The cross-sectional areas of a tube `T_(1)` and the hole in the vessel at B are a and `a//2` respectively. There is a hole in the tube at C (at the level of A) through which liquid in the vessel rises by a height h in the tube. The other liquid heights are shown in the diagram. The plugs at A and B are removed simultaneously. How much horizontal force is required to keep vessel in equilibrium if p is the pressure in the tube and `p_(0)` is the atmospheric pressure? Hole C is closed when plugs are removed.

Options

A. `a(p_(0)-p)`
B. `(a)/(2)(p_(0)-p)`
C. `2a(p_(0)-p)`
D. `4a(p_(0)-p)`

Answer

C

Explanation

`p_(0)=p+rhogh,h=((p_(0)-p))/(rhog)`
`F_(A)=rhoaV^(2)=rhoa(2gxx(3h)/(2))`
`=3rho a g xx ((p_(0) -p))/(rho g) = 3a (p_(0) -p)`
`F_(B) = rho (a)/(2) V_(B)^(2) = a (p_(0) - p)`
`(as V_(B)^(2) = 2g.((p_(0)-p))/(rho g))`
`F_(A) - F_(b) = 2a (p_(0) - p)`