A spherical ball of radius $1 \times 10^{-4} \mathrm{~m}$ and of density $10^4 \mathrm{kgm}^{-3}$ falls freely under gravity through a distance $h$ before entering a tank of water. After entering water if the velocity of the ball does not change, then $h$ is (The coefficient of viscosity of water $9.8 \times 10^{-6} \mathrm{Nsm}^{-2}$ )

A large tank filled water to a height $h$ is to be emptied through a small hole at the bottom. The ratio of the time taken for the level to fall from $h$ to $\frac{h}{2}$ and that taken for the level to fall from $\frac{h}{2}$ to 0 is

Depth of a river is 100 m magnitude of compressibility of the water is $05 \times 10^{-9} \mathrm{~N}^{-1} \mathrm{~m}^2$. The fractional compression in water at the bottom of the river is (acceleration due to gravity $=10 \mathrm{~m} / \mathrm{s}^2$ )

Two mercury drops, each with same radius $r$ merged to form a bigger drop. $T$ is the surface tension of single drop, then the surface energy of bigger drop is given by