Spheres of iron and lead having the same mass are completely immersed in water. The density of lead is greater than that of iron. If the apparent loss of weight is $w_{1}$ for the iron sphere and $w_{2}$ for the lead sphere,then the ratio $\frac{w_{1}}{w_{2}}$ is:

$A$ soft plastic bottle,filled with water of density $1 \text{ g/cc}$,contains an inverted glass test-tube with some air (ideal gas) trapped inside,as shown in the figure. The test-tube has a mass of $5 \text{ g}$,and it is made of thick glass with a density of $2.5 \text{ g/cc}$. Initially,the bottle is sealed at atmospheric pressure $P_0 = 10^5 \text{ Pa}$,such that the volume of the trapped air is $V_0 = 3.3 \text{ cc}$. When the bottle is squeezed from the outside at a constant temperature,the pressure inside increases and the volume of the trapped air decreases. It is observed that the test-tube begins to sink at a pressure $P_0 + \Delta P$ without changing its orientation. At this pressure,the volume of the trapped air is $V_0 - \Delta V$.
Let $\Delta V = X \text{ cc}$ and $\Delta P = Y \times 10^3 \text{ Pa}$.
$(1)$ The value of $X$ is
$(2)$ The value of $Y$ is

$A$ wooden cube first floats inside water when a $200\,g$ mass is placed on it. When the mass is removed,the cube rises $2\,cm$ above the water level. The side of the cube is ........ $cm$.

$A$ piece of steel has a weight $W$ in air,$W_1$ when completely immersed in water,and $W_2$ when completely immersed in an unknown liquid. The relative density (specific gravity) of the liquid is:

$A$ hollow sphere of volume $V$ is floating on the water surface with half immersed in it. What should be the minimum volume of water poured inside the sphere so that the sphere now sinks into the water?

$A$ log of wood of mass $120 \ kg$ floats in water. The mass that can be put on the raft to make it just sink should be ....... $kg$ (density of wood = $600 \ kg/m^3$,density of water = $1000 \ kg/m^3$).