Fill in the blanks

solution:

(a) $1 \mathrm{~cm}=\frac{1}{100} \mathrm{~m}$

Volume of the cube $=1 \mathrm{~cm}^{3}$

But, $1 \mathrm{~cm}^{3}=1 \mathrm{~cm} \times 1 \mathrm{~cm} \times 1 \mathrm{~cm}=\left(\frac{1}{100}\right) \mathrm{m} \times\left(\frac{1}{100}\right) \mathrm{m} \times\left(\frac{1}{100}\right) \mathrm{m}$

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

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

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

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

Given that,

$r=2 \mathrm{~cm}=2 \times 1 \mathrm{~cm}=2 \times 10 \mathrm{~mm}=20 \mathrm{~mm}$

$h=10 \mathrm{~cm}=10 \times 10 \mathrm{~mm}=100 \mathrm{~mm}$

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

(c) Using the conversion,

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

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

Therefore, distance can be obtained using the relation:

Distance $=$ Speed $\times$ Time $=5 \times 1=5 \mathrm{~m}$

Hence, the vehicle covers $5 \mathrm{~m}$ in $1 \mathrm{~s}$.

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

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

Density of water $=1 \mathrm{~g} / \mathrm{cm}^{3}$

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

Density of water $=1 \mathrm{~g} / \mathrm{cm}^{3}$

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

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

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