A cylindrical tank is filled with water to level of 3 ~m . A hole is opened at height of 52.5 ~cm from bottom. The ratio of the area of the hole to that of cross-sectional area of the cylinder is 0.1 . The square of the speed with which water is coming out from the orifice is (Take g=10 ~m ~s ⁻² )

Step-by-step solution

Suppose A be the area of cross section of tank, a be the area of hole, v_e be the velocity of efflux, h be the height of liquid above the hole, Let v be the speed with which the level decreases in the container. Using equation of continuity, we get a v_e=A v or v= a v_e A Using Bernoulli's theorem, we have aligned & P₀+h g+ 1 2 v^2=P₀+ 1 2 v_e^2 & h g+ 1 2 ( a v_e A )^2= 1 2 v_e^2 & or v_e^2= 2 h g 1- (a^2 / A^2 ) = 2 (3-0.525) 10 1-(0.1)^2 & =50 ~m ^2 ~s ⁻² aligned