Question Detail

For ideal fluid, velocity profile is uniform: v = v₀, so flow rate Q = v₀ × πr². For viscous fluid in a pipe, the flow is Poiseuille flow with parabolic profile: v(r) = v_max (1 - r²/R²). The average velocity v_avg = v_max/2. Flow rate Q = v_avg × πr² = (v_max/2) πr². Setting Q equal: (v_max/2) πr² = v₀ πr² => v_max/2 = v₀ => v_max = 2v₀? Wait, that gives v_max = 2v₀, not in options. If Q is same, then v_avg for viscous = v₀ (ideal speed). So v_max/2 = v₀ => v_max = 2v₀. Not in options. If they meant the viscous liquid has the same volume flow rate as the ideal fluid, and v₀ is the speed of ideal fluid, then v_max = 2v₀. Option B is 1/2 v₀, which is v_avg. Maybe they ask for average speed? The question asks for maximum speed. Possibly they meant v₀ is the average speed for ideal? If v₀ is the max speed for ideal? Rethink: For ideal, v₀ is the speed (uniform). For viscous, max speed is 2 × average speed. If average speed is same as ideal speed v₀, then max speed = 2v₀. Not in options. If they've swapped definitions, maybe v₀ is the max speed for ideal? Then for viscous, max speed would be? The only logical from given options is 1/2 v₀ if they meant the average speed for viscous is half of something. Possibly a misprint. The correct physics: v_max(viscous) = 2 × v_avg = 2 × v₀(ideal) if flow rates equal.