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Class 8 NCERT Solutions – Chapter 12 Exponents and Powers – Exercise 12.2
  • Last Updated : 23 Feb, 2021

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 × 10-13

= 8.5 × 10 × 10-13 = 8.5 × 101-13 = 8.5 × 10-12 

So, the required standard form is 8.5 × 10-12

(ii) 0.00000000000942

= 942/100000000000000 = 942/1014 = 942 × 10-14

= 9.42 × 102 × 10-14 = 9.42 × 102-14 = 9.42 × 10-12



So, the required standard form is 9.42 × 10-12

(iii) 6020000000000000

= 602 × 10000000000000 

= 6.02 × 102 × 1013 = 6.02 × 1015

So, the required standard form is 6.02 × 1015

(iv) 0.00000000837

= 837/100000000000 = 837/1011 = 837 × 10-11

= 8.37 × 102 × 10-11 = 8.37 ×102-11 = 8.37 × 10-9 

 So, the required standard form is 8.37 × 10-9

(v) 31860000000

= 3186 × 10000000

= 3.186 × 103 × 107 = 3.186 × 1010

So, the required standard form is 3.186 × 1010 

Question 2. Express the following numbers in the usual form

(i) 3.02 × 10-6

(ii) 4.5 × 104

(iii) 3 × 10-8 

(iv) 1.0001 × 109

(v) 5.8 × 1012 

(vi) 3.61492 × 106 

Solution:

(i) 3.02 × 10-6 

= 3.02/106

= 302 ×  1/102 × 1/106 = 302/102+6

= 302/108 = 0.00000302

So, the usual form is 0.00000302

(ii) 4.5 × 104 

= 45/101 × 104

= 45 × 10-1 × 104 = 45 × 10-1+4

= 45 × 103 = 45000

So, the usual form is 45000

(iii) 3 × 10-8 

= 3/108

= 3/100000000 = 0.00000003

So, the usual form is 0.00000003

(iv) 1.0001 × 109 

= 10001/10000 × 109

= 10001 × 10-4 × 109 = 10001 × 10-4+9

= 10001 × 105 = 1000100000

So, the usual form is 1000100000

(v) 5.8 × 1012 

= 58 X 10-1 × 1012

= 58 × 10-1+12 = 58 × 1011

= 5800000000000

So, the usual form is 5800000000000

(vi) 3.61492 × 106 

= 361492/100000 × 106

= 361492 × 10-5 ×106 = 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

= 1/106 = 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 × 10 X10-20

= 1.6 × 101-20 = 1.6 × 10-19

So, the standard from is 1.6 × 10-19 C

(iii) Size of Bacteria = 0.0000005 m

= 5/10000000 = 5/107 = 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 = 1275/108 = 1.275 × 103 × 10-8

= 1.275 × 103-8 = 1.275 × 10-5 

So, the standard from is 1.275 × 10-5 m

(v) Thickness of a Paper = 0.07mm

= 7/100 = 7/102 = 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.

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