(a) The volume of a cube of side 1 cm is given by, $V = (1\,\text{cm})^{3}$ or $V = (10^{-2}\,\text{m})^{3}=10^{-6}\,\text{m}^{3}$. (b) The surface area of a solid cylinder of radius r and height h is givenby : $A=$ Area of two caps $+$ curved surface area $=2\pi r^{2}+2\pi r h=2\pi r(r+h)$ here $r = 2\,\text{cm} = 20\,\text{mm},$ $h = 10\,\text{cm} = 100\,\text{mm}$ $\therefore A = 2\times\frac{22}{7}\times 20(20+100)\,\text{mm}^{2}$ $=15086\,\text{mm}^{2}$ $=1.5086\times 10^{4}\,\text{mm}^{2}$ $=1.5\times10^{4}\,\text{mm}^{2}$ (c) Here $v=18\,\text{km}\,\text{h}^{-1}$ $=\frac{18\times1000\,\text{m}}{3600\,\text{s}}=5\,\text{m}\,\text{s}^{-1}$ $t=1\,\text{s},\ x=v t=5\times1=5\,\text{m}$ (d) Relative density of lead = 11.3, density of water = 1 g cm$^{-3}$ We know that relative density of lead $=\frac{\text{density of lead}}{\text{density of water}}$ $\therefore$ density of lead = relative density of lead × density of water$=11.3\times 1\,\text{g}\,\text{cm}^{-3}$$=11.3\,\text{g}\,\text{cm}^{-3}$Also in S.I. system density of water = 10$^{3}$kg m$^{-3}$$\therefore$ density of lead $=11.3\times 10^{3}\,\text{kg}\,\text{m}^{-3}$$=1.13\times10^{4}\,\text{kg}\,\text{m}^{-3}$