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IIT-JEE 2009 · Paper 2, Physics question 51

A cylindrical vessel of height 500 mm has an orifice (small hole) at its bottom. The orifice is initially closed and water is filled in it up to height H. Now the top is completely sealed with a cap and the orifice at the bottom is opened. Some water comes out from the orifice and the water level in the vessel becomes steady with height of water column being 200 mm. Find the fall in height (in mm) of water level due to opening of the orifice. [Take atmospheric pressure =1.0×105 N/m2=1.0\times10^5\ \mathrm{N/m^2}, density of water =1000 kg/m3=1000\ \mathrm{kg/m^3} and g=10 m/s2g=10\ \mathrm{m/s^2}. Neglect any effect of surface tension.]

Answer and worked solution

Official answer: 6

Final state. Water stops flowing when the pressure at the orifice equals atmospheric:
P+ρg(0.2)=P0  ⟹  P=105−2000=9.8×104 N/m2.P+\rho g(0.2)=P_0\implies P=10^5-2000=9.8\times10^4\ \mathrm{N/m^2}.

Boyle's law for the trapped air (initially at P0P_0, length 500−H500-H mm; finally length 300 mm):
105(500−H)=9.8×104(300)  ⟹  500−H=294  ⟹  H=206 mm.10^5(500-H)=9.8\times10^4(300)\implies500-H=294\implies H=206\ \text{mm}.

Fall in level: 206−200=6206-200=6 mm.

Answer: 6

Topics:Pressure and Pascal's lawIdeal gas equation
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