A long cylindrical tank of cross-sectional area `0.5m^(2)` is filled with water. It has a small hole at a height `50cm` from the bottom. A movable piston of cross-sectional area almost equal to `0.5m^(2)` is fitted on the top of the tank such that it can slide in the tank freely. A load of `20 kg` is applied on the top of the water by piston, as shown in the figure. Calculate the speed of the water jet with which it hits the surface when piston is 1m above the bottom. (Ignore the mass of the piston)

Answer

A, D

Explanation

`1/2 rho v^(2)=rho gh +(mg)/(A)`
Here, `h=1.0-0.5=0.5m`
`A =`area of piston `=0.5m^(2)`
`:.v=sqrt(2gh +(2mg)/(rho A))`
`=sqrt(2xx9.8xx0.5+(2xx20xx9.8)/(10^(3)xx0.5))`
`=3.25m//s`
`:.` Speed with which it hits the surface is
`v'=sqrt(v^(2)=2gh')
`=sqrt((3.25)^(2)+(2xx9.8xx0.5))`
`=4.51m//s`