ML Aggarwal class 8 chapter 2 Exponents and powers Exercise 2.2

  Exercise 2.2

Question 1

Sol :
(i)
 $\begin{aligned} 0.0000000000085 &=\frac{85}{(10)^{13}} . \\ &=8.5 \times 10^{-12} . \end{aligned}$

(ii)
 $\begin{aligned} 0.000000000000942 &=\frac{942}{(10)^{15}} \\ &=9.42 \times 10^{-13} \end{aligned}$
                                                                     
(iii) 
$6020000000000000=6.02 \times 10^{15}$

(iv) 
$0.00000000837=8.37 \times 10^{-9}$\

Question 2

Sol :
(i) $3.02 \times 10^{-6}$
0.00000302

(ii) $1.007 \times 10^{11}$
100700000000

(ii) $5.375 \times 10^{14}$
537500000000000

(iv) $7.579 \times 10^{-14}-7.579 \times 10^{-14}$
0.00000000000007579

Question 3

Sol :
(i)The mass of a proton is $1.673 \times 10^{-24} \mathrm{gram}$
(ii) thickness of a piece of paper is $1.6 \times 10^{-3} \mathrm{~cm}$
(iii) Diameter of a wire on a Compurer dip is $3 \times 10^{-6} \mathrm{~m}$
(iv) A helium atom has a diameter of $32 \times 10^{-11} \mathrm{~m}$
(v) Mass of a molecule of hydrogen gas is about
                           $3.34 \times 10^{-21}$ tons                          
(vi) Human body has $10^{+12}$ of cells which vary
in shapes and sizes
(vii) The distance from earth to the sun is $149.6 \times 10^{9} \mathrm{~m}$
(viii) the speed of light is $3 \times 10^{8} \mathrm{~m} / \mathrm{sec}$

(ix) Mass of the  Earth is $5.97 \times 10^{24} \mathrm{~kg}$


$\begin{aligned}x\rangle \quad 3 \text { years } &=3 \times 365 \text { doys } \\ &=3 \times 365 \times 24 \text { Hows } \\ &=3 \times 365 \times 24 \times 60 \text { Minutes } \\ &=3 \times 365 \times 24 \times 60 \times 60 \text { seconds } \\ &=94608000 \text { seconds } \\ &=9.4608 \times 10^{7} \mathrm{~seconds} \end{aligned}$

$\begin{aligned}\left.x_{i}\right\rangle \quad 7 \text { hectares } &=7 \times 10,000 \mathrm{~m}^{2} \\ &=7 \times 10000 \times 10000 \mathrm{~cm}^{2} \\ 7 \text { hectares } &=7 \times 10^{8} \mathrm{~cm}^{2} \end{aligned}$

xii) A sugar factory has annual sales of 3,720,000,000 Kilograms of sugar
A sugar factory has annual sales of $3.72 \times 10^{9} \mathrm{~kg}$ of sugar

Question 4

Sol :
size of plant cell. $=0.00001275 \mathrm{~m}=1.275 \times 10^{-5} \mathrm{~m}$

thickness of piece of paper $=0.0016 \mathrm{~cm}=1.6 \times 10^{-5}$

Diameter of a wire on a computer chip

$=0.000003 \mathrm{~m}=3 \times 10^{-6} \mathrm{~m}$

(i) $\frac{\text { Size of plant cell }}{\text { thickness of piece of paper }}$

       = $\frac{1.275 \times 10^{-5}}{1.6 \times 10^{-5}}=0.796 \approx 0.8$

$\therefore$ Size of plant cell 0.8 times the Thickness of
piece of paper

(ii) $\frac{\text { size of plant cell }}{\text { Diameter of wire on computerchip }}$

$\frac{1.275 \times 10^{-5}}{3 \times 10^{-6}}$

$\underline{4.25}$

$\therefore$ size of plant cell is 4.25 times bigger than

the diameter of wire on Computer chip.

iii) $\frac{\text { thickness of piece of paper }}{\text { Diameter of wire on Computer chip }}$

=$\frac{1.6 \times 10^{-5}}{3 \times 10^{-6}}$

=5.33

             $\therefore$ thickness of piece of paper is 5.33 times bigger then
                   the diameter of wire on Computer chip

Question 5

Sol :
$\begin{aligned} \text { no. ot red blood cells per cubic millimeter } \\ &=5.5 \times 10^{6} / \mathrm{mm}^{3} \\ &=5.5 \times 10^{6} / \mathrm{m} \cdot \mathrm{m}^{3} \end{aligned}$

total no. of red blood cells in 5 liters of blood
is $=5.5 \times 10^{6} / \mathrm{mm}^{3} 5 \mathrm{kt}$

$=5.5 \times 106 / \mathrm{mm} 3 \mathrm{x} 105 \mathrm{~mm}^{3}$

$=5.5 \times 5 \times 10^{6} \times 10^{5}$

$=27.5 \times 10^{11} \mathrm{H}$

Question 6

Sol :
Mass of Mars $=6.42 \times 10^{29} \mathrm{~kg}$

Mass of sun $=1.99 \times 10^{30} \mathrm{~kg}$

Total Mass $=$ Mass of Mars $+$ Mass of Sun

$=6.42 \times 10^{29}+1.99 \times 10^{30}$

$=0.642 \times 10^{30}+1.99 \times 10^{30}$

$\therefore$ (:: make powers equal in both  terms)

$\therefore \quad$ Total mass $=2.632 \times 10^{30} \mathrm{~kg}$

Question 7

Sol : 
 Distance between  star and earth = $8.1 \times 10^{13} \mathrm{~km}$

=$8.1 \times 10^{13} \times 10^{3} \mathrm{~m}$

=$81 \times 10^{6} \mathrm{~m}$

$\begin{aligned} \text { Speed ot light } &=3 \times 10^{8} \mathrm{~m} / \mathrm{sec} \\ \text { Time } &=\frac{\text { distance }}{\text { Speed }} \\ &=\frac{8.1 \times 10^{16}}{3 \times 108} \\ \text { Time } &=2.7 \times 10^{8} \mathrm{sec} \end{aligned}$

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