So, 1 cm = 1/100 m=${10}^{-2} m$

The volume of the cube = 1 cm × 1 cm × 1 cm = $={10}^{-2}\times {10}^{-2}\times {10}^{-2} {\mathrm{m}}^3={10}^{-6} {\mathrm{m}}^3$

Hence, the volume of a cube of side 1 cm is equal to ${10}^{-6} {\mathrm{m}}^3$.

(b) The total surface area of a cylinder of radius r and height h,

S = 2$\pi$r(r + h).

Given that, r = 2 cm = 2 × 1 cm = 2 × 10 mm = 20 mm

h = 10 cm = 10 × 10 mm = 100 mm

$\Rightarrow \mathrm{S} = 2 \times 3.14 \times 20 \times \left(20 + 100\right)=15072=1.5\times {10}^4 {\mathrm{mm}}^2$

(c) Using the conversion,

1 km/h = $\frac{5}{18} \mathrm{m}/\mathrm{s}$

$18 \mathrm{km}/\mathrm{h} = 18 \times \frac{5}{18} = 5 \mathrm{m}/\mathrm{s}$

Therefore, distance can be obtained using the relation:

Distance = Speed × Time = 5 × 1 = 5 m

Hence, the vehicle covers 5 m in 1 s.

(d)The relative density of a substance is given by the relation,

Relative density = $\frac{\mathrm{Density} \mathrm{of} \mathrm{substance}}{\mathrm{Density} \mathrm{of} \mathrm{water}}$

The density of water = 1 $\mathrm{g}/{\mathrm{cm}}^3$

The density of lead = Relative density of lead $\times$ Density of water
                            = 11.3 x 1 = 11.3 $\mathrm{g}/{\mathrm{cm}}^3$

Again, 1g = $\frac{1}{1000} \mathrm{kg}$

$1 {\mathrm{cm}}^3 = {10}^{-6} {\mathrm{m}}^3$

$1 \mathrm{g}/{\mathrm{cm}}^3 = \frac{{10}^{-3}}{{10}^{-6}} \mathrm{kg}/{\mathrm{m}}^3 = {10}^3 \mathrm{kg}/{\mathrm{m}}^3$

$\Rightarrow 11.3 \mathrm{g}/{\mathrm{cm}}^3 = 11.3 \times {10}^3 \mathrm{kg}/{\mathrm{m}}^3$