$A$ body weighs $50 \,g$ in air and $40 \,g$ in water. How much would it weigh in a liquid of specific gravity $1.5$ (in $\,g$)?

$A$ cube having a side of $10 \ cm$ with unknown mass $m$ and a $200 \ g$ mass were hung at two ends of a uniform rigid rod of $27 \ cm$ length. The rod along with the masses was placed on a wedge,keeping the distance between the wedge point and the $200 \ g$ weight as $25 \ cm$. Initially,the masses were not in balance. $A$ beaker is placed beneath the unknown mass and water is added slowly to it. At a given point,the masses were in balance and half the volume of the unknown mass was submerged in the water. (Take the density of the unknown mass to be greater than that of water,the mass did not absorb water,and the water density is $1 \ g/cm^3$). The unknown mass $m$ is . . . . . . $kg$.

Two bodies are in equilibrium when suspended in water from the arms of a balance. The mass of one body is $36 \ g$ and its density is $9 \ g/cm^3$. If the mass of the other is $48 \ g$,its density in $g/cm^3$ is:

Two solids $P$ and $Q$ float in water. It is observed that $P$ floats with half of its volume immersed and $Q$ floats with $\frac{2}{3}$ of its volume immersed. The ratio of densities of $P$ and $Q$ is

$A$ uniform rod is suspended horizontally from its mid-point. $A$ piece of metal whose weight is $w$ is suspended at a distance $l$ from the mid-point. Another weight $w_{1}$ is suspended on the other side at a distance $l_{1}$ from the mid-point to bring the rod to a horizontal position. When $w$ is completely immersed in water,$w_{1}$ needs to be kept at a distance $l_{2}$ from the mid-point to get the rod back into a horizontal position. The specific gravity of the metal piece is

The weight of a sphere in air is $50 \ g$. Its weight is $40 \ g$ in a liquid at temperature $20^{\circ} C$. When the temperature increases to $70^{\circ} C$,its weight becomes $45 \ g$. The ratio of the densities of the liquid at the two given temperatures is: (in $: 1$)