of outflow.Rearranging gives:$$P_1 - P_2 = \frac{1}{2}\varrho v^2$$Solving for $v$:$$v^2 = \frac{2(P_1 - P_2)}{\varrho}$$$$v = \left[\frac{2(P_1 - P_2)}{\varrho}\right]^{\frac{1}{2}}$$Answer:The speed of water flowing out is:$$v = \left[\frac{2(P_1 - P_2)}{\varrho}\right]^{\frac{1}{2}}$$Therefore, the correct option is Option B.