NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers

*According to the CBSE Syllabus 2023-24, this chapter has been renumbered as Chapter 11.

NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers are given here in detail. Maths is a subject which requires understanding and reasoning skills accompanied by logic. Along with this, students should practise solving Maths problems regularly. Students of Class 7 are suggested to solve NCERT Solutions Chapter 13 to strengthen their fundamentals and be able to solve questions that are usually asked in the examination.

A total of 3 exercises are present in Chapter 13 – Exponents and Powers of NCERT Solutions for Class 7 Maths. These exercises cover various concepts mentioned in the chapter, which are given below.

• Exponents
• Laws of Exponents
• Multiplying Powers with the Same Base
• Dividing Powers with the Same Base
• Taking Power of a Power
• Multiplying Powers with the Same Base
• Dividing Powers with the Same Exponents
• Miscellaneous Examples Using the Laws of Exponents
• Decimal Number System
• Expressing Large Numbers in the Standard Form

NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers

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Exercise 13.1 Solutions

Exercise 13.2 Solutions

Exercise 13.3 Solutions

Access Answers to Maths NCERT Solutions for Class 7 Chapter 13 – Exponents and Powers

Exercise 13.1 Page: 252

1. Find the value of:

(i) 2⁶

Solution:-

The above value can be written as,

= 2 × 2 × 2 × 2 × 2 × 2

= 64

(ii) 9³

Solution:-

The above value can be written as,

= 9 × 9 × 9

= 729

(iii) 11²

Solution:-

The above value can be written as,

= 11 × 11

= 121

(iv) 5⁴

Solution:-

The above value can be written as,

= 5 × 5 × 5 × 5

= 625

2. Express the following in exponential form:

(i) 6 × 6 × 6 × 6

Solution:-

The given question can be expressed in the exponential form as 6⁴.

(ii) t × t

Solution:-

The given question can be expressed in the exponential form as t².

(iii) b × b × b × b

Solution:-

The given question can be expressed in the exponential form as b⁴.

(iv) 5 × 5× 7 × 7 × 7

Solution:-

The given question can be expressed in the exponential form as 5² × 7³.

(v) 2 × 2 × a × a

Solution:-

The given question can be expressed in the exponential form as 2² × a².

(vi) a × a × a × c × c × c × c × d

Solution:-

The given question can be expressed in the exponential form as a³ × c⁴ × d.

3. Express each of the following numbers using the exponential notation:

(i) 512

Solution:-

The factors of 512 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2

So it can be expressed in the exponential form as 2⁹.

(ii) 343

Solution:-

The factors of 343 = 7 × 7 × 7

So it can be expressed in the exponential form as 7³.

(iii) 729

Solution:-

The factors of 729 = 3 × 3 × 3 × 3 × 3 × 3

So it can be expressed in the exponential form as 3⁶.

(iv) 3125

Solution:-

The factors of 3125 = 5 × 5 × 5 × 5 × 5

So it can be expressed in the exponential form as 5⁵.

4. Identify the greater number, wherever possible, in each of the following.

(i) 4³ or 3⁴

Solution:-

The expansion of 4³ = 4 × 4 × 4 = 64

The expansion of 3⁴ = 3 × 3 × 3 × 3 = 81

Clearly,

64 < 81

So, 4³ < 3⁴

Hence, 3⁴ is the greater number.

(ii) 5³ or 3⁵

Solution:-

The expansion of 5³ = 5 × 5 × 5 = 125

The expansion of 3⁵ = 3 × 3 × 3 × 3 × 3= 243

Clearly,

125 < 243

So, 5³ < 3⁵

Hence, 3⁵ is the greater number.

(iii) 2⁸ or 8²

Solution:-

The expansion of 2⁸ = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 256

The expansion of 8² = 8 × 8= 64

Clearly,

256 > 64

So, 2⁸ > 8²

Hence, 2⁸ is the greater number.

(iv) 100² or 2¹⁰⁰

Solution:-

The expansion of 100² = 100 × 100 = 10000

The expansion of 2¹⁰⁰

2¹⁰ = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 1024

Then,

2¹⁰⁰ = 1024 × 1024 ×1024 × 1024 ×1024 × 1024 × 1024 × 1024 × 1024 × 1024 = (1024)¹⁰

Clearly,

100² < 2¹⁰⁰

Hence, 2¹⁰⁰ is the greater number.

(v) 2¹⁰ or 10²

Solution:-

The expansion of 2¹⁰ = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 1024

The expansion of 10² = 10 × 10= 100

Clearly,

1024 > 100

So, 2¹⁰ > 10²

Hence, 2¹⁰ is the greater number.

5. Express each of the following as a product of powers of their prime factors:

(i) 648

Solution:-

Factors of 648 = 2 × 2 × 2 × 3 × 3 × 3 × 3

= 2³ × 3⁴

(ii) 405

Solution:-

Factors of 405 = 3 × 3 × 3 × 3 × 5

= 3⁴ × 5

(iii) 540

Solution:-

Factors of 540 = 2 × 2 × 3 × 3 × 3 × 5

= 2² × 3³ × 5

(iv) 3,600

Solution:-

Factors of 3600 = 2 × 2 × 2 × 2 × 3 × 3 × 5 × 5

= 2⁴ × 3² × 5²

6. Simplify:

(i) 2 × 10³

Solution:-

The above question can be written as,

= 2 × 10 × 10 × 10

= 2 × 1000

= 2000

(ii) 7² × 2²

Solution:-

The above question can be written as,

= 7 × 7 × 2 × 2

= 49 × 4

= 196

(iii) 2³ × 5

Solution:-

The above question can be written as,

= 2 × 2 × 2 × 5

= 8 × 5

= 40

(iv) 3 × 4⁴

Solution:-

The above question can be written as,

= 3 × 4 × 4 × 4 × 4

= 3 × 256

= 768

(v) 0 × 10²

Solution:-

The above question can be written as,

= 0 × 10 × 10

= 0 × 100

= 0

(vi) 5² × 3³

Solution:-

The above question can be written as,

= 5 × 5 × 3 × 3 × 3

= 25 × 27

= 675

(vii) 2⁴ × 3²

Solution:-

The above question can be written as,

= 2 × 2 × 2 × 2 × 3 × 3

= 16 × 9

= 144

(viii) 3² × 10⁴

Solution:-

The above question can be written as,

= 3 × 3 × 10 × 10 × 10 × 10

= 9 × 10000

= 90000

7. Simplify:

(i) (– 4)³

Solution:-

The expansion of -4³

= – 4 × – 4 × – 4

= – 64

(ii) (–3) × (–2)³

Solution:-

The expansion of (-3) × (-2)³

= – 3 × – 2 × – 2 × – 2

= – 3 × – 8

= 24

(iii) (–3)² × (–5)²

Solution:-

The expansion of (-3)² × (-5)²

= – 3 × – 3 × – 5 × – 5

= 9 × 25

= 225

(iv) (–2)³ × (–10)³

Solution:-

The expansion of (-2)³ × (-10)³

= – 2 × – 2 × – 2 × – 10 × – 10 × – 10

= – 8 × – 1000

= 8000

8. Compare the following numbers:

(i) 2.7 × 10¹² ; 1.5 × 10⁸

Solution:-

By observing the question

Comparing the exponents of base 10,

Clearly,

2.7 × 10¹² > 1.5 × 10⁸

(ii) 4 × 10¹⁴ ; 3 × 10¹⁷

Solution:-

By observing the question

Comparing the exponents of base 10,

Clearly,

4 × 10¹⁴ < 3 × 10¹⁷

Exercise 13.2 Page: 260

1. Using laws of exponents, simplify and write the answer in exponential form:

(i) 3² × 3⁴ × 3⁸

Solution:-

By the rule of multiplying the powers with the same base = a^m × aⁿ = a^{m + n}

Then,

= (3)^{2 + 4 + 8}

= 3¹⁴

(ii) 6¹⁵ ÷ 6¹⁰

Solution:-

By the rule of dividing the powers with the same base = a^m ÷ aⁿ = a^{m – n}

Then,

= (6)^{15 – 10}

= 6⁵

(iii) a³ × a²

Solution:-

By the rule of multiplying the powers with the same base = a^m × aⁿ = a^{m + n}

Then,

= (a)^{3 + 2}

= a⁵

(iv) 7^x × 7²

Solution:-

By the rule of multiplying the powers with the same base = a^m × aⁿ = a^{m + n}

Then,

= (7)^{x + 2}

(v) (5²)³ ÷ 5³

Solution:-

By the rule of taking the power of as power = (a^m)ⁿ = a^mn

(5²)³ can be written as = (5)^{2 × 3}

= 5⁶

Now, 5⁶ ÷ 5³

By the rule of dividing the powers with the same base = a^m ÷ aⁿ = a^{m – n}

Then,

= (5)^{6 – 3}

= 5³

(vi) 2⁵ × 5⁵

Solution:-

By the rule of multiplying the powers with the same exponents = a^m × b^m = ab^m

Then,

= (2 × 5)⁵

= 10⁵

(vii) a⁴ × b⁴

Solution:-

By the rule of multiplying the powers with the same exponents = a^m × b^m = ab^m

Then,

= (a × b)⁴

= ab⁴

(viii) (3⁴)³

Solution:-

By the rule of taking the power of as power = (a^m)ⁿ = a^mn

(3⁴)³ can be written as = (3)^{4 × 3}

= 3¹²

(ix) (2²⁰ ÷ 2¹⁵) × 2³

Solution:-

By the rule of dividing the powers with the same base = a^m ÷ aⁿ = a^{m – n}

(2²⁰ ÷ 2¹⁵) can be simplified as,

= (2)^{20 – 15}

= 2⁵

Then,

By the rule of multiplying the powers with the same base = a^m × aⁿ = a^{m + n}

2⁵ × 2³ can be simplified as,

= (2)^{5 + 3}

= 2⁸

(x) 8^t ÷ 8²

Solution:-

By the rule of dividing the powers with the same base = a^m ÷ aⁿ = a^{m – n}

Then,

= (8)^{t – 2}

2. Simplify and express each of the following in exponential form:

(i) (2³ × 3⁴ × 4)/ (3 × 32)

Solution:-

Factors of 32 = 2 × 2 × 2 × 2 × 2

= 2⁵

Factors of 4 = 2 × 2

= 2²

Then,

= (2³ × 3⁴ × 2²)/ (3 × 2⁵)

= (2^{3 + 2} × 3⁴) / (3 × 2⁵) … [∵a^m × aⁿ = a^{m + n}]

= (2⁵ × 3⁴) / (3 × 2⁵)

= 2^{5 – 5} × 3^{4 – 1} … [∵a^m ÷ aⁿ = a^{m – n}]

= 2⁰ × 3³

= 1 × 3³

= 3³

(ii) ((5²)³ × 5⁴) ÷ 5⁷

Solution:-

(5²)³ can be written as = (5)^{2 × 3} … [∵(a^m)ⁿ = a^mn]

= 5⁶

Then,

= (5⁶ × 5⁴) ÷ 5⁷

= (5^{6 + 4}) ÷ 5⁷ … [∵a^m × aⁿ = a^{m + n}]

= 5¹⁰ ÷ 5⁷

= 5^{10 – 7} … [∵a^m ÷ aⁿ = a^{m – n}]

= 5³

(iii) 25⁴ ÷ 5³

Solution:-

(25)⁴ can be written as = (5 × 5)⁴

= (5²)⁴

(5²)⁴ can be written as = (5)^{2 × 4} … [∵(a^m)ⁿ = a^mn]

= 5⁸

Then,

= 5⁸ ÷ 5³

= 5^{8 – 3} … [∵a^m ÷ aⁿ = a^{m – n}]

= 5⁵

(iv) (3 × 7² × 11⁸)/ (21 × 11³)

Solution:-

Factors of 21 = 7 × 3

Then,

= (3 × 7² × 11⁸)/ (7 × 3 × 11³)

= 3^1-1 × 7^2-1 × 11^{8 – 3}

= 3⁰ × 7 × 11⁵

= 1 × 7 × 11⁵

= 7 × 11⁵

(v) 3⁷/ (3⁴ × 3³)

Solution:-

= 3⁷/ (3⁴⁺³) … [∵a^m × aⁿ = a^{m + n}]

= 3⁷/ 3⁷

= 3^{7 – 7} … [∵a^m ÷ aⁿ = a^{m – n}]

= 3⁰

= 1

(vi) 2⁰ + 3⁰ + 4⁰

Solution:-

= 1 + 1 + 1

= 3

(vii) 2⁰ × 3⁰ × 4⁰

Solution:-

= 1 × 1 × 1

= 1

(viii) (3⁰ + 2⁰) × 5⁰

Solution:-

= (1 + 1) × 1

= (2) × 1

= 2

(ix) (2⁸ × a⁵)/ (4³ × a³)

Solution:-

(4)³ can be written as = (2 × 2)³

= (2²)³

(2²)³ can be written as = (2)^{2 × 3} … [∵(a^m)ⁿ = a^mn]

= 2⁶

Then,

= (2⁸ × a⁵)/ (2⁶ × a³)

= 2^{8 – 6} × a^{5 – 3} … [∵a^m ÷ aⁿ = a^{m – n}]

= 2² × a² … [∵(a^m)ⁿ = a^mn]

= 2a²

(x) (a⁵/a³) × a⁸

Solution:-

= (a^{5 -3}) × a⁸ … [∵a^m ÷ aⁿ = a^{m – n}]

= a² × a⁸

= a^{2 + 8} … [∵a^m × aⁿ = a^{m + n}]

= a¹⁰

(xi) (4⁵ × a⁸b³)/ (4⁵ × a⁵b²)

Solution:-

= 4^{5 – 5} × (a^{8 – 5} × b^{3 – 2}) … [∵a^m ÷ aⁿ = a^{m – n}]

= 4⁰ × (a³b)

= 1 × a³b

= a³b

(xii) (2³ × 2)²

Solution:-

= (2^{3 + 1})² … [∵a^m × aⁿ = a^{m + n}]

= (2⁴)²

(2⁴)² can be written as = (2)^{4 × 2} … [∵(a^m)ⁿ = a^mn]

= 2⁸

3. Say true or false and justify your answer:

(i) 10 × 10¹¹ = 100¹¹

Solution:-

Let us consider Left Hand Side (LHS) = 10 × 10¹¹

= 10^{1 + 11} … [∵a^m × aⁿ = a^{m + n}]

= 10¹²

Now, consider Right Hand Side (RHS) = 100¹¹

= (10 × 10)¹¹

= (10^{1 + 1})¹¹

= (10²)¹¹

= (10)^{2 × 11} … [∵(a^m)ⁿ = a^mn]

= 10²²

By comparing LHS and RHS,

LHS ≠ RHS

Hence, the given statement is false.

(ii) 2³ > 5²

Solution:-

Let us consider LHS = 2³

Expansion of 2³ = 2 × 2 × 2

= 8

Now, consider RHS = 5²

Expansion of 5² = 5 × 5

= 25

By comparing LHS and RHS,

LHS < RHS

2³ < 5²

Hence, the given statement is false.

(iii) 2³ × 3² = 6⁵

Solution:-

Let us consider LHS = 2³ × 3²

Expansion of 2³ × 3²= 2 × 2 × 2 × 3 × 3

= 72

Now, consider RHS = 6⁵

Expansion of 6⁵ = 6 × 6 × 6 × 6 × 6

= 7776

By comparing LHS and RHS,

72 ≠ 7776

LHS ≠ RHS

Hence, the given statement is false.

(iv) 3⁰ = (1000)⁰

Solution:-

Let us consider LHS = 3⁰

= 1

Now, consider RHS = 1000⁰

= 1

By comparing LHS and RHS,

LHS = RHS

3⁰ = 1000⁰

Hence, the given statement is true.

4. Express each of the following as a product of prime factors only in exponential form:

(i) 108 × 192

Solution:-

The factors of 108 = 2 × 2 × 3 × 3 × 3

= 2² × 3³

The factors of 192 = 2 × 2 × 2 × 2 × 2 × 2 × 3

= 2⁶ × 3

Then,

= (2² × 3³) × (2⁶ × 3)

= 2^{2 + 6} × 3^{3 + 1} … [∵a^m × aⁿ = a^{m + n}]

= 2⁸ × 3⁴

(ii) 270

Solution:-

The factors of 270 = 2 × 3 × 3 × 3 × 5

= 2 × 3³ × 5

(iii) 729 × 64

The factors of 729 = 3 × 3 × 3 × 3 × 3 × 3

= 3⁶

The factors of 64 = 2 × 2 × 2 × 2 × 2 × 2

= 2⁶

Then,

= (3⁶ × 2⁶)

= 3⁶ × 2⁶

(iv) 768

Solution:-

The factors of 768 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3

= 2⁸ × 3

5. Simplify:

(i) ((2⁵)² × 7³)/ (8³ × 7)

Solution:-

8³ can be written as = (2 × 2 × 2)³

= (2³)³

We have,

= ((2⁵)² × 7³)/ ((2³)³ × 7)

= (2^{5 × 2} × 7³)/ ((2^{3 × 3} × 7) … [∵(a^m)ⁿ = a^mn]

= (2¹⁰ × 7³)/ (2⁹ × 7)

= (2^{10 – 9} × 7^{3 – 1}) … [∵a^m ÷ aⁿ = a^{m – n}]

= 2 × 7²

= 2 × 7 × 7

= 98

(ii) (25 × 5² × t⁸)/ (10³ × t⁴)

Solution:-

25 can be written as = 5 × 5

= 5²

10³ can be written as = 10³

= (5 × 2)³

= 5³ × 2³

We have,

= (5² × 5² × t⁸)/ (5³ × 2³ × t⁴)

= (5^{2 + 2} × t⁸)/ (5³ × 2³ × t⁴) … [∵a^m × aⁿ = a^{m + n}]

= (5⁴ × t⁸)/ (5³ × 2³ × t⁴)

= (5^{4 – 3} × t^{8 – 4})/ 2³ … [∵a^m ÷ aⁿ = a^{m – n}]

= (5 × t⁴)/ (2 × 2 × 2)

= (5t⁴)/ 8

(iii) (3⁵ × 10⁵ × 25)/ (5⁷ × 6⁵)

Solution:-

10⁵ can be written as = (5 × 2)⁵

= 5⁵ × 2⁵

25 can be written as = 5 × 5

= 5²

6⁵ can be written as = (2 × 3)⁵

= 2⁵ × 3⁵

Then we have,

= (3⁵ × 5⁵ × 2⁵ × 5²)/ (5⁷ × 2⁵ × 3⁵)

= (3⁵ × 5^{5 + 2} × 2⁵)/ (5⁷ × 2⁵ × 3⁵) … [∵a^m × aⁿ = a^{m + n}]

= (3⁵ × 5⁷ × 2⁵)/ (5⁷ × 2⁵ × 3⁵)

= (3^{5 – 5} × 5^{7 – 7} × 2^{5 – 5})

= (3⁰ × 5⁰ × 2⁰) … [∵a^m ÷ aⁿ = a^{m – n}]

= 1 × 1 × 1

= 1

Exercise 13.3 Page: 263

1. Write the following numbers in the expanded forms:

(a) 279404

Solution:-

The expanded form of the number 279404 is,

= (2 × 100000) + (7 × 10000) + (9 × 1000) + (4 × 100) + (0 × 10) + (4 × 1)

Now we can express it using powers of 10 in the exponent form,

= (2 × 10⁵) + (7 × 10⁴) + (9 × 10³) + (4 × 10²) + (0 × 10¹) + (4 × 10⁰)

(b) 3006194

Solution:-

The expanded form of the number 3006194 is,

= (3 × 1000000) + (0 × 100000) + (0 × 10000) + (6 × 1000) + (1 × 100) + (9 × 10) + (4 × 1)

Now we can express it using powers of 10 in the exponent form,

= (3 × 10⁶) + (0 × 10⁵) + (0 × 10⁴) + (6 × 10³) + (1 × 10²) + (9 × 10¹) + (4 × 10⁰)

(c) 2806196

Solution:-

The expanded form of the number 2806196 is,

= (2 × 1000000) + (8 × 100000) + (0 × 10000) + (6 × 1000) + (1 × 100) + (9 × 10) + (6 × 1)

Now we can express it using powers of 10 in the exponent form,

= (2 × 10⁶) + (8 × 10⁵) + (0 × 10⁴) + (6 × 10³) + (1 × 10²) + (9 × 10¹) + (6 × 10⁰)

(d) 120719

Solution:-

The expanded form of the number 120719 is,

= (1 × 100000) + (2 × 10000) + (0 × 1000) + (7 × 100) + (1 × 10) + (9 × 1)

Now we can express it using powers of 10 in the exponent form,

= (1 × 10⁵) + (2 × 10⁴) + (0 × 10³) + (7 × 10²) + (1 × 10¹) + (9 × 10⁰)

(e) 20068

Solution:-

The expanded form of the number 20068 is,

= (2 × 10000) + (0 × 1000) + (0 × 100) + (6 × 10) + (8 × 1)

Now we can express it using powers of 10 in the exponent form,

= (2 × 10⁴) + (0 × 10³) + (0 × 10²) + (6 × 10¹) + (8 × 10⁰)

2. Find the number from each of the following expanded forms:

(a) (8 × 10)⁴ + (6 × 10)³ + (0 × 10)² + (4 × 10)¹ + (5 × 10)⁰

Solution:-

The expanded form is,

= (8 × 10000) + (6 × 1000) + (0 × 100) + (4 × 10) + (5 × 1)

= 80000 + 6000 + 0 + 40 + 5

= 86045

(b) (4 × 10)⁵ + (5 × 10)³ + (3 × 10)² + (2 × 10)⁰

Solution:-

The expanded form is,

= (4 × 100000) + (0 × 10000) + (5 × 1000) + (3 × 100) + (0 × 10) + (2 × 1)

= 400000 + 0 + 5000 + 300 + 0 + 2

= 405302

(c) (3 × 10)⁴ + (7 × 10)² + (5 × 10)⁰

Solution:-

The expanded form is,

= (3 × 10000) + (0 × 1000) + (7 × 100) + (0 × 10) + (5 × 1)

= 30000 + 0 + 700 + 0 + 5

= 30705

(d) (9 × 10)⁵ + (2 × 10)² + (3 × 10)¹

Solution:-

The expanded form is,

= (9 × 100000) + (0 × 10000) + (0 × 1000) + (2 × 100) + (3 × 10) + (0 × 1)

= 900000 + 0 + 0 + 200 + 30 + 0

= 900230

3. Express the following numbers in standard form:

(i) 5,00,00,000

Solution:-

The standard form of the given number is 5 × 10⁷

(ii) 70,00,000

Solution:-

The standard form of the given number is 7 × 10⁶

(iii) 3,18,65,00,000

Solution:-

The standard form of the given number is 3.1865 × 10⁹

(iv) 3,90,878

Solution:-

The standard form of the given number is 3.90878 × 10⁵

(v) 39087.8

Solution:-

The standard form of the given number is 3.90878 × 10⁴

(vi) 3908.78

Solution:-

The standard form of the given number is 3.90878 × 10³

4. Express the number appearing in the following statements in standard form.

(a) The distance between Earth and Moon is 384,000,000 m.

Solution:-

The standard form of the number appearing in the given statement is 3.84 × 10⁸m.

(b) Speed of light in a vacuum is 300,000,000 m/s.

Solution:-

The standard form of the number appearing in the given statement is 3 × 10⁸m/s.

(c) Diameter of the Earth is 1,27,56,000 m.

Solution:-

The standard form of the number appearing in the given statement is 1.2756 × 10⁷m.

(d) Diameter of the Sun is 1,400,000,000 m.

Solution:-

The standard form of the number appearing in the given statement is 1.4 × 10⁹m.

(e) In a galaxy, there are, on average, 100,000,000,000 stars.

Solution:-

The standard form of the number appearing in the given statement is 1 × 10¹¹ stars.

(f) The universe is estimated to be about 12,000,000,000 years old.

Solution:-

The standard form of the number appearing in the given statement is 1.2 × 10¹⁰ years old.

(g) The distance of the Sun from the centre of the Milky Way Galaxy is estimated to be 300,000,000,000,000,000,000 m.

Solution:-

The standard form of the number appearing in the given statement is 3 × 10²⁰m.

(h) 60,230,000,000,000,000,000,000 molecules are contained in a drop of water weighing 1.8 gm.

Solution:-

The standard form of the number appearing in the given statement is 6.023 × 10²² molecules.

(i) The Earth has 1,353,000,000 cubic km of seawater.

Solution:-

The standard form of the number appearing in the given statement is 1.353 × 10⁹ cubic km.

(j) The population of India was about 1,027,000,000 in March 2001.

Solution:-

The standard form of the number appearing in the given statement is 1.027 × 10⁹.