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Exponents and Powers - Laws of exponents with negative integral powers and expressing numbers in standard form
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MathematicsClass 8Exponents and Powers

Exponents and Powers - Laws of exponents with negative integral powers and expressing numbers in standard form

2026-10-1713 min readRHS Academic Faculty
Overview & Key Summary:Exponents and Powers Laws of Exponents with Negative Integral Powers and Standard Form In mathematics, we frequently encounter quantities that are either extremely large, such a...

Exponents and Powers - Laws of Exponents with Negative Integral Powers and Standard Form

In mathematics, we frequently encounter quantities that are either extremely large, such as the distance between celestial bodies, or unimaginably small, such as the mass of an electron or the size of a bacterial cell. Writing and computing with numbers containing long strings of zeros is inefficient and prone to error.

The concept of Exponents and Powers provides a systematic mathematical notation to express these numbers concisely. While you have previously worked with positive integral exponents (such as 103=100010^3 = 1000103=1000), this guide extends these principles to negative integral exponents and demonstrates how the fundamental Laws of Exponents apply universally. Furthermore, we will explore how to express numbers in Standard Form (scientific notation), a critical skill used extensively across mathematics, physics, chemistry, and engineering.


1. In-Depth Conceptual Breakdown

Understanding Negative Exponents

To understand what a negative exponent represents, let us analyze the pattern formed by decreasing powers of 101010:

103=100010^3 = 1000103=1000 102=100(divided by 10)10^2 = 100 \quad \text{(divided by 10)}102=100(divided by 10) 101=10(divided by 10)10^1 = 10 \quad \text{(divided by 10)}101=10(divided by 10) 100=1(divided by 10)10^0 = 1 \quad \text{(divided by 10)}100=1(divided by 10)

Continuing this division pattern beyond zero:

10−1=110=110110^{-1} = \frac{1}{10} = \frac{1}{10^1}10−1=101​=1011​ 10−2=1100=110210^{-2} = \frac{1}{100} = \frac{1}{10^2}10−2=1001​=1021​ 10−3=11000=110310^{-3} = \frac{1}{1000} = \frac{1}{10^3}10−3=10001​=1031​

General Rule for Negative Exponents

For any non-zero rational number aaa and any positive integer mmm:

a−m=1ama^{-m} = \frac{1}{a^m}a−m=am1​

Here, a−ma^{-m}a−m is the multiplicative inverse (or reciprocal) of ama^mam.

Conversely:

1a−m=am\frac{1}{a^{-m}} = a^ma−m1​=am

If we have a fractional base (ab)\left(\frac{a}{b}\right)(ba​) raised to a negative exponent −m-m−m, we take the reciprocal of the base and change the sign of the exponent:

(ab)−m=(ba)m=bmam\left(\frac{a}{b}\right)^{-m} = \left(\frac{b}{a}\right)^m = \frac{b^m}{a^m}(ba​)−m=(ab​)m=ambm​


The Comprehensive Laws of Exponents

The laws of exponents established for positive integers hold true for any integer exponent (positive, negative, or zero). Let aaa and bbb be non-zero rational numbers, and let mmm and nnn be any integers.

Law NameMathematical FormulaConceptual MeaningExample
Product Lawam×an=am+na^m \times a^n = a^{m+n}am×an=am+nWhen multiplying powers with the same base, add the exponents.2−3×25=2−3+5=22=42^{-3} \times 2^5 = 2^{-3+5} = 2^2 = 42−3×25=2−3+5=22=4
Quotient Lawam÷an=am−na^m \div a^n = a^{m-n}am÷an=am−nWhen dividing powers with the same base, subtract the exponents.52÷5−3=52−(−3)=555^2 \div 5^{-3} = 5^{2 - (-3)} = 5^552÷5−3=52−(−3)=55
Power of a Power Law(am)n=am⋅n(a^m)^n = a^{m \cdot n}(am)n=am⋅nRaising a power to another exponent multiplies the exponents.(3−2)3=3−2×3=3−6(3^{-2})^3 = 3^{-2 \times 3} = 3^{-6}(3−2)3=3−2×3=3−6
Power of a Product Lawam×bm=(a×b)ma^m \times b^m = (a \times b)^mam×bm=(a×b)mPowers with different bases but the same exponent can combine bases.2−3×3−3=(2×3)−3=6−32^{-3} \times 3^{-3} = (2 \times 3)^{-3} = 6^{-3}2−3×3−3=(2×3)−3=6−3
Power of a Quotient Lawambm=(ab)m\frac{a^m}{b^m} = \left(\frac{a}{b}\right)^mbmam​=(ba​)mQuotient of powers with the same exponent equals the power of the quotient.4−22−2=(42)−2=2−2\frac{4^{-2}}{2^{-2}} = \left(\frac{4}{2}\right)^{-2} = 2^{-2}2−24−2​=(24​)−2=2−2
Zero Exponent Lawa0=1a^0 = 1a0=1Any non-zero base raised to the power zero equals 111.(−7)0=1(-7)^0 = 1(−7)0=1
Negative Exponent Lawa−m=1ama^{-m} = \frac{1}{a^m}a−m=am1​A negative exponent indicates the reciprocal of the base raised to a positive power.4−3=143=1644^{-3} = \frac{1}{4^3} = \frac{1}{64}4−3=431​=641​

Expressing Numbers in Standard Form

A number is said to be in Standard Form (or scientific notation) when it is written as a product of a number between 1.01.01.0 and 10.010.010.0 (including 1.01.01.0, but strictly less than 10.010.010.0) and an integral power of 101010.

Standard Form=m×10n\text{Standard Form} = m \times 10^nStandard Form=m×10n

Where:

  • 1≤m<101 \le m < 101≤m<10 (mmm is a decimal number with one non-zero digit to the left of the decimal point).
  • nnn is an integer (positive for large numbers, negative for small numbers).
                      ┌── Number between 1.0 and 10.0 (1 ≤ m < 10)
                      │
                   m  ×  10ⁿ ─── Integer exponent (positive or negative)

Rule 1: Expressing Very Large Numbers (n>0n > 0n>0)

  1. Place the decimal point after the first non-zero digit from the left.
  2. Count how many places the decimal point moved to the left.
  3. That count becomes the positive exponent nnn of 101010.

Example: Express 150,000,000,000150,000,000,000150,000,000,000 in standard form.

  • The first non-zero digit is 111. So m=1.5m = 1.5m=1.5.
  • The decimal moved 11 places to the left.
  • Standard form: 1.5×10111.5 \times 10^{11}1.5×1011.

Rule 2: Expressing Very Small Numbers (n<0n < 0n<0)

  1. Place the decimal point after the first non-zero digit from the left.
  2. Count how many places the decimal point moved to the right.
  3. That count becomes the negative exponent −n-n−n of 101010.

Example: Express 0.00000780.00000780.0000078 in standard form.

  • The first non-zero digit is 777. So m=7.8m = 7.8m=7.8.
  • The decimal moved 6 places to the right.
  • Standard form: 7.8×10−67.8 \times 10^{-6}7.8×10−6.

Rule 3: Converting Standard Form back to Usual Form

  • If the power of 101010 is positive, move the decimal point to the right by nnn places.
  • If the power of 101010 is negative, move the decimal point to the left by nnn places.

Comparing Numbers in Standard Form

To compare two numbers written in standard form:

  1. Compare the Exponents of 101010: The number with the larger exponent of 101010 is greater.
    • Example: 3.2×10−4>8.9×10−63.2 \times 10^{-4} > 8.9 \times 10^{-6}3.2×10−4>8.9×10−6 because −4>−6-4 > -6−4>−6.
  2. If the Exponents are Equal: Compare the decimal factors (mmm).
    • Example: 5.4×108>2.7×1085.4 \times 10^8 > 2.7 \times 10^85.4×108>2.7×108 because 5.4>2.75.4 > 2.75.4>2.7.

2. Real-World Applications

1. Nanoscience and Cell Biology

Cellular structures and nanotechnology deal with objects far smaller than a millimeter.

  • The size of a human red blood cell is approximately 0.000007 m=7×10−6 m0.000007\text{ m} = 7 \times 10^{-6}\text{ m}0.000007 m=7×10−6 m (7 micrometers7\text{ micrometers}7 micrometers).
  • The diameter of a DNA molecule is about 0.000000002 m=2×10−9 m0.000000002\text{ m} = 2 \times 10^{-9}\text{ m}0.000000002 m=2×10−9 m (2 nanometers2\text{ nanometers}2 nanometers). Without negative exponents, scientific comparison of cell sizes would require counting long chains of zero digits.

2. Astrophysics and Space Exploration

Distances in space are incomprehensibly vast.

  • The distance from Earth to the Sun is roughly 149,600,000,000 m=1.496×1011 m149,600,000,000\text{ m} = 1.496 \times 10^{11}\text{ m}149,600,000,000 m=1.496×1011 m.
  • The mass of the Sun is approximately 1,989,000,000,000,000,000,000,000,000,000 kg=1.989×1030 kg1,989,000,000,000,000,000,000,000,000,000\text{ kg} = 1.989 \times 10^{30}\text{ kg}1,989,000,000,000,000,000,000,000,000,000 kg=1.989×1030 kg.

3. Computing and Digital Storage

Data transfers and computer processor clock speeds rely on sub-second measurements.

  • One nanosecond (the time taken for a high-speed processor to execute a single basic command) is 0.000000001 seconds=1×10−9 s0.000000001\text{ seconds} = 1 \times 10^{-9}\text{ s}0.000000001 seconds=1×10−9 s.

3. Step-by-Step Solved Textbook Examples

Example 1: Simplifying Expressions with Negative Exponents

Problem: Simplify and express the result in exponential form with a positive exponent: (i) (−4)5÷(−4)8(ii) (2−7÷2−10)×2−5\text{(i) } (-4)^5 \div (-4)^8 \qquad \text{(ii) } \left(2^{-7} \div 2^{-10}\right) \times 2^{-5}(i) (−4)5÷(−4)8(ii) (2−7÷2−10)×2−5

Solution:

(i) Using the Quotient Law am÷an=am−na^m \div a^n = a^{m-n}am÷an=am−n: (−4)5÷(−4)8=(−4)5−8(-4)^5 \div (-4)^8 = (-4)^{5 - 8}(−4)5÷(−4)8=(−4)5−8 =(−4)−3= (-4)^{-3}=(−4)−3

Converting to a positive exponent using a−m=1ama^{-m} = \frac{1}{a^m}a−m=am1​: (−4)−3=1(−4)3(-4)^{-3} = \mathbf{\frac{1}{(-4)^3}}(−4)−3=(−4)31​

(ii) First, simplify inside the brackets using am÷an=am−na^m \div a^n = a^{m-n}am÷an=am−n: 2−7÷2−10=2−7−(−10)=2−7+10=232^{-7} \div 2^{-10} = 2^{-7 - (-10)} = 2^{-7 + 10} = 2^32−7÷2−10=2−7−(−10)=2−7+10=23

Now, substitute this result back into the main expression: 23×2−52^3 \times 2^{-5}23×2−5

Apply the Product Law am×an=am+na^m \times a^n = a^{m+n}am×an=am+n: 23+(−5)=2−22^{3 + (-5)} = 2^{-2}23+(−5)=2−2

Convert to a positive exponent: 2−2=1222^{-2} = \mathbf{\frac{1}{2^2}}2−2=221​


Example 2: Finding an Unknown Variable in an Exponential Equation

Problem: Find the value of mmm such that: (−3)m+1×(−3)5=(−3)7(-3)^{m+1} \times (-3)^5 = (-3)^7(−3)m+1×(−3)5=(−3)7

Solution:

Step 1: Apply the Product Law of exponents (am×an=am+na^m \times a^n = a^{m+n}am×an=am+n) on the Left-Hand Side (LHS): LHS=(−3)(m+1)+5=(−3)m+6\text{LHS} = (-3)^{(m+1) + 5} = (-3)^{m+6}LHS=(−3)(m+1)+5=(−3)m+6

Step 2: Equate LHS to Right-Hand Side (RHS): (−3)m+6=(−3)7(-3)^{m+6} = (-3)^7(−3)m+6=(−3)7

Step 3: Since the bases on both sides are equal and non-zero (both are −3-3−3), their exponents must be equal: m+6=7m + 6 = 7m+6=7

Step 4: Solve for mmm: m=7−6m = 7 - 6m=7−6 m=1\mathbf{m = 1}m=1


Example 3: Expressing Numbers in Standard Form

Problem: Express the following numbers in standard form: (i) 0.0000000000085(ii) 6020000000000000\text{(i) } 0.0000000000085 \qquad \text{(ii) } 6020000000000000(i) 0.0000000000085(ii) 6020000000000000

Solution:

(i) 0.00000000000850.00000000000850.0000000000085

  • Move the decimal point to the right to place it after the first non-zero digit (888).
  • The non-zero digit factor m=8.5m = 8.5m=8.5.
  • Count the places moved to the right: 121212 places.
  • Since the movement was to the right, the exponent of 101010 is negative: −12-12−12.

Standard Form=8.5×10−12\text{Standard Form} = \mathbf{8.5 \times 10^{-12}}Standard Form=8.5×10−12

(ii) 602000000000000060200000000000006020000000000000

  • The decimal point is implicitly at the end: 6020000000000000.06020000000000000.06020000000000000.0
  • Move the decimal point to the left to place it after the first digit (666).
  • The factor m=6.02m = 6.02m=6.02.
  • Count the places moved to the left: 151515 places.
  • Since the movement was to the left, the exponent of 101010 is positive: 151515.

Standard Form=6.02×1015\text{Standard Form} = \mathbf{6.02 \times 10^{15}}Standard Form=6.02×1015


Example 4: Real-World Comparison Problem

Problem: The size of a plant cell is 0.00001275 m0.00001275\text{ m}0.00001275 m and that of a red blood cell is 0.000007 m0.000007\text{ m}0.000007 m. Compare their sizes by finding their ratio.

Solution:

Step 1: Convert both measurements into standard form.

  • Plant cell size =0.00001275 m=1.275×10−5 m= 0.00001275\text{ m} = 1.275 \times 10^{-5}\text{ m}=0.00001275 m=1.275×10−5 m
  • Red blood cell size =0.000007 m=7×10−6 m= 0.000007\text{ m} = 7 \times 10^{-6}\text{ m}=0.000007 m=7×10−6 m

Step 2: Express the ratio of plant cell size to red blood cell size: Ratio=Plant cell sizeRed blood cell size=1.275×10−57×10−6\text{Ratio} = \frac{\text{Plant cell size}}{\text{Red blood cell size}} = \frac{1.275 \times 10^{-5}}{7 \times 10^{-6}}Ratio=Red blood cell sizePlant cell size​=7×10−61.275×10−5​

Step 3: Separate the numerical decimal division and the powers of 101010: Ratio=(1.2757)×(10−510−6)\text{Ratio} = \left(\frac{1.275}{7}\right) \times \left(\frac{10^{-5}}{10^{-6}}\right)Ratio=(71.275​)×(10−610−5​)

Apply am÷an=am−na^m \div a^n = a^{m-n}am÷an=am−n for the powers of 101010: 10−5−(−6)=10−5+6=101=1010^{-5 - (-6)} = 10^{-5 + 6} = 10^1 = 1010−5−(−6)=10−5+6=101=10

Now calculate: Ratio=1.2757×10=12.757≈1.82\text{Ratio} = \frac{1.275}{7} \times 10 = \frac{12.75}{7} \approx 1.82Ratio=71.275​×10=712.75​≈1.82

Conclusion: The plant cell is approximately 1.821.821.82 times larger (or roughly twice the size) of a red blood cell.


4. Common Student Mistakes to Avoid

Misconception 1: Treating Negative Exponents as Negative Numbers

  • Wrong: 2−3=−62^{-3} = -62−3=−6 or 2−3=−82^{-3} = -82−3=−8
  • Correct: 2−3=123=182^{-3} = \frac{1}{2^3} = \frac{1}{8}2−3=231​=81​
  • Key Takeaway: A negative exponent indicates a reciprocal, not a negative value.

Misconception 2: Confusing Direction of Decimal Shift in Standard Form

  • Wrong: 0.00053=5.3×1040.00053 = 5.3 \times 10^40.00053=5.3×104
  • Correct: 0.00053=5.3×10−40.00053 = 5.3 \times 10^{-4}0.00053=5.3×10−4
  • Key Takeaway: Moving the decimal point to the right creates a negative exponent; moving it to the left creates a positive exponent.

Misconception 3: Incorrectly Applying Laws to Addition or Subtraction

  • Wrong: 23+24=23+4=272^3 + 2^4 = 2^{3+4} = 2^723+24=23+4=27
  • Correct: 23+24=8+16=242^3 + 2^4 = 8 + 16 = 2423+24=8+16=24
  • Key Takeaway: The laws of exponents apply only to multiplication and division. You cannot add exponents when adding or subtracting terms!

Misconception 4: Ignoring Bracket Placement with Negative Bases

  • Wrong: −32=(−3)×(−3)=9-3^2 = (-3) \times (-3) = 9−32=(−3)×(−3)=9
  • Correct: −32=−(3×3)=−9-3^2 = -(3 \times 3) = -9−32=−(3×3)=−9, whereas (−3)2=(−3)×(−3)=9(-3)^2 = (-3) \times (-3) = 9(−3)2=(−3)×(−3)=9.
  • Key Takeaway: Pay close attention to parentheses. The exponent applies only to what is directly inside the parentheses or immediately attached to it.

5. Practice Questions for Self-Assessment

Question 1

Evaluate the expression: [(13)−2−(12)−3]÷(14)−2\left[\left(\frac{1}{3}\right)^{-2} - \left(\frac{1}{2}\right)^{-3}\right] \div \left(\frac{1}{4}\right)^{-2}[(31​)−2−(21​)−3]÷(41​)−2

Solution:

Using the rule (ab)−m=(ba)m\left(\frac{a}{b}\right)^{-m} = \left(\frac{b}{a}\right)^m(ba​)−m=(ab​)m:

  • (13)−2=32=9\left(\frac{1}{3}\right)^{-2} = 3^2 = 9(31​)−2=32=9
  • (12)−3=23=8\left(\frac{1}{2}\right)^{-3} = 2^3 = 8(21​)−3=23=8
  • (14)−2=42=16\left(\frac{1}{4}\right)^{-2} = 4^2 = 16(41​)−2=42=16

Substitute these values back into the expression: [9−8]÷16[9 - 8] \div 16[9−8]÷16 =1÷16=116= 1 \div 16 = \mathbf{\frac{1}{16}}=1÷16=161​


Question 2

Simplify the expression and write the answer with positive exponents (t≠0t \neq 0t=0): 25×t−45−3×10×t−8\frac{25 \times t^{-4}}{5^{-3} \times 10 \times t^{-8}}5−3×10×t−825×t−4​

Solution:

First, express numbers in terms of prime factors (25=5225 = 5^225=52 and 10=2×510 = 2 \times 510=2×5): =52×t−45−3×(2×51)×t−8= \frac{5^2 \times t^{-4}}{5^{-3} \times (2 \times 5^1) \times t^{-8}}=5−3×(2×51)×t−852×t−4​

Group like bases together in the denominator: Denominator=21×5−3+1×t−8=21×5−2×t−8\text{Denominator} = 2^1 \times 5^{-3 + 1} \times t^{-8} = 2^1 \times 5^{-2} \times t^{-8}Denominator=21×5−3+1×t−8=21×5−2×t−8

The expression becomes: =52×t−421×5−2×t−8= \frac{5^2 \times t^{-4}}{2^1 \times 5^{-2} \times t^{-8}}=21×5−2×t−852×t−4​

Apply the Quotient Law (am÷an=am−na^m \div a^n = a^{m-n}am÷an=am−n): =52−(−2)×t−4−(−8)21= \frac{5^{2 - (-2)} \times t^{-4 - (-8)}}{2^1}=2152−(−2)×t−4−(−8)​ =52+2×t−4+82= \frac{5^{2 + 2} \times t^{-4 + 8}}{2}=252+2×t−4+8​ =54×t42= \frac{5^4 \times t^4}{2}=254×t4​

Since 54=6255^4 = 62554=625: =625t42= \mathbf{\frac{625 t^4}{2}}=2625t4​


Question 3

Write 3.61492×10−63.61492 \times 10^{-6}3.61492×10−6 in usual form.

Solution:

Since the exponent of 101010 is negative (−6-6−6), move the decimal point 666 places to the left.

  • Original: 3.614923.614923.61492
  • Moving 111 place left gives: .361492.361492.361492
  • Moving 555 more places left requires inserting 555 zeros before 333:

Usual Form=0.00000361492\text{Usual Form} = \mathbf{0.00000361492}Usual Form=0.00000361492


Question 4

Compare 2.7×10122.7 \times 10^{12}2.7×1012 and 1.5×1081.5 \times 10^{8}1.5×108.

Solution:

Look at the powers of 101010:

  • The exponent in 2.7×10122.7 \times 10^{12}2.7×1012 is 121212.
  • The exponent in 1.5×1081.5 \times 10^{8}1.5×108 is 888.

Since 12>812 > 812>8, 101210^{12}1012 is much larger than 10810^8108.

Therefore: 2.7×1012>1.5×108\mathbf{2.7 \times 10^{12} > 1.5 \times 10^{8}}2.7×1012>1.5×108

To see how many times larger, divide them: 2.7×10121.5×108=(2.71.5)×1012−8=1.8×104=18,000\frac{2.7 \times 10^{12}}{1.5 \times 10^8} = \left(\frac{2.7}{1.5}\right) \times 10^{12-8} = 1.8 \times 10^4 = 18,0001.5×1082.7×1012​=(1.52.7​)×1012−8=1.8×104=18,000

The first number is 18,00018,00018,000 times larger than the second number.


6. Exam Revision & FAQs

FAQ 1: What is the multiplicative inverse of a−ma^{-m}a−m?

Answer: The multiplicative inverse of a−ma^{-m}a−m is ama^mam (or 1a−m\frac{1}{a^{-m}}a−m1​). By definition, two numbers are multiplicative inverses if their product equals 111:

a−m×am=a−m+m=a0=1a^{-m} \times a^m = a^{-m + m} = a^0 = 1a−m×am=a−m+m=a0=1


FAQ 2: Why is a0=1a^0 = 1a0=1 for any non-zero number aaa?

Answer: Using the Quotient Law of exponents am÷an=am−na^m \div a^n = a^{m-n}am÷an=am−n, let m=nm = nm=n:

amam=am−m=a0\frac{a^m}{a^m} = a^{m-m} = a^0amam​=am−m=a0

Since any non-zero number divided by itself equals 111, we have:

amam=1  ⟹  a0=1\frac{a^m}{a^m} = 1 \implies a^0 = 1amam​=1⟹a0=1

(Note: 000^000 is undefined in elementary mathematics).


FAQ 3: What is the difference between standard form and expanded form?

Answer:

  • Standard Form expresses a number as m×10nm \times 10^nm×10n where 1≤m<101 \le m < 101≤m<10. Example: 4.5×1034.5 \times 10^34.5×103.
  • Expanded Form breaks down a number into the sum of the values of its individual digits. Example: 4500=(4×103)+(5×102)+(0×101)+(0×100)4500 = (4 \times 10^3) + (5 \times 10^2) + (0 \times 10^1) + (0 \times 10^0)4500=(4×103)+(5×102)+(0×101)+(0×100).

FAQ 4: How do you add or subtract two numbers written in standard form with different exponents?

Answer: You cannot add or subtract the factors directly if the powers of 101010 are different. You must first convert them so that they have the same power of 101010:

Example: Calculate (3×104)+(5×103)(3 \times 10^4) + (5 \times 10^3)(3×104)+(5×103)

  1. Rewrite 5×1035 \times 10^35×103 as 0.5×1040.5 \times 10^40.5×104.
  2. Factor out 10410^4104: (3+0.5)×104=3.5×104(3 + 0.5) \times 10^4 = \mathbf{3.5 \times 10^4}(3+0.5)×104=3.5×104
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  • Chemical Reactions and Equations
  • Introduction to Trigonometry
  • Quadratic Equations
  • Real Numbers
Class 9 NCERT Guides20 chapters
  • Quadrilaterals
  • Linear Equations in Two Variables
  • Circles
  • Improvement in Food Resources
  • Triangles
  • Lines and Angles
  • Is Matter Around Us Pure?
  • Heron's Formula
  • Tissues
  • Structure of the Atom
  • Atoms and Molecules
  • Work and Energy
  • Gravitation
  • Force and Laws of Motion
  • Motion
  • The Fundamental Unit of Life
  • Matter in Our Surroundings
  • Coordinate Geometry
  • Number Systems
  • Polynomials
Class 8 NCERT Guides23 chapters
  • Mensuration
  • Direct and Inverse Proportions
  • Cubes and Cube Roots
  • Light
  • Reaching the Age of Adolescence
  • Factorisation
  • → Exponents and Powers (Mathematics)
  • Chemical Effects of Electric Current
  • Force and Pressure
  • Practical Geometry
  • Practical Geometry
  • Crop Production and Management
  • Comparing Quantities
  • Algebraic Expressions and Identities
  • Friction
  • Squares and Square Roots
  • Sound
  • Combustion and Flame
  • Coal and Petroleum
  • Microorganisms: Friend and Foe
  • Linear Equations in One Variable
  • Understanding Quadrilaterals
  • Rational Numbers
Class 7 NCERT Guides16 chapters
  • Data Handling
  • Reproduction in Plants
  • Transportation in Animals and Plants
  • The Triangle and Its Properties
  • Electric Current and Its Effects
  • Respiration in Organisms
  • Physical and Chemical Changes
  • Motion and Time
  • Acids, Bases and Salts
  • Heat
  • Nutrition in Animals
  • Nutrition in Plants
  • Perimeter and Area
  • Integers
  • Rational Numbers
  • Simple Equations
Class 6 NCERT Guides14 chapters
  • Basic Geometrical Ideas
  • Body Movements
  • Ratio and Proportion
  • Fun with Magnets
  • Light, Shadows and Reflections
  • Playing With Numbers
  • Algebra
  • Decimals
  • Fractions
  • Knowing Our Numbers
  • Electricity and Circuits
  • Components of Food
  • Getting to Know Plants
  • Separation of Substances
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Ravindra Higher Secondary School

Waidhan, Singrauli (M.P.)

We Serve Society By Serving People

Established in 1988, Ravindra Higher Secondary School (RHS Waidhan) is dedicated to delivering excellence in education, character building, and holistic growth for students in Waidhan, Singrauli (MP).

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