There are two identical small holes of area of cross-section a on the opposite sides of a tank containing a liquid of density $\rho $. The difference in height between the holes is $h$. Tank is resting on a smooth horizontal surface. Horizontal force which will have to be applied on the tank to keep it in equilibrium is
$g\,h\,\rho a$
$\frac{{2gh}}{{\rho a}}$
$2g\,h\,\rho a$
$\frac{{\rho gh}}{a}$
A liquid is kept in a cylindrical vessel which is being rotated about a vertical axis through the centre of the circular base. If the radius of the vessel is $r$ and angular velocity of rotation is $\omega $ , then the difference in the heights of the liquid at the centre of the vessel and the edge is
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 both holes are the same. Then, $R$ is equal to
Two small drops of mercury, each of radius $R$ coalesce to form a single large drop. The radio of the total surface energies before and after the change is
The work done in splitting a drop of water of $1\, mm$ radius into $10^6$ droplets is (surface tension of water $72\times10^{-3}\, N/m$) :
If the terminal speed of a sphere of gold (density $19.5 \,kg / m ^2$ ) is $0.2 \,m / s$ in a viscous liquid (density $=1.5 \,kg / m ^3$ ), find the terminal speed of a sphere of silver (density $=10.5 \,kg / m ^3$ ) of the same size in the same liquid is ............ $m / s$