The area of cross section of a large tank is `0.5m^2`. It has an opening near the bottom having area of cross section `1cm^2`. A load of 20 kg is applied on the water at the top. Find the velocity of the water coming out of the opening at the time when the height of water level is 50 cm above the bottom. Take `g=10ms^-2`

Explanation

as the area of cross section of the tank is large compared to that of the opening, the speed of water in the tank will be very smal as compared to the speed at the opening. The pressure at the surface of water in the tank is that due to the atmosphere plus due to the load.
`P_A=P_0+((20kg)(10ms^-2))/(0.5m^2)=P_0+400Nm^-2`
At the opening the pressure is that due to the atmosphere,.
Using Bernoulli equation
`P_A+rhogh+1/2rhov_A^2=P_B+1/2rhov_B^2`
`or, P_0+400Nm^-2+(1000kgm^-3)(10ms^-2)(0.5m)+0`
`=P_0+/2(1000kgm^-3)v_B^2`
`5400Nm^-2=(500kgm^-1)v_B^2`
`or `v_B=3.3ms^-1`