A tank of height 15 m and cross-section area 10 `m^(2)` is filled with water. There is a small hole of cross-section area a which is much smaller than the container, located at a height of 12 m from the base of the container. How much force should be applied with coming out of the top level, so that water coming out of the holwe hits the ground at a distance of 16 m? `("Take of water =1000 kg m"^(-3))`

Options

A. 233 kN
B. 200 kN
C. 320 kN
D. 400 kN

Answer

A

Explanation

As, range of water jet = 16 m.
If velocity of efflux is v, then
`v xx sqrt((2H)/(g))=` 16, where H = 12 m
Also, `v = sqrt(2gh+(2(p-p_(a)))/(P)),P = 1000 " kg-m"^(-3)` and h = 3 m
So combining these, we get
`(2gh+(2(p-p_(a)))/(P))((2H)/(g))=256`
`rArr 2 xx 10 xx 3 + (2Deltap)/(1000)=(2560)/(2xx12)rArr Delta p = 23 3"k" (N)/(m^(2))`
So, force applied `=Deltap xx A = 233` kN