GSEB Class 8 Maths Solutions Chapter 12 Exponents and Powers Exercise 12.2

Question 1. Express the following numbers in standard form?
1. \( 0.0000000000085 \)
2. \( 0.00000000000942 \)
3. \( 6020000000000000 \)
4. \( 0.00000000837 \)
5. \( 31860000000 \)
Answer:
1. \( 0.0000000000085 = \frac{85}{10000000000000} \)
\( = \frac{85}{10^{13}} \)
\( = \frac{8.5 \times 10}{10^{13}} \)
\( = 8.5 \times 10 \times 10^{-13} \)
\( = 8.5 \times 10^{-12} \)
2. \( 0.00000000000942 = \frac{942}{100000000000000} \)
\( = \frac{942}{10^{14}} \)
\( = \frac{9.42 \times 10^{2}}{10^{14}} \)
\( = 9.42 \times 10^{2} \times 10^{-14} \)
\( = 9.42 \times 10^{-12} \)
3. \( 6020000000000000 \)
\( = 602 \times 10000000000000 \)
\( = 602 \times 10^{13} \)
\( = 6.02 \times 10^{2} \times 10^{13} \)
\( = 6.02 \times 10^{15} \)
4. \( 0.00000000837 = \frac{837}{100000000000} \)
\( = \frac{837}{10^{11}} \)
\( = \frac{8.37 \times 10^{2}}{10^{11}} \)
\( = 8.37 \times 10^{2} \times 10^{-11} \)
\( = 8.37 \times 10^{2+(-11)} \)
\( = 8.37 \times 10^{-9} \)
5. \( 31860000000 \)
\( = 3186 \times 1000000 \)
\( = 3186 \times 10^{7} \)
\( = 3.186 \times 10^{3} \times 10^{7} \)
\( = 3.186 \times 10^{3+7} \)
\( = 3.186 \times 10^{10} \)
In simple words: To write a number in standard form, move the decimal point so there's only one non-zero digit before it. Then, multiply by 10 raised to the power of how many places you moved the decimal. If you move it right, the power is negative; if left, it's positive.

Exam Tip: Remember that standard form always has one non-zero digit before the decimal point. Pay close attention to the sign of the exponent, which depends on the direction the decimal point was moved.

Question 2. Express the following numbers in usual form?
1. \( 3.02 \times 10^{-6} \)
2. \( 4.5 \times 10^{4} \)
3. \( 3 \times 10^{-8} \)
4. \( 1.0001 \times 10^{9} \)
5. \( 5.8 \times 10^{12} \)
6. \( 3.61492 \times 10^{6} \)
Answer:
1. \( 3.02 \times 10^{-6} = 0.00000302 \)
2. \( 4.5 \times 10^{4} = 45000 \)
3. \( 3 \times 10^{-8} = 0.00000003 \)
4. \( 1.0001 \times 10^{9} = 1000100000 \)
5. \( 5.8 \times 10^{12} = 5800000000000 \)
6. \( 3.61492 \times 10^{6} = 3614920 \)
In simple words: To write a number in its regular or usual form, you move the decimal point based on the power of 10. If the power is positive, move it to the right; if it's negative, move it to the left. Fill any empty spaces with zeros.

Exam Tip: Always count the number of places to move the decimal point carefully, and remember that positive exponents mean larger numbers (move right) while negative exponents mean smaller numbers (move left).

Question 3. Express the number appearing in the following statements in standard form?
1. 1 micron is equal to \( \frac{1}{1000000} \) m.
2. Charge of an electron is 0.000,000,000,000,000,000,16 coulomb.
3. Size of a bacteria is 0.0000005 m.
4. Size of a plant cell is 0.00001275 m.
5. Thickness of a thick paper is 0.07 mm.
Answer:
1. \( 1 \text{ micron} = \frac{1}{1000000} \text{ m} \)
\( = \frac{1}{10^{6}} \text{ m} \)
\( = 1 \times 10^{-6} \text{ m} \)
2. \( 0.00000000000000000016 \text{ coulomb} \)
\( = \frac{16}{100000000000000000000} \)
\( = \frac{1.6 \times 10}{10^{20}} \)
\( = \frac{1.6}{10^{19}} \)
\( = 1.6 \times 10^{-19} \text{ coulomb} \)
3. \( 0.0000005 \text{ m} \)
\( = \frac{5}{10000000} \text{ m} \)
\( = \frac{5}{10^{7}} \text{ m} \)
\( = 5 \times 10^{-7} \text{ m} \)
4. \( 0.00001275 \text{ m} \)
\( = \frac{1275}{100000000} \text{ m} \)
\( = \frac{1275}{10^{8}} \text{ m} \)
\( = \frac{1.275 \times 10^{3}}{10^{8}} \text{ m} \)
\( = 1.275 \times 10^{3-8} \text{ m} \)
\( = 1.275 \times 10^{-5} \text{ m} \)
5. \( 0.07 \text{ mm} \)
\( = \frac{7}{100} \text{ mm} \)
\( = \frac{7}{10^{2}} \text{ mm} \)
\( = 7 \times 10^{-2} \text{ mm} \)
In simple words: To write these measurements in standard form, you adjust the decimal point so that there is only one non-zero digit before it. Then, you show how many places the decimal was moved using a power of 10.

Exam Tip: When converting values from real-world contexts to standard form, clearly identify the numeric value and then apply the rules for powers of 10. Always include the correct units in your final answer.

Question 4. In a stack there are 5 books each of thickness 20 mm and 5 paper sheets each of thickness 0.016 mm. What is the total thickness of the stack?
Answer: The thickness of one book is 20 mm.
The total thickness of 5 books is \( 5 \times 20 \text{ mm} = 100 \text{ mm} \).
Each paper sheet has a thickness of 0.016 mm.
The total thickness of 5 paper sheets is \( 5 \times 0.016 \text{ mm} = 0.080 \text{ mm} \).
Therefore, the combined thickness of the stack is \( 100 \text{ mm} + 0.080 \text{ mm} = 100.08 \text{ mm} \).
In standard form, this is \( 1.0008 \times 10^{2} \text{ mm} \).
In simple words: First, find the total thickness of all the books. Then, find the total thickness of all the paper sheets. Finally, add these two amounts together to get the complete thickness of the entire stack.

Exam Tip: When solving problems involving multiple items, always calculate the total for each type of item separately before adding them together to avoid errors. Double-check your units throughout the calculation.