MHTCET 19 Apr 2026 Shift 1 Paper

Concept:Steady flow of an incompressible fluid obeys the equation of continuity: the volume flow rate remains constant along the pipe.Explanation:For a horizontal cylindrical pipe, the equation of continuity is$$A_A v_A = A_B v_B$$where $A$ is the cross-sectional area and $v$ is the flow velocity.The cross-sectional area of a pipe of radius $r$ is$$A = \pi r^2$$At point A, the radius is $3R$, so$$A_A = \pi (3R)^2 = 9\pi R^2$$At point B, the radius is $1.5R$, so$$A_B = \pi (1.5R)^2 = 2.25\pi R^2$$Given that the velocity at A is $v$, substitute into the continuity equation:$$9\pi R^2 \cdot v = 2.25\pi R^2 \cdot v_B$$Cancel $\pi R^2$ from both sides:$$9v = 2.25v_B$$$$v_B = \frac{9}{2.25}v = 4v$$Thus, the velocity at the narrower point B becomes four times the velocity at A.Answer:The velocity at point B is $4v$.Correct option: D.