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NCERT Class 8 Maths Ch. 2 Power Play Other Exercises:-

Exercise 2.1

Exercise 2.2

NCERT Solutions Class 8 Maths All Chapters:-

Chapter 1 - A Square and a Cube

Chapter 2 - Power play

Chapter 3 - A story of Numbers

Chapter 4 - Quadrilaterals

Chapter 5 - Number Play

Chapter 6 - We distibute, yet thinmgs multiply

Chapter 7 - Proportional Reasoning

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NCERT Class 8 Maths Chapter 2 Exercise 2.2 Solutions - Scientific Notation & Standard Form

Mastering the representation of very large and very small numbers is simple with our NCERT Solutions for Class 8 Maths Ch 2: Power Play – Exercise 2.2. This exercise focuses on "Exponents and Powers" and specifically teaches students how to use scientific notation to make complex numbers easier to read, write, and compare. Exercise 2.2 concentrates on converting numbers between standard form and scientific notation (also known as usual form).

Learn how to easily work with big and small numbers in Class 8 Math Lessons (Ch 2). The solutions to Exercise 2.2 of Class 8 Maths will be presented in a step-by-step manner to help students understand how to move the decimal point and determine the correct power of 10. These NCERT Solutions are in accordance with the CBSE Guidelines to provide students with the best available path to improve their calculation speed and accuracy. Additionally, these solutions will aid students in building a solid foundation for future studies in physics and Chemistry, where scientific notation is used daily.

1.0Download Class 8 Maths Chapter 2 Ex 2.2 NCERT Solutions PDF

You can get a PDF of our simple NCERT Solutions for Class 8 Maths Chapter 2 Exercise 2.2. Ideal for rapid reference prior to tests and offline study.

NCERT Solutions Class 8 Maths Chapter 2 Ex 2.2

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2.0NCERT Solutions for Class 8 Maths Chapter 2 Power Play : All Exercises

NCERT Solutions for Class 8 Maths Chapter 2 Power Play - Exercise 2.1

NCERT Solutions for Class 8 Maths Chapter 2 Power Play - Exercise 2.2

3.0Detailed NCERT Class 8 Maths Chapter 2 Solutions of Exercise 2.2

1.Find out the units digit in the value of $2^{224} \div 4^{32}$ ? [Hint: $4 = 2^{2}$ ]

Sol. The expression is $2^{224} \div 4^{32}$ . We can rewrite the base 4 as $2^{2}$ . $4^{32} = (2^{2})^{32}$ Using the exponent rule $(a^{m})^{n} = a^{m \times n}$ : $4^{32} = 2^{2 \times 32} = 2^{64}$ Now substitute this back into the division: $2^{224} \div 4^{32} = \frac{2 ^{224}}{2 ^{64}}$ Using the exponent rule $\frac{a ^{m}}{a ^{n}} = a^{m - n}$ : $\frac{2 ^{224}}{2 ^{64}} = 2^{224 - 64} = 2^{160}$ The units digits of powers of 2 follow a repeating cycle of length 4 as $2, 4, 8, 6$ $(∵ 2^{1} = 2, 2^{2} = 4, 2^{3} = 8, 2^{4} = 16, 2^{5} = 32)$ To find the units digit of $2^{160}$ , we need to find the remainder of the exponent (160) when divided by the cycle length (4). Remainder $= 160 \div 4$ We can calculate this: $160 \div 4 = 40$ with a remainder of 0 When the remainder is 0 , the units digit is the last digit in the cycle, which corresponds to the units digit of 24 . The last digit in the cycle $(2, 4, 8, 6)$ is 6 . Therefore, the units digit of $2^{160}$ is 6 . The units digit in the value of $2^{224} \div 4^{32}$ is 6 .

2.There are 5 bottles in a container. Every day, a new container is brought in. How many bottles would be there after 40 days?

Sol. Start with 1 container having 5 bottles, 1 new container is added every day, so after 40 days. Number of containers after 40 days $= 40$ containers $\times 5$ bottles each $= 200$ bottles

3.Write the given number as the product of two or more powers in three different ways. The powers can be any integers. (i) $6 4^{3}$ (ii) $19 2^{8}$ (iii) $3 2^{- 5}$

Sol.

	No.	Way-1	Way-2	Way-3
(i)	$6 4^{3}$	$2^{18}$	$8^{6}$	$4^{9}$
(ii)	$19 2^{8}$	$2^{48} \times 3^{8}$	$6 4^{8} \times 3^{8}$	$1 6^{8} \times 1 2^{8}$
(iii)	$3 2^{- 5}$	$2^{- 25}$	$8^{- 5} \times 4^{- 5}$	$\frac{1}{2 ^{25}}$

4.Examine each statement below and find out if it is ‘Always True’, ‘Only Sometimes True', or ‘Never True’. Explain your reasoning. (i) Cube numbers are also square numbers. (ii) Fourth powers are also square numbers. (iii) The fifth power of a number is divisible by the cube of that number. (iv) The product of two cube numbers is a cube number. (v) $q^{46}$ is both a $4^{th}$ power and a $6^{th}$ power ( $q$ is a prime number).

Sol. (i) Cube numbers are also square numbers only sometimes true reason: (a) $64 = 4^{3} = 8^{2} \to$ both cube and square (b) $8 = 2^{3} \to$ not a square.

A number must be both a square and a cube, i.e., a sixth power, to satisfy both.

Not all cubes are sixth powers. (ii) Fourth powers are also square numbers - Always True Reason: Any fourth power is the form of $a^{4} = (a^{2})^{2}$ , which is clearly a square. Fourth powers are squares because squaring a square gives a fourth power. (iii) The fifth power of a number is divisible by the cube of that number.

Always True

Reason: $a^{5} \div a^{3} = a^{5 - 3} = a^{2}$ , which is valid for any $a \neq = 0$ . The fifth power contains at least three powers of the base, so it's divisible by its cube. (iv) The product of two cube numbers is a cube number. - Always True Reason: The product of two cubes is also a cube, just raise the product of their bases to the third power. (v) $q^{46}$ is both a $4^{th}$ power and a $6^{th}$ power ( $q$ is a prime number)- Never True

Reason: 46 is not divisible by 4 or 6 → can't be a $4^{th}$ or $6^{th}$ power. 5. Simplify and write these in the exponential form. (i) $1 0^{- 2} \times 1 0^{- 5}$ (ii) $5^{7} \div 5^{4}$ (iii) $9^{- 7} \div 9^{4}$ (iv) $(1 3^{- 2})^{- 3}$ (v) $m^{5} n^{12} (mn)^{9}$

Sol. (i) $1 0^{- 2} \times 1 0^{- 5} = 1 0^{- 2 - 5} = 1 0^{- 7}$ $(a^{m} \times a^{n} = a^{m + n})$ (ii) $5^{7} \div 5^{4} = 5^{7 - 4} = 5^{3}$ $(a^{m} \div a^{n} = a^{m - n})$ (iii) $9^{- 7} \div 9^{4} = 9^{- 7 - 4} = 9^{- 11}$ $(a^{m} \div a^{n} = a^{m - n})$ (iv) $(1 3^{- 2})^{- 3} = (13)^{- 2 \times (- 3)} = (13)^{6}$

$[(a^{m})^{n} = a^{mn}]$

(v) $m^{5} n^{12} (mn)^{9} = m^{5} n^{12} m^{9} n^{9}$

6.If $1 2^{2} = 144$ , what is (i) $(1.2)^{2}$ (ii) $(0.12)^{2}$ (iii) $(0.012)^{2}$ (iv) $12 0^{2}$

Sol. (i) $(1.2)^{2} = (\frac{12}{10})^{2} = \frac{144}{100} = 1.44$ (ii) $(0.12)^{2} = (\frac{12}{100})^{2} = \frac{144}{10000} = 0.0144$ (iii) $(0.012)^{2} = (\frac{12}{1000})^{2} = \frac{144}{1000000}$

$= 0.000144$

(iv) $12 0^{2} = (12 \times 10)^{2}$

7.Circle the numbers that are the same

$2^{4} \times 3^{6}$
$6^{4} \times 3^{2}$
$6^{10}$
$1 8^{2} \times 6^{2}$
$6^{24}$

Sol.

8.Identify the greater number in each of the following: (i) $4^{3}$ or $3^{4}$ (ii) $2^{8}$ or $8^{2}$ (iii) $10 0^{2}$ or $2^{100}$

Sol. (i) $4^{3} = 64; 3^{4} = 81$ $81 > 64$ $∴ 3^{4} > 4^{3}$ (ii) $2^{8} = 256; 8^{2} = (2^{3})^{2} = 2^{6} = 64$ $256 > 64$ $∴ 2^{8} > 8^{2}$ (iii) $10 0^{2} = 10, 000$ $2^{100} \sim 1.27 \times 1 0^{30}$ A number with 30 zeros is much larger than 10,000. $∴ 2^{100} > 10 0^{2}$ 9. A dairy plans to produce 8.5 billion packets of milk in a year. They want a unique ID (identifier) code for each packet. If they choose to use the digits 0 9, how many digits should the code consist of?

Sol. Total no. of packets produced in a year $= 8.5$ billion ∴ Number of codes required $= 8.5 \times 1 0^{9}$ codes

For ID digits to be taken from 0 to 9 . So, each digit has 10 choices. To make code with n digits, the total number of possible codes $= 1 0^{n}$

$∴ 1 0^{n} \geq 8.5 \times 1 0^{9} 1 0^{9} = 1, 00, 00, 00, 000$

The smallest value of $n$ that satisfies $1 0^{n} \geq 8.5 \times 1 0^{9}$ is 10 .

Hence number of digits the code should consist of $1 0^{10}$ . 10. 64 is a square number ( $8^{2}$ ) and a cube number ( $4^{3}$ ). Are there other numbers that are both squares and cubes? Is there a way to describe such numbers in general?

Sol.

No.	Form	Square of	Cube of
1	$1^{6}$	$1^{2}$	$1^{3}$
64	$2^{6}$	$8^{2}$	$4^{3}$
729	$3^{6}$	$2 7^{2}$	$9^{3}$
4096	$4^{6}$	$6 4^{2}$	$1 6^{3}$
15625	$5^{6}$	$12 5^{2}$	$2 5^{3}$

Yes, other numbers are both squares and cubes. General Rule: A number is both a perfect square and a perfect cube if and only if it is a perfect sixth power, i.e., it can be written as $x^{6}$ for some integer x . 11. A digital locker has an alphanumeric (it can have both digits and letters) pass code of length 5. Some example codes are G89P0, 38098, BRJKW, and 003AZ. How many such codes are possible?

Sol. Number of letters (A-Z) = 26 Number of digits (0-9) = 10 So, each character in the passcode can be any of the 36 alphanumeric characters. Also, each of the 5 positions has 36 options. Total codes $= 3 6^{5} = 6, 04, 66, 176$ Hence number of possible 5-character alphanumeric passcodes is 6,04,66,176. 12. The worldwide population of sheep (2024) is about $1 0^{9}$ , and that of goats is also about the same. What is the total population of sheep and goats? (i) $2 0^{9}$ (ii) $1 0^{11}$ (iii) $1 0^{10}$ (iv) $1 0^{18}$ (v) $2 \times 1 0^{9}$ (vi) $1 0^{9} + 1 0^{9}$

Sol. Sheep population $= 1 0^{9}$ Goat population $= 1 0^{9}$ Total population of sheep and goats = $1 0^{9} + 1 0^{9} = 2 \times 1 0^{9}$ 13. Calculate and write the answer in scientific notation: (i) If each person in the world had 30 pieces of clothing, find the total number of pieces of clothing. (ii) There are about 100 million bee colonies in the world. Find the number of honeybees if each colony has about 50,000 bees. (iii) The human body has about 38 trillion bacterial cells. Find the bacterial population residing in all humans in the world. (iv) Total time spent eating in a lifetime in seconds.

Sol. (i) World population $= 8.2$ billion $= 8.2 \times 1 0^{9}$ Pieces of clothing $= 30$ pieces per person

Total pieces of clothing $= 8.2 \times 1 0^{9} \times 30$ $= 246 \times 1 0^{9}$ $= 2.46 \times 1 0^{11}$ pieces of clothing (ii) No. of bee colonies = 100 million $= 1 \times 1 0^{8}$ No. of bee per colony $= 50, 000$ $= 5 \times 1 0^{4}$ Total no. of bees $= 1 \times 1 0^{8} \times 5 \times 1 0^{4}$ $= 5 \times 1 0^{8 + 4}$ $= 5 \times 1 0^{12}$ honeybees (iii) No. of bacterial cells per human body $= 38$ trillion $= 3.8 \times 1 0^{13}$ World population (approx.) $= 8.2$ billion $= 8.2 \times 1 0^{9}$ Total bacterial population $= 3.8 \times 1 0^{13} \times 8.2 \times 1 0^{9}$ $= 31.16 \times 1 0^{13 + 9}$ $= 31.16 \times 1 0^{22}$ (iv) Let average eating time per day = 1.5 hours In seconds $= 1.5 \times 60 \times 60 = 5400$ $= 5.4 \times 1 0^{3}$ and an average person's lifetime (approx.) = 70 years In seconds $= 70 \times 365 \times 24 \times 60 \times 60$ = 161,148,960,000 $= 1.61 \times 1 0^{11}$ Total time spent in eating $= 5.4 \times 1 0^{3} \times 1.61 \times 1 0^{11}$ $= 8.694 \times 1 0^{3 + 11}$ $= 8.7 \times 1 0^{14}$ seconds 14. What was the date 1 arab/ 1 billion seconds ago?

Sol. 1 arab / 1 billion = $1 0^{9}$ minutes $= 1, 000, 000, 000/60$ hours $= 1, 000, 000, 000/ (60 \times 60)$ days $= 1, 000, 000, 000/ (60 \times 60 \times 24)$ years $= 1, 000, 000, 000/ (365 \times 60 \times 60 \times 24)$ Now, we go back to the calendar. So, it is approx. 31 years, 8 months, and 15 days. Let today's date = 1 Jan, 2026 After 31 years and 8 months and 15 days before the date was $24^{th}$ April, 1994 (approx.)

Very short answer type questions

1.Evaluate: (i) $4^{- 3}$ (ii) $(\frac{1}{2})^{- 5}$

2. Express as rational number: (i) $(\frac{- 3}{4})^{- 1}$ (ii) $(\frac{- 5}{7})^{- 3}$

3. Express as a power of a rational number with positive exponent: (i) $8^{6} \times 8^{- 5}$ (ii) $[(\frac{4}{5})^{- 2}]^{4}$

4. Express as a power of a rational number with negative exponent: (i) $(\frac{3}{4})^{2}$ (ii) $[(\frac{- 7}{3})^{2}]^{5}$

5. Express each of the following with positive indices: (i) $x^{\frac{- 1}{2}}$ (ii) $x^{\frac{- 2}{5}}$ (iii) $\frac{7}{x ^{- 5/6}}$ (iv) $(x^{- 3})^{4}$

6. Simplify and express in exponential form: (i) $(\frac{5}{2})^{3} \times (\frac{5}{2})^{7}$ (ii) $(\frac{- 2}{3})^{4} \times (\frac{- 2}{3})^{3} \times (\frac{- 2}{3})$ (iii) $(\frac{4}{9})^{6} \div (\frac{4}{9})^{4}$

7. State the exponential form of each expression : (i) $- 6 \times - 6 \times - 6 \times - 6$ (ii) $X \times X \times X \times X \times X \times X \times X$ (iii) $\frac{2}{3} \times \frac{2}{3} \times \frac{2}{3} \times \frac{2}{3} \times \frac{2}{3}$

8. Simplify : (i) $\frac{5 a ^{2} b ^{2}}{15 a ^{3} b ^{3}}$ (ii) $(6 x^{3} y^{2})^{2}$ (iii) $(4^{5} \div 4^{4}) \div 4^{6}$ (iv) $15 x^{3} \div 3 x^{5}$ (v) $18 x^{2} y^{2} \div 9 x^{3} y^{4}$

9. Simplify : (i) $(6 a^{2} bc^{2})^{2} \times (3 abc)^{3}$ (ii) $(15 p^{2} q^{2} r^{2})^{4} \div (25 pq r)^{2}$ (iii) $(6^{8} \div 6^{4}) \div (6^{5} \div 6^{3})$ (iv) $[(8 a^{2} b^{3} c^{2})^{3} \times (6 ab c)^{2}] \div 48 a^{4} b^{4} c^{4}$ (v) $[(9 a^{2} b^{4} c^{3})^{3} \times (4 ab c)^{2}] \div 36 a^{6} b^{6} c^{6}$

10. Express the following with a single exponent $((- 7)^{5} \times (- 7)^{4})^{2}$

11. Write in scientific notation (i) $8, 040, 000, 000$ (ii) $31, 408, 000, 000$

12. Express the following numbers in standard form: (i) 0.0000000000085 (ii) 0.00000000000942

13. Express the following numbers in usual form: (i) $3.02 \times 1 0^{- 6}$ (ii) $4.5 \times 1 0^{4}$ (iii) $3 \times 1 0^{- 8}$

Short answer type questions

14. Evaluate: (i) $(\frac{5}{3})^{2} \times (\frac{5}{3})^{2}$ (ii) $(\frac{5}{6})^{6} \times (\frac{5}{6})^{- 4}$

15. Evaluate: $(\frac{5}{9})^{- 2} \times (\frac{3}{5})^{- 3} \times (\frac{3}{5})^{0}$

16. Simplify: (i) $(- \frac{2}{5})^{- 3} \times (- \frac{2}{5})^{4}$ (ii) $(\frac{- 3}{4})^{- 5} \div (\frac{- 3}{4})^{- 3}$

17. Simplify: (i) $(\frac{3}{5})^{3} \times (\frac{15}{2})^{3}$ (ii) $((\frac{4}{7})^{8})^{0}$ (iii) $[(\frac{- 2}{3})^{3} \times (\frac{- 2}{3})] \div (\frac{- 2}{3})^{2}$

18. Evaluate: ${(\frac{1}{3})^{- 3} - (\frac{1}{2})^{- 3}} \div (\frac{1}{4})^{- 3}$

19. Evaluate: $[(5^{- 1} \times 3^{- 1})^{- 1} \div 6^{- 1}]$

20. Find the value of $[(- 2)^{- 2}]^{- 4}$

21. Simplify: $\frac{9 \times t ^{- 5}}{3 ^{- 2} \times 15 \times t ^{- 15}}; (t \neq = 0)$

22. Simplify : (i) $\frac{( 3 ^{3} ) ^{2} \times 5 ^{2}}{( 9 ) ^{2} \times 5}$ (ii) $\frac{( \frac{3}{5} ) ^{3} \times ( \frac{1}{7} ) ^{3}}{( \frac{3}{5} ) ^{2} \times ( \frac{1}{7} ) ^{4}}$

23. Simplify: $\frac{2 ^{2} \times 64 \times 8}{2 ^{3} \times 2 ^{5}}$

24. Simplify $\frac{3 ^{5} \times 1 0 ^{5} \times 25}{5 ^{7} \times 6 ^{5}}$

25. Simplify and write the terms with positive index. (i) $(- 5 x)^{2} (5 x)^{0} (- 1)^{3}$ (ii) $(8 a^{3}) (6 a)^{3} + (2 a)^{6}$ (iii) $(6 x^{3} y^{2})^{3} + (4 x^{3} y^{2})^{3}$ (iv) $\frac{( 5 x ^{5} ) ^{3} \times ( 15 x ^{2} ) ^{2}}{( 5 ) ^{4} ( 3 x ^{6} ) ^{4}}$ (v) $\frac{( 6 x ^{2} y ^{4} z ^{3} ) ^{3} \times ( 6 x ^{2} y ^{2} z ^{2} ) ^{3}}{( 24 x yz ) ^{3}}$

26. Simplify: (i) $(5^{- 1} \div 4^{- 1})^{3}$ (ii) $[(\frac{2}{3})^{- 1} \times (\frac{3}{4})^{- 1}]^{- 1}$

27. Find the reciprocal of the rational number $(\frac{1}{3})^{- 1} \div (\frac{4}{5})^{- 3}$ .

28. Simplify: (i) $3^{0} + 2^{- 2}$ (ii) $(\frac{3}{2})^{0} \times (\frac{4}{5})^{- 2}$

29. Insects fire ants are $2^{- 3}$ inches long. The ants are in a line across a porch that is $2^{6}$ inches long. How many fire ants are there?

30. Astronomy: The star Betelgeuse, in the constellation of Orion is approximately $3.36 \times 1 0^{15}$ miles from Earth. This is approximately $1.24 \times 1 0^{6}$ times as far as Pluto's minimum distance from Earth. What is Pluto's approximate minimum distance from Earth? Write your answer in scientific notation.

Long answer type questions

31. If $a + b + c = 0$ . Find the value of $\frac{( x ^{a} ) ^{3}}{x ^{- 3 b} \cdot x ^{- 3 c}}$

32. An adult person has roughly $24 \times 1 0^{12}$ red blood cells. The average diameter of each is $8 \times 1 0^{- 9} km$ . If these cells are placed end to end, how long would the line be?

33. A star in the Milky Way galaxy is $7 \times 1 0^{4}$ light years away from the earth and a light year is approximately $5.8 \times 1 0^{12}$ miles. Find out in scientific form the distance of the star from the earth in miles.

34. Biology: Escherichia coli is a type of bacterium that is sometimes found in swimming pools. Each E. coli bacterium has a mass of $2 \times 1 0^{- 12}$ gram. The number of bacteria increase so that, after 30 hours, one bacterium has been replaced by a population of $4.8 \times 1 0^{8}$ bacteria. (i) Suppose a pool begins with a population of only 1 bacterium. What would be the mass of the population after 30 hours? (ii) A small paper clip has a mass of about 1 gram. The paper clip has how many times the mass of the $4.8 \times 1 0^{8}$ E. coli bacteria?

35. Physical Science: The speed of light is about $2 \times 1 0^{5}$ miles per second. (i) On average, it takes light about 500 seconds to travel from the sun to Earth. What is the average distance from Earth to the sun? Write your answer in scientific notation. (ii) The star Alpha Centauri is approximately $2.5 \times 1 0^{13}$ miles from Earth. How many seconds does it take light to travel between Alpha Centauri and Earth?

ANSWER KEY

4.0Very short answer type questions

1. (i) $\frac{1}{64}$ (ii) 32 2. (i) $\frac{- 4}{3}$ (ii) $\frac{- 343}{125}$ 3. (i) 8 (ii) $(\frac{5}{4})^{8}$ 4. (i) $(\frac{4}{3})^{- 2}$ (ii) $(\frac{- 3}{7})^{- 10}$ 5. (i) $\frac{1}{x ^{1/2}}$ (ii) $\frac{1}{x ^{2/5}}$ (iii) $7 x^{5/6}$ (iv) $\frac{1}{x ^{12}}$ 6. (i) $(\frac{5}{2})^{10}$ (ii) $(\frac{- 2}{3})^{8}$ (iii) $(\frac{2}{3})^{4}$ 7. (i) $(- 6)^{4}$ (ii) $x^{7}$ (iii) $(\frac{2}{3})^{5}$ 8. (i) $\frac{1}{3 ab}$ (ii) $36 x^{6} y^{4}$ (iii) $(4)^{- 5}$ (iv) $\frac{5}{x ^{2}}$ (v) $\frac{2}{xy ^{2}}$ 9. (i) $972 a^{7} b^{5} c^{7}$ (ii) $81 p^{6} q^{6} r^{6}$ (iii) 36 (iv) $384 a^{4} b^{7} c^{4}$ (v) $324 a^{2} b^{8} c^{5}$ 10. $(- 7)^{18}$ 11. (i) $8.04 \times 1 0^{9}$ (ii) $3.1408 \times 1 0^{10}$ 12. (i) $8.5 \times 1 0^{- 12}$ (ii) $9.42 \times 1 0^{- 12}$ 13. (i) 0.00000302 (ii) 45000 (iii) 0.00000003

Short answer type questions 14. (i) $\frac{625}{81}$ (ii) $\frac{25}{36}$ 15. 15 16. (i) $\frac{- 2}{5}$ (ii) $\frac{16}{9}$ 17. (i) $\frac{729}{8}$ (ii) 1 (iii) $\frac{4}{9}$ 18. $\frac{19}{64}$ 19. 90 20. $(- 2)^{8} = 256$ 21. $\frac{27 t ^{10}}{5}$ 22. (i) 45 (ii) $\frac{21}{5}$ 23.8 24.1 25. (i) $- 25 x^{2}$ (ii) $1792 a^{6}$ (iii) $280 x^{9} y^{6}$ (iv) $\frac{5}{9 x ^{5}}$ (v) $\frac{27}{8} x^{9} y^{15} z^{12}$ 26. (i) $\frac{64}{125}$ (ii) $\frac{1}{2}$ 27. $\frac{125}{192}$ 28. (i) $\frac{5}{4}$ (ii) $\frac{25}{16}$ 29. $512 = 2^{9}$ 30. $2.7 \times 1 0^{9}$ miles

Long answer type questions 31. 1 32. $1, 92, 000 km$ 33. $4.06 \times 1 0^{17}$ miles 34. (i) $9.6 \times 1 0^{- 4}$ grams (ii) $1.04 \times 1 0^{3}$ 35. (i) $1 0^{8}$ miles (ii) $1.25 \times 1 0^{8} sec$

5.0Key Concepts of Chapter 2 Exercise 2.2

Standard Form (Scientific Notation): Expressing numbers as $a \times 1 0^{n}$ where $1 \leq a < 10$ and n is an integer.
Positive Exponents: Used when converting very large numbers into standard form (moving the decimal to the left).
Negative Exponents: Used when converting very small decimals into standard form (moving the decimal to the right).
Usual Form: Converting a number back from scientific notation to its expanded decimal or whole-number format.
Comparing Numbers: Using powers of 10 to quickly determine which of two very large or very small numbers is greater.

6.0Benefits of NCERT Solutions for Class 8 Maths Chapter 2 Exercise 2.2

Conceptual Clarity: Detailed explanations of how moving the decimal point affects the exponent of 10.
Exam-Oriented: Covers all major conversion types found in CBSE school tests.
Logical Steps: Every solution shows the count of decimal places moved to ensure no zeros are missed.
Error Prevention: Teaches students a consistent direction-based rule (Left = Positive, Right = Negative) to avoid exponent sign errors.

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NCERT Class 8 Maths Chapter 2 Exercise 2.2 Solutions - Scientific Notation & Standard Form

1.0Download Class 8 Maths Chapter 2 Ex 2.2 NCERT Solutions PDF

2.0NCERT Solutions for Class 8 Maths Chapter 2 Power Play : All Exercises

3.0Detailed NCERT Class 8 Maths Chapter 2 Solutions of Exercise 2.2

Very short answer type questions

Short answer type questions

Long answer type questions

ANSWER KEY

4.0Very short answer type questions

5.0Key Concepts of Chapter 2 Exercise 2.2

6.0Benefits of NCERT Solutions for Class 8 Maths Chapter 2 Exercise 2.2