$A$ cylindrical vessel is filled with water up to a height of $1 \,m$ from the base. $A$ small orifice is opened at some height in the cylinder, and the water level is reduced to the height of the orifice in $20 \,s$. If the base area of the cylinder is $100$ times the area of the orifice, then the height of the orifice from the base is (take $g = 10 \,m/s^2$): (in $\,cm$)

$A$ large tank is filled with water to a height $H$. $A$ small hole is made at the base of the tank. It takes $T_1$ time to decrease the height of water to $H/\eta$ (where $\eta > 1$),and it takes $T_2$ time to drain the rest of the water. If $T_1 = T_2$,then the value of $\eta$ is:

$A$ cylinder of height $h$ is filled with water and is kept on a block of height $h/2$. The level of water in the cylinder is kept constant. Four holes numbered $1, 2, 3$ and $4$ are at the side of the cylinder and at heights $0, h/4, h/2$ and $3h/4$ from the base of the cylinder, respectively. When all four holes are opened together, the hole from which water will reach the farthest distance on the plane $PQ$ is the hole number:

$A$ rectangular tank is placed on a horizontal ground and is filled with water to a height $H$ above the base. $A$ small hole is made on one vertical side at a depth $D$ below the level of the water in the tank. The distance $x$ from the bottom of the tank at which the water jet from the tank will hit the ground is

$A$ closed pipe containing liquid showed a pressure $P_{1}$ by gauge. When the valve is opened,the pressure was reduced to $P_{2}$. The speed of water flowing out of the pipe is $[\rho = \text{density of water}]$

There are two holes, one each along the opposite sides of a wide rectangular tank. The cross-section of each hole is $0.01 \,m^2$ and the vertical distance between the holes is $1 \,m$. The tank is filled with water. The net force on the tank in newton when the water flows out of the holes is (Density of water $= 1000 \,kg/m^3$, $g = 10 \,m/s^2$)