Exercise 2.2
Q1 | Ex-2.2 | Class 8 | ML Aggarwal | Exponents and Powers | Chapter 2 | myhelper
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Question 1
Express the following numbers in standard form:
(i) 0.0000000000085
(ii) 0.000000000000942
(iii) 6020000000000000
(iv) 0.00000000837
Sol :
(i)
$\begin{aligned} 0.0000000000085 &=\frac{85}{(10)^{13}} . \\ &=8.5 \times 10^{-12} . \end{aligned}$
(ii)
$\begin{aligned} 0.000000000000942 &=\frac{942}{(10)^{13}} \\ &=9.42 \times 10^{-13} \end{aligned}$
(iii)
$6020000000000000=6.02 \times 10^{15}$
(iv)
$0.00000000837=8.37 \times 10^{-9}$
Question 2
Express the following numbers in the usual form:
(i) 3.02 × 10-6
(ii) 1.007 × 1011
(iii) 5.375 × 1014
(iv) 7.579 × 10-14
(i) 3.02 × 10-6
(ii) 1.007 × 1011
(iii) 5.375 × 1014
(iv) 7.579 × 10-14
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
Express the number appearing in the following statements in standard form:
(i) The mass of a proton is 0.000000000000000000000001673 gram.
(ii) The thickness of a piece of paper is 0.0016 cm.
(iii) The diameter of a wire on a computer chip is 0.000003 m.
(iv) A helium atom has a diameter of $\frac{22}{100000000000}$ m
(v) Mass of a molecule of hydrogen gas is about 0.00000000000000000000334 tons.
(vi) The human body has 1 trillion cells which vary in shapes and sizes.
(vii) The distance from the Earth of the Sun is 149,600,000,000 m.
(viii) The speed of light is 300,000,000 m/sec.
(ix) Mass of the Earth is 5,970,000,000,000,000,000,000,000 kg.
(x) Express 3 years in seconds.
(xi) Express 7 hectares in cm2.
(xii) A sugar factory has annual sales of 3 billion 720 million kilograms of sugar.
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 $22 \times 10^{-11} \mathrm{~m}$
(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 $1.496 \times 10^{11} \mathrm{~m}$
(viii) the speed of light is $3 \times 10^{8} \mathrm{~m} / \mathrm{sec}$
(ix) Mass of the Earth is $5.97 \times 10^{24} \mathrm{~kg}$
$\begin{aligned}x\rangle \quad 3 \text { years } &=3 \times 365 \text { days } \\ &=3 \times 365 \times 24 \text { Hours } \\ &=3 \times 365 \times 24 \times 60 \text { Minutes } \\ &=3 \times 365 \times 24 \times 60 \times 60 \text { seconds } \\ &=94608000 \text { seconds } \\ &=9.4608 \times 10^{7} \mathrm{~seconds} \end{aligned}$
$\begin{aligned}\left.x_{i}\right\rangle \quad 7 \text { hectares } &=7 \times 10,000 \mathrm{~m}^{2} \\ &=7 \times 10000 \times 10000 \mathrm{~cm}^{2} \\ 7 \text { hectares } &=7 \times 10^{8} \mathrm{~cm}^{2} \end{aligned}$
(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
Compare the following:
(i) Size of a plant cell to the thickness of a piece of paper.
(ii) Size of a plant cell to the diameter of a wire on a computer chip.
(iii) The thickness of a piece of paper to the diameter of a wire on a computer chip.
Given size of plant cell = 0.00001275 m
Thickness of a piece of paper = 0.0016 cm
Diameter of a wire on a computer chip = 0.000003 m
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
Q5 | Ex-2.2 | Class 8 | ML Aggarwal | Exponents and Powers | Chapter 2 | myhelper
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Question 5
The number of red blood cells per cubic millimetre of blood is approximately 5.5 million. If the average body contains 5 litres of blood, what is the total number of red cell in the body? (1 litre = 1,00,000 $mm^3$)
Sol :
Number of red blood cells per cubic millimeter $=5.5 \times 10^{6}\mathrm{mm}^{3}$
Number of red blood cells per cubic millimeter $=5.5 \times 10^{6}\mathrm{mm}^{3}$
Total no. of red blood cells in 5 liters of blood is
1 litre = 1,00,000 $mm^3$
∴$5 l =5 \times 100000~mm^3$
$5~l=500000~mm^3$
Given : $1mm^3$=5.5 million red blood cell
$1mm^3=5.5 \times 1000000$
$1mm^3=\frac{55}{10} \times 1000000$
$1mm^3=5500000$ red blood cell
Multiplying 500000 both sides
∴$(500000 \times 1)~mm^3=500000 \times 5500000$ red blood cell
$500000~mm^3=2750000000000$ red blood cell
$5~l=500000~mm^3$
∴$5~l=275\times 10^{10}$ red blood cell
or 5~l=\frac{275}{100} \times 10^{10} \times 10^{2}$ red blood cell
$5~l=2.75 \times 10^{12}$ red blood cell
Question 6
Mass of Mars is $6.42 × 10^{29}$ kg and the mass of the sun is $1.99 × 10^{30}$ kg. What is the total mass?
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
A particular star is at a distance of about $8.1 × 10^{13}$ km from the Earth. Assuming that the light travels at $3 × 10^8$ m/sec, find how long does light take from that star to reach the Earth.
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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