MHTCET 16 Apr 2026 Shift 1 Paper

Concept:For an incompressible fluid, the equation of continuity states that the volume flow rate remains constant.$A_1 V = A_2 V_1$where $A_1$ and $A_2$ are the cross-sectional areas of the pipe and the nozzle, respectively.Explanation:Since water is incompressible, the volume of water passing per second through the pipe must equal that through the nozzle.The internal diameter of the pipe is $d$, so its area of cross-section is:$$A_1 = \frac{\pi d^2}{4}$$Let the diameter of the nozzle be $D$.Then the area of cross-section of the nozzle is:$$A_2 = \frac{\pi D^2}{4}$$Using the equation of continuity:$$\frac{\pi d^2}{4} V = \frac{\pi D^2}{4} V_1$$Cancel the common factor $\frac{\pi}{4}$ from both sides.$$d^2 V = D^2 V_1$$Rearranging for $D^2$:$$D^2 = \frac{d^2 V}{V_1}$$Taking the square root on both sides:$$D = d \sqrt{\frac{V}{V_1}}$$Thus, the diameter of the nozzle is directly proportional to the pipe diameter and depends on the ratio of speeds.Answer:The diameter of the nozzle is $d \sqrt{\frac{V}{V_1}}$.Hence, the correct option is A.