# Class 8 NCERT Solutions – Chapter 12 Exponents and Powers – Exercise 12.2

### 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 = 85/1013 = 85 Ã— 10-13

= 8.5 Ã— 10 Ã— 10-13 = 8.5 Ã— 101-13 = 8.5 Ã— 10-12

So, the required standard form is 8.5 Ã— 10-12

(ii) 0.00000000000942

= 942/100000000000000 = 942/1014 = 942 Ã— 10-14

= 9.42 Ã— 102 Ã— 10-14 = 9.42 Ã— 102-14 = 9.42 Ã— 10-12

So, the required standard form is 9.42 Ã— 10-12

(iii) 6020000000000000

= 602 Ã— 10000000000000

= 6.02 Ã— 102 Ã— 1013 = 6.02 Ã— 1015

So, the required standard form is 6.02 Ã— 1015

(iv) 0.00000000837

= 837/100000000000 = 837/1011 = 837 Ã— 10-11

= 8.37 Ã— 102 Ã— 10-11 = 8.37 Ã—102-11 = 8.37 Ã— 10-9

So, the required standard form is 8.37 Ã— 10-9

(v) 31860000000

= 3186 Ã— 10000000

= 3.186 Ã— 103 Ã— 107 = 3.186 Ã— 1010

So, the required standard form is 3.186 Ã— 1010

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

(i) 3.02 Ã— 10-6

(ii) 4.5 Ã— 104

(iii) 3 Ã— 10-8

(iv) 1.0001 Ã— 109

(v) 5.8 Ã— 1012

(vi) 3.61492 Ã— 106

Solution:

(i) 3.02 Ã— 10-6

= 3.02/106

= 302 Ã—  1/102 Ã— 1/106 = 302/102+6

= 302/108 = 0.00000302

So, the usual form is 0.00000302

(ii) 4.5 Ã— 104

= 45/101 Ã— 104

= 45 Ã— 10-1 Ã— 104 = 45 Ã— 10-1+4

= 45 Ã— 103 = 45000

So, the usual form is 45000

(iii) 3 Ã— 10-8

= 3/108

= 3/100000000 = 0.00000003

So, the usual form is 0.00000003

(iv) 1.0001 Ã— 109

= 10001/10000 Ã— 109

= 10001 Ã— 10-4 Ã— 109 = 10001 Ã— 10-4+9

= 10001 Ã— 105 = 1000100000

So, the usual form is 1000100000

(v) 5.8 Ã— 1012

= 58 X 10-1 Ã— 1012

= 58 Ã— 10-1+12 = 58 Ã— 1011

= 5800000000000

So, the usual form is 5800000000000

(vi) 3.61492 Ã— 106

= 361492/100000 Ã— 106

= 361492 Ã— 10-5 Ã—106 = 361492 Ã— 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

= 1/106 = 10-6 m

So, the standard from is 10-6 m

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

= 16/100000000000000000000 = 16/1020 = 1.6 Ã— 10 X10-20

= 1.6 Ã— 101-20 = 1.6 Ã— 10-19

So, the standard from is 1.6 Ã— 10-19 C

(iii) Size of Bacteria = 0.0000005 m

= 5/10000000 = 5/107 = 0.5 Ã— 10 Ã— 10-7

= 0.5 Ã— 10-6

So, the standard from is 0.5 Ã— 10-6 m

(iv) Size of plant cell = 0.00001275 m

= 1275/100000000 = 1275/108 = 1.275 Ã— 103 Ã— 10-8

= 1.275 Ã— 103-8 = 1.275 Ã— 10-5

So, the standard from is 1.275 Ã— 10-5 m

(v) Thickness of a Paper = 0.07mm

= 7/100 = 7/102 = 0.7 Ã— 10 Ã— 10-2

= 0.7 Ã— 10-1

So, the standard from is 0.7 Ã— 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 Ã— 20mm = 100mm or 10cm

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

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

= 100mm + 0.080mm = 100.080mm

or

1.0008 Ã— 10-2 mm

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