$A$ cylindrical container containing water has a small hole at a height of $H = 8 \text{ cm}$ from the bottom and at a depth of $h = 2 \text{ cm}$ from the top surface of the liquid. The horizontal distance (range) travelled by the water before it hits the ground is:

There are two identical small holes on the opposite sides of a tank filled with a liquid. The tank is open at the top. The difference in height between the two holes is $h$. As the liquid comes out of the two holes,the tank will experience a net horizontal force proportional to

$A$ tank is filled with water up to a height $H$. Water is allowed to come out of a hole in one of the walls at a depth $D$ below the surface of water. Express the horizontal distance $x$ in terms of $H$ and $D$.

$A$ cylindrical tank having a large diameter is filled with water to a height $H$. $A$ hole of cross-sectional area $5 \,cm^2$ in the tank allows water to drain out. If the water drains out at the rate of $2 \times 10^{-3} \,m^3 \,s^{-1}$, then the value of $H$ is (acceleration due to gravity $= 10 \,ms^{-2}$) (in $\,cm$)

$A$ large open tank has two holes in the wall. One is a square hole of side $L$ at a depth $y$ from the top and the other is a circular hole of radius $R$ at a depth $4y$ from the top. When the tank is completely filled with water, the quantities of water flowing out per second from the two holes are the same. Then the value of $R$ is

$A$ cubical box of wine has a small spout located in one of the bottom corners. When the box is full and placed on a level surface,opening the spout results in a flow of wine with an initial speed of $v_0$ (see figure). When the box is half empty,someone tilts it at $45^{\circ}$ so that the spout is at the lowest point (see figure). When the spout is opened,the wine will flow out with a speed of