**Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...****Sol. ** (i) The S.I. unit of distance is metre ( $m$ )

\[

\begin{array}{l}

{1 \text { mile }=1.60 \text { kilometre }=1.60 \times 1000 \mathrm{m}} \\

{\text { Unit factor }=\frac{1.60 \times 1000(\mathrm{m})}{1 \mathrm{mile}}=\frac{1.6 \times 10^{3}(\mathrm{m})}{1 \mathrm{mile}}}

\end{array}

\]

93 millionmiles $=\frac{93 \times 10^{6}(\text { miles }) \times 1.6 \times 10^{3}(\mathrm{m})}{1 \text { mile }}$

$\therefore=93 \times 1.6 \times 10^{9} \mathrm{m}=148.8 \times 10^{9} \mathrm{m}=1.49 \times 10^{11} \mathrm{m}$

**(ii) 5 feet 2 inches $=62$ inches**

\[

\begin{array}{l}

{1 \text { inch }=2.54 \times 10^{-2} \mathrm{m}} \\

{\text { Unit factor }=\frac{2.54 \times 10^{-2}(\mathrm{m})}{1 \mathrm{inch}}} \\

{62 \text { inches }=\frac{62(\text { inches }) \times 2.54 \times 10^{-2} \mathrm{m}}{1 \mathrm{inch}}} \\

{=62 \times 2.54 \times 10^{-2} \mathrm{m}} \\

{=157.48 \times 10^{-2} \mathrm{m}=1.57 \mathrm{m}}

\end{array}

\]

(iii) 1 mile $=1.60 \mathrm{km}=1.60 \times 10^{3} \mathrm{m}$

\[

\begin{array}{l}

{\text { Unit factor }=\frac{1.60 \times 10^{3} \mathrm{m}}{1 \mathrm{mile}}} \\

{1 \mathrm{hr}=60 \times 60 \mathrm{s}=3.6 \times 10^{3} \mathrm{s}} \\

{\text { Unit factor }=\frac{3.6 \times 10^{3} \mathrm{s}}{1 \mathrm{hr}}}

\end{array}

\]

$\therefore \quad$ Speed $=\frac{100 \text { miles }}{h r}$

$=\frac{100 \text { miles }}{h r} \times \frac{1.60 \times 10^{3} \mathrm{m}}{1 \mathrm{mile}} \times \frac{1 \mathrm{hr}}{3.6 \times 10^{3} \mathrm{s}}=44 \mathrm{ms}^{-1}$

(iv) $1 \dot{A}=10^{-10} \mathrm{m}$

\[

\text { Unit factor }=\frac{10^{-10} \mathrm{m}}{1(\hat{A})}

\]

$\therefore \quad 0.74 \mathrm{A}=\frac{0.74 \hat{A} \times 10^{-10}(m)}{1(\hat{A})}$

\[

=0.74 \times 10^{-10} \mathrm{m} \text { or }=7.4 \times 10^{-11} \mathrm{m}

\]

(v) $\quad 0^{\circ} \mathrm{C}=273.15 \mathrm{K}$

$\quad 46^{\circ} \mathrm{C}=273.15 \mathrm{K}+46 \mathrm{K}=319.15 \mathrm{K}$

(vi) 1 pound $=454 \times 10^{-3} \mathrm{kg}$

\[

\text { Unit factor }=\frac{454 \times 10^{-3}(k g)}{1(\text { pound })}

\]

$\begin{aligned} \therefore \quad 150 \text { pound }=\frac{150(\text { pound }) \times 454 \times 10^{-3}(k g)}{1(\text { pound })} & \\=& 150 \times 454 \times 10^{-3} \mathrm{kg}=68.1 \mathrm{kg} \end{aligned}$