Fill in the blanks:
$(a)$ The volume of a cube of side $1\; cm$ is equal to $\ldots \ldots\, m ^{3}$
$(b)$ The surface area of a solid cylinder of radius $2.0 \;cm$ and height $10.0\; cm$ is equal to $\ldots(mm)^{2}$
$(c)$ $A$ vehicle moving with a speed of $18\; km\, h ^{-1}$ covers $\ldots ..m$ in $1\; s$
$(d)$ The relative density of lead is $11.3$. Its density is $\ldots ..\, g\, cm ^{-3}$ or $\ldots . .\, kg\, m ^{-3}$.

(N/A) Since $1\; cm = 10^{-2}\; m$,the volume of a cube of side $1\; cm$ is $(10^{-2}\; m)^{3} = 10^{-6}\; m^{3}$.
$(b)$ The total surface area of a cylinder is $S = 2\pi r(r + h)$. Given $r = 2.0\; cm = 20\; mm$ and $h = 10.0\; cm = 100\; mm$. Thus,$S = 2 \times 3.14 \times 20\; mm \times (20\; mm + 100\; mm) = 2 \times 3.14 \times 20 \times 120\; mm^{2} = 1.5072 \times 10^{4}\; mm^{2}$.
$(c)$ Speed $= 18\; km/h = 18 \times (5/18)\; m/s = 5\; m/s$. Distance covered in $1\; s$ is $5\; m/s \times 1\; s = 5\; m$.
$(d)$ Relative density is the ratio of the density of a substance to the density of water $(1\; g/cm^{3})$. Thus,density of lead $= 11.3 \times 1\; g/cm^{3} = 11.3\; g/cm^{3}$. Since $1\; g/cm^{3} = 10^{3}\; kg/m^{3}$,the density is $11.3 \times 10^{3}\; kg/m^{3}$.