**Question 1. Express the following numbers in standard form:**

**(i) 0.0000000000085**

**(ii) 0.00000000000942**

**(iii) 6020000000000000**

**(iv) 0.00000000837**

**(v) 31860000000**

**Solution:**

(i) 0.0000000000085= 85/10000000000000 = 85/10

^{13}= 85 × 10^{-13}= 8.5 × 10 × 10

^{-13}= 8.5 × 10^{1-13}= 8.5 × 10^{-12}So, the required standard form is

8.5 × 10^{-12}

(ii) 0.00000000000942= 942/100000000000000 = 942/10

^{14}= 942 × 10^{-14}= 9.42 × 10

^{2}× 10^{-14}= 9.42 × 10^{2-14}= 9.42 × 10^{-12}So, the required standard form is

9.42 × 10^{-12}

(iii) 6020000000000000= 602 × 10000000000000

= 6.02 × 10

^{2}× 10^{13}= 6.02 × 10^{15}So, the required standard form is

6.02 × 10^{15}

(iv) 0.00000000837= 837/100000000000 = 837/10

^{11}= 837 × 10^{-11}= 8.37 × 10

^{2}× 10^{-11}= 8.37 ×10^{2-11}= 8.37 × 10^{-9}So, the required standard form is

8.37 × 10^{-9}

(v) 31860000000= 3186 × 10000000

= 3.186 × 10

^{3}× 10^{7}= 3.186 × 10^{10}So, the required standard form is

3.186 × 10^{10 }

**Question 2. ****Express the following numbers in ****the ****usual form**

**(i) 3.02 × 10 ^{-6}**

**(ii) 4.5 × 10 ^{4}**

**(iii) 3 × 10 ^{-8} **

**(iv) 1.0001 × 10 ^{9}**

**(v) 5.8 × 10 ^{12} **

**(vi) 3.61492 × 10 ^{6} **

**Solution:**

(i)3.02 × 10^{-6}= 3.02/10

^{6}= 302 ×

^{ }1/10^{2}× 1/10^{6}= 302/10^{2+6}= 302/10

^{8}= 0.00000302So, the usual form is

0.00000302

(ii)4.5 × 10^{4}= 45/10

^{1}× 10^{4}= 45 × 10

^{-1}× 10^{4}= 45 × 10^{-1+4}= 45 × 10

^{3}= 45000So, the usual form is

45000

(iii)3 × 10^{-8}= 3/10

^{8}= 3/100000000 = 0.00000003

So, the usual form is

0.00000003

(iv)1.0001 × 10^{9}^{ }= 10001/10000 × 10

^{9}= 10001 × 10

^{-4}× 10^{9}= 10001 × 10^{-4+9}= 10001 × 10

^{5}= 1000100000So, the usual form is

1000100000

(v)5.8 × 10^{12}= 58 X 10

^{-1}× 10^{12}= 58 × 10

^{-1+12}= 58 × 10^{11}= 5800000000000

So, the usual form is

5800000000000

(vi)3.61492 × 10^{6}= 361492/100000 × 10

^{6}= 361492 × 10

^{-5}×10^{6}= 361492 × 10^{-5+6}= 3614920

So, the usual form is

3614920

**Question 3. ****Express the number appearing in the following statements in standard form.**

**(i) 1 micron is equal to 1/1000000 m.****(ii) Charge of an electron is 0.000,000,000,000,000,000,16 coulombs****(iii) Size of bacteria is 0.0000005 m****(iv) Size of a plant cell is 0.00001275 m****(v) Thickness of a thick paper is 0.07 mm.**

**Solution:**

(i) 1 micron = 1/1000000 m= 1/10

^{6}= 10^{-6}mSo, the standard from is

10^{-6}m

(ii)Charge of an electron = 0.000,000,000,000,000,000,16 C= 16/100000000000000000000 = 16/10

^{20}= 1.6 × 10 X10^{-20}= 1.6 × 10

^{1-20}= 1.6 × 10^{-19}So, the standard from is

1.6 × 10^{-19}C

(iii) Size of Bacteria =0.0000005 m= 5/10000000 = 5/10

^{7}= 0.5 × 10 × 10^{-7}= 0.5 × 10

^{-6}So, the standard from is

0.5 × 10^{-6}m

(iv) Size of plant cell =0.00001275 m= 1275/100000000 = 1275/10

^{8}= 1.275 × 10^{3 }× 10^{-8}= 1.275 × 10

^{3-8}= 1.275 × 10^{-5}So, the standard from is

1.275 × 10^{-5 }m

(v) Thickness of a Paper = 0.07mm= 7/100 = 7/10

^{2}= 0.7 × 10 × 10^{-2}= 0.7 × 10

^{-1}So, the standard from is

0.7 × 10^{-1 }mm

**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?**** **

**Solution:**

The thickness of Books will be 5 × 20mm = 100mm or 10cm

The thickness of Paper sheets will be 5 × 0.016mm = 0.080mm

Hence, the total thickness is = thickness of books + thickness of paper sheets

= 100mm + 0.080mm = 100.080mm

or

1.0008 × 10

^{-2 }mm

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the **DSA Self Paced Course** at a student-friendly price and become industry ready.