An ideal liquid of density ρ is filled in a horizontally fixed syringe fitted with piston. There is no friction between the piston and the inner surface of the syringe. Cross-section area of the syringe is A. At one end of the syringe, an orifice of negligible cross-section area is made. When the piston is pushed into the syringe, the liquid comes out of the orifice following parabolic path and falls on the ground. With what speed the liquid strikes the ground? Neglect the air drag :-
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Question

An ideal liquid of density ρ is filled in a horizontally fixed syringe fitted with piston. There is no friction between the piston and the inner surface of the syringe. Cross-section area of the syringe is A. At one end of the syringe, an orifice of negligible cross-section area is made. When the piston is pushed into the syringe, the liquid comes out of the orifice following parabolic path and falls on the ground. With what speed the liquid strikes the ground? Neglect the air drag :-

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G1 Physics · Accepted Answer

Step 1: Pushing the piston creates gauge pressure F/A, so Bernoulli from inside the syringe to the orifice gives the exit speed u with u² = 2F/(ρA). Step 2: The jet then falls through height h, gaining vertical speed √(2gh). Hence impact speed v = √(u²+2gh) = √[2(F+ρghA)/(ρA)]. Correct answer: D.