$A$ large tank is filled with water (density $= 10^3 \ kg/m^3$). $A$ small hole is made at a depth $10 \ m$ below the water surface. The range of water issuing out of the hole is $R$ on the ground. What extra pressure must be applied on the water surface so that the range becomes $2R$? (Take $1 \ atm = 10^5 \ Pa$ and $g = 10 \ m/s^2$)

$A$ 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:

If in a stationary lift,a man is standing with a bucket full of water,having a hole at its bottom. The rate of flow of water through this hole is $R_0$. If the lift starts to move up and down with the same acceleration $a$,and the rates of flow of water are $R_u$ and $R_d$ respectively,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

For the shown diagram,find the ratio of $R_1: R_2$.

$A$ 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 $\,cm$)