$A$ body of density $d_1$ is counterpoised by $M$ of weights of density $d_2$ in air of density $d$. Then the true mass of the body is

$A$ wooden block of volume $1000\, cm^3$ is suspended from a spring balance. It weighs $12\, N$ in air. It is suspended in water such that half of the block is below the surface of water. The reading of the spring balance is ...... $N$.

$A$ body of density $\rho$ is dropped from rest from a height $h$ into a lake of density $\sigma$,where $\sigma > \rho$. Neglecting all dissipative forces,the maximum depth to which the body sinks before returning to float on the surface is:

$A$ wooden block floating in a bucket of water has $\frac{4}{5}$ of its volume submerged. When a certain amount of oil is poured into the bucket,it is found that the block is just under the oil surface with half of its volume under water and half in oil. The density of oil relative to that of water is

$A$ cylindrical block floats vertically in a liquid of density $\rho_{1}$ kept in a container such that the fraction of volume of the cylinder inside the liquid is $x_{1}$. Then some amount of another immiscible liquid of density $\rho_{2} (\rho_{2} < \rho_{1})$ is added to the liquid in the container so that the cylinder now floats just fully immersed in the liquids with $x_{2}$ being the fraction of volume of the cylinder inside the liquid of density $\rho_{1}$. The ratio $\rho_{1} / \rho_{2}$ will be

The tension in a massless cable connected to an iron ball of $100 \,kg$ when it is submerged in sea water is (Given: $\rho_{\text{iron}} = 8 \times 10^3 \,kg/m^3$, $\rho_{\text{sea water}} = 1000 \,kg/m^3$, $g = 10 \,m/s^2$). (in $\,N$)