8类NCERT解决方案–第12章指数和幂–练习12.2

问题1.以标准形式表示以下数字：

(i)0.0000000000085

(ii)0.00000000000942

(iii)6020000000000000

(iv)0.00000000837

(v)31860000000

解决方案：

(i) 0.0000000000085

= 85/10000000000000 = 85/10¹³ = 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 = 942/10¹⁴ = 942 × 10^-14

= 9.42 × 10² × 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 × 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 ×10^2-11 = 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 

问题2.以通常的形式表示以下数字

(i)3.02×10 ^-6

(ii)4.5×10 ⁴

(iii)3×10 ^-8

(iv)1.0001×10 ⁹

(v)5.8×10 ¹²

(六)3.61492×10 ⁶

解决方案：

(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

问题3.以标准形式表示以下语句中出现的数字。

(i)1微米等于1/1000000 m。
(ii)电子的电荷为0.000,000,000,000,000,000,16库仑
(iii)细菌大小为0.0000005 m
(iv)植物室的大小为0.00001275 m
(v)厚纸的厚度为0.07毫米。

解决方案：

(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 × 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 = 5/10⁷ = 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 × 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 = 7/10² = 0.7 × 10 × 10^-2 

= 0.7 × 10^-1

So, the standard from is 0.7 × 10^-1 mm

问题4.在一堆中，有5本书，每本的厚度为20毫米，每本有5张纸，每本的厚度为0.016毫米。堆叠的总厚度是多少?  

解决方案：

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