So, 1 cm = 1/100 m=${10}^{-2}<mspace\; width="1em"></mspace>m$

The volume of the cube = 1 cm × 1 cm × 1 cm = $={10}^{-2}\times {10}^{-2}\times {10}^{-2}<mspace\; width="1em"></mspace>{\mathrm{m}}^3={10}^{-6}<mspace\; width="1em"></mspace>{\mathrm{m}}^3$

Hence, the volume of a cube of side 1 cm is equal to ${10}^{-6}<mspace\; width="1em"></mspace>{\mathrm{m}}^3$.

(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

$\Rightarrow <mspace\; width="1em"></mspace>\mathrm{S}<mspace\; width="1em"></mspace>=<mspace\; width="1em"></mspace>2<mspace\; width="1em"></mspace>\times <mspace\; width="1em"></mspace>3.14<mspace\; width="1em"></mspace>\times 20<mspace\; width="1em"></mspace>\times <mspace\; width="1em"></mspace>\left(20<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>100\right)=15072=1.5\times {10}^4<mspace\; width="1em"></mspace>{\mathrm{mm}}^2$

(c) Using the conversion,

1 km/h = $\frac{5}{18}<mspace\; width="1em"></mspace>\mathrm{m}/\mathrm{s}$

$18<mspace\; width="1em"></mspace>\mathrm{km}/\mathrm{h}<mspace\; width="1em"></mspace>=<mspace\; width="1em"></mspace>18<mspace\; width="1em"></mspace>\times <mspace\; width="1em"></mspace>\frac{5}{18}<mspace\; width="1em"></mspace>=<mspace\; width="1em"></mspace>5<mspace\; width="1em"></mspace>\mathrm{m}/\mathrm{s}$

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 = $\frac{\mathrm{Density}<mspace\; width="1em"></mspace>\mathrm{of}<mspace\; width="1em"></mspace>\mathrm{substance}}{\mathrm{Density}<mspace\; width="1em"></mspace>\mathrm{of}<mspace\; width="1em"></mspace>\mathrm{water}}$

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 = $\frac{1}{1000}<mspace\; width="1em"></mspace>\mathrm{kg}$

$1<mspace\; width="1em"></mspace>{\mathrm{cm}}^3<mspace\; width="1em"></mspace>=<mspace\; width="1em"></mspace>{10}^{-6}<mspace\; width="1em"></mspace>{\mathrm{m}}^3$

$1<mspace\; width="1em"></mspace>\mathrm{g}/{\mathrm{cm}}^3<mspace\; width="1em"></mspace>=<mspace\; width="1em"></mspace>\frac{{10}^{-3}}{{10}^{-6}}<mspace\; width="1em"></mspace>\mathrm{kg}/{\mathrm{m}}^3<mspace\; width="1em"></mspace>=<mspace\; width="1em"></mspace>{10}^3<mspace\; width="1em"></mspace>\mathrm{kg}/{\mathrm{m}}^3$

$\Rightarrow <mspace\; width="1em"></mspace>11.3<mspace\; width="1em"></mspace>\mathrm{g}/{\mathrm{cm}}^3<mspace\; width="1em"></mspace>=<mspace\; width="1em"></mspace>11.3<mspace\; width="1em"></mspace>\times <mspace\; width="1em"></mspace>{10}^3<mspace\; width="1em"></mspace>\mathrm{kg}/{\mathrm{m}}^3$