A cylindrical tank has a hole of area 2 ~cm ^2 at its bottom, If water is poured into the tank from a tube above it at the rate of 100 ~cm ^3 ~s ⁻¹ , then the maximum height upto which water can rise in the tank is (Acceleration due to gravity, g=10 ~ms ⁻² )

Step-by-step solution

Given, area of hole in tank, A=2 ~cm ⁻² aligned & A=2 10⁻⁴ ~m ⁻² & Volume flow rate =100 ~cm ^2 / sec & =100 10⁻⁶ ~m ^3 / sec =10⁻⁴ ~m ^2 sec ⁻¹ aligned At maximum height h , velocity of water flowing through hole, v= 2 g h From principle of continuity of flow of liquid, volume flow rate =A v array rlrl & & 10⁻² & =A v=2 10⁻⁴ 2 g h v= 2 g h & & 10⁻⁴ 2 10⁻⁴ & = 2 g h & & 1 2 & = 2 g h & & . 1 2 )⁻⁴ & =2 g h & 1 4 1 2 g & =h & h & = 1 80 =0.125 ~m & & h & =1.25 ~cm array