$A$ cylindrical tank of height $0.4\,m$ is open at the top and has a diameter $0.16\,m$. Water is filled in it up to a height of $0.16\,m$. How long will it take to empty the tank through a hole of radius $5 \times 10^{-3}\,m$ in its bottom? (in seconds)

There are two identical small holes of cross-sectional area $a$ on the opposite sides of a tank containing a liquid of density $\rho$. The difference in height between the holes is $h$. The tank is resting on a smooth horizontal surface. The horizontal force that must be applied to the tank to keep it in equilibrium is

$A$ vertical tank,open at the top,is filled with a liquid and rests on a smooth horizontal surface. $A$ small hole is opened at the centre of one side of the tank. The area of cross-section of the tank is $N$ times the area of the hole,where $N$ is a large number. Neglect the mass of the tank itself. The initial acceleration of the tank is:

$A$ light cylindrical vessel is kept on a horizontal surface. The area of the base is $A$. $A$ hole of cross-sectional area $a$ is made just at its bottom side. The minimum coefficient of friction necessary to prevent the vessel from sliding due to the impact force of the emerging liquid is $(a \ll A)$.

Consider a water jar of radius $R$ that has water filled up to height $H$ and is kept on a stand of height $h$ (see figure). Through a hole of radius $r$ $(r << R)$ at its bottom,the water leaks out and the stream of water coming down towards the ground has a shape like a funnel as shown in the figure. If the radius of the cross-section of water stream when it hits the ground is $x$,then:

$A$ large tank filled with water to a height $h$ is to be emptied through a small hole at the bottom. The ratio of times taken for the level of water to fall from $h$ to $h/2$ and from $h/2$ to zero is