Water is filled in a container upto height 3m. A small hole of area 'a' is punched in the wall of the container at a height 52.5 cm from the bottom. The cross sectional area of the container is A. If `a//A=0.1` then `v^2` is (where v is the velocity of water coming out of the hole)

Options

A. `50 m^(2)//s^(2)`
B. `51 m^(2)//s^(2)`
C. `50.2 m^(2)//s^(2)`
D. `52 m^(2)//s^(2)`

Answer

A

Explanation

Let speed of the liquid at the top is v and the same at the orifice is u. Let a is the area of orifice and A is the area of container. We can write the following using equation of continuity.
`v=a/A u ..........(i)`
We can apply Bernoulli.s equation as follows: (At the point of top surface and orifice)
`p+pgh+1/2 pv^(2)=p+0+1/2 pu^(2)`
Substituting from equation (i) we get the following
`u^(2)=(2gh)/(1-(a/A)^(2))=(2 xx 10 xx (3-0.525))/(1-(0.1)^(2)`
`=50 m^(2)//s^(2)`