$A$ sphere of solid material of relative density $9$ has a concentric spherical cavity and floats just submerged in water. If the radius of the sphere is $R$,then the radius of the cavity $(r)$ is related to $R$ as:

$A$ cubical block is floating in a liquid with half of its volume immersed in the liquid. When the whole system accelerates upwards with a net acceleration of $g/3$,the fraction of volume immersed in the liquid will be:

$A$ body having volume $V$ and density $\rho$ is attached to the bottom of a container as shown. The density of the liquid is $d$ (where $d > \rho$). The container has a constant upward acceleration $a$. The tension in the string is:

$A$ homogeneous solid cylinder of length $L$ $(L < H/2)$, cross-sectional area $A$ is immersed such that it floats with its axis vertical at the liquid-liquid interface with length $L/4$ in the denser liquid (density $2d$) and $3L/4$ in the lighter liquid (density $d$). The lower density liquid is open to the atmosphere having pressure $P_0$. Then, the density $D$ of the solid is given by:

$A$ machine is blowing spherical soap bubbles of different radii filled with helium gas. It is found that,if the bubbles have a radius smaller than $1 \, cm$,then they sink to the floor in still air. Larger bubbles float in the air. Assume that the thickness of the soap film in all bubbles is uniform and equal. Assume that the density of soap solution is same as that of water $\left(= 1000 \, kg \, m^{-3}\right)$. The density of helium inside the bubbles and air are $0.18 \, kg \, m^{-3}$ and $1.23 \, kg \, m^{-3}$,respectively. Then,the thickness of the soap film of the bubbles is .......... $\mu m$ (Note: $1 \, \mu m = 10^{-6} \, m$)

$A$ cube of side $40 \ cm$ is floating with $\frac{1}{4}$th of its volume immersed in water. When a circular disc is placed on the cube, it floats with $\frac{2}{5}$th of its volume immersed in water. The mass of the disc is (in $kg$)