## ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 2 Exponents and Powers Ex 2.2

Question 1.

Express the following numbers in standard form:

(i) 0.0000000000085

(ii) 0.000000000000942

(iii) 6020000000000000

(iv) 0.00000000837

Solution:

(i) 0.0000000000085 = 8.5 × 10^{-12}

(ii) 0.000000000000942 = 9.42 × 10^{-13}

(iii) 6020000000000000 = 6.02 × 10^{15}

(iv) 0.00000000837 = 8.37 × 10^{-9}

Question 2.

Express the following numbers in usual form:

(i) 3.02 × 10^{-6}

(ii) 1-007 × 10^{11}

(iii) 5.375 × 10^{14}

(iv) 7.579 × 10^{-14}

Solution:

(i) 3.02 × 10^{-6} = 0.00000302

(ii) 1.007 × 10^{11} = 100700000000

(iii) 5.375 × 10^{14} = 537500000000000

(iv) 7.579 × 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 cm^{2}.

(xii) A sugar factory has annual sales of 3 billion 720 million kilograms of sugar.

Solution:

(i) The mass of a proton is

0.000000000000000000000001673 gram = 1.673 × 10^{-24} gram.

(ii) Thickness of a piece of paper is 0.0016 cm = 1.6 × 10^{-3}

(iii) Diameter of a wire on a computer chip is 0.000003 m = 3.0 × 10^{-6} m

(iv) A helium atom has a diameter of \(\frac { 22 }{ 100000000000 }\) m = 22 × 10^{-12} = 2.2 × 10^{-10}

(v) Mass of a molecule of hydrogen gas is about

0.00000000000000000000334 tons = 3.34 × 10^{-21} tons

(vi) Human body has 1 trillion of cells which vary in shapes and sizes

= 1,000,000,000,000 = 10^{12}

(vii) The distance from the Earth of the Sun is

149,600,000,000 m = 1.496 × 10^{11}

(viii) The speed of light is 300,000,000 m/sec = 3.0 × 10^{8} m/sec

(ix) Mass of the Earth is

5,970,000,000,000,000,000,000,000 kg = 5.97 × 10^{24} kg

(x) Express 3 years in seconds 3 years = 3 × 365 days

= 3 × 365 × 24 hours

= 3 × 365 × 24 × 3600 seconds

= 1040688000 seconds

= 1.040688 × 10^{9} seconds

(xi) Express 7 hectares in cm^{2}

7 hectares = 7 × 10000 m^{2}

= 7 × 10000 × 100 × 100 cm^{2}

= 700000000 cm^{2}

= 7.0 × 10^{8} cm^{2}

(xii) A sugar factory has annual sales of

3 billion 720 million kilograms of sugar >

Annual sale of a sugar factory = 3 billion

720 million kilograms sugar = 3,720,000,000 kg = 3.72 × 10^{9} kg

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

Solution:

(i) Size of plant cell= 0.00001275 m = 1.275 × 10^{-5} m

Thickness of a piece of paper = 0.0016 cm = 1.6 × 10^{-3} cm

Diameter of a wire on a computer chip = 0.000003 m = 3.0 × 10^{-6} m

(i) Size of plant cell : thickness of a piece of paper

= 1.275 × 10^{-5} : 1.6 × 10^{-3}

Size of plant cell = \(\frac { 1.2 }{ 1.6 }\) = \(\frac { 3 }{ 4 }\) times of thickness of paper

(ii) Comparison between size of plant cell : diameter of wire on a computer chip

= 1.275 × 10^{-5} : 3.0 × 10^{-6}

= 12.75 : 3.00

Size of plant cell is 4 times of diameter of wire.

(iii) Thickness of a piece of paper: diameter of a wire on a computer chip

= 1.6 × 10^{-3} : 3.0 × 10^{-6} × 100 cm

= 1.6 × 1000 : 300

= 16.1 : 3

Approximately 5 times is the thickness of paper to diameter of wire.

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})

Solution:

Red blood per cubic millimeter = 5.5 million = 5.5 × 10^{6}

Red blood in 5 litres of blood

= 5.5 × 10^{6} × 5 × 10^{5} (1 litre = 10^{5} mm)

= 27.5 × 10^{6+5}

= 27.5 × 10^{11}

= 2.75 × 10 × 10^{11}

= 2.75 × 10^{12}

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?

Solution:

Mass of Mars = 6.42 × 10^{29} kg

and mass of sun = 1.99 × 10^{30}

Total mass = 6.42 × 10^{29} + 1.99 × 10^{30}

= 10^{29} (6.42 + 1.99 × 10)

= 10^{29} (6.42 + 19.9)

= 26.32 × 10^{29}

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.

Solution:

Distance between earth and a particular star = 8.1 × 10^{13} km

Speed of light = 3 × 10^{8} m/sec.

Time is taken to reach the earth