A cylindrical tank having cross-sectional area `A = 0.5 m^(2)` is filled with two liquids of density `rho_(1)=900 kg m^(-3)` and `rho_(2)=600 kgm^(-3)`, to a height `h=60cm` each as shown in the figure. A small hole having area `a=5cm^(2)` is made in right vertical wall at a height `y = 20 cm` from the bottom. A horizontal force `F` is applied on the tank to keep it in static equilibrium. The tank is lying on a horizontal surface. Neglect mass of the cylindrical tank in comparison to mass of the liquids (`ake `g = 10 ms^(-2)`).

The horizontal force required to keep the cylinder in static equilibrium, on a smooth horizontal plane is

Options

A. 10.5N
B. 7.2N
C. 6N
D. 11.2N

Explanation

Since area of a hole is very small in comparison to base area A of the cylinder .
Therefore , velocity of liquid inside the cylinder is negligible .
Let velocity of efflux be v and atmosphere pressure `P_(0)`
Consider two points A ( inside the cylinder ) and B ( just outside the hole ) in the same horizontal line as shown in figure .
Pressure at `A_(1) P_(A) = P_(0) + h rho_(2) + " " (h-y)rho_(1) g`
Pressure at `B_(1) " " P_(B) = P_(0)`
According to Bernoulli.s theorem
Pressure energy at `A_(1) ` = Pressure energy at B + Kinetic energy at B
` :. P_(A) = P_(B) + 1/2 rho_(1)v^(2)`
` :. v = 4 ms^(-1)`