Ncert Solutions For Class 8 Maths Ex 12.2

Ncert Solutions For Class 8 Maths Chapter 12 Ex 12.2

Q.1. Converts the numbers into standard form:

(i)0.0000000000085

(ii)0.00000000000942

(iii)6020000000000000

(iv)0.00000000837

(v)31860000000

Solution:

(i)0.0000000000085

0.0000000000085=0.0000000000085×10121012=8.5×1012$0.0000000000085=0.0000000000085\times \frac{10^{12}}{10^{12}}=8.5\times 10^{-12}$

(ii)0.00000000000942

0.00000000000942=0.00000000000942×10121012=9.42×1012$0.00000000000942 =0.00000000000942 \times \frac{10^{12}}{10^{12}}=9.42\times 10^{-12}$

(iii)6020000000000000

6020000000000000=6020000000000000×10151015=6.02×1015$6020000000000000 =6020000000000000 \times \frac{10^{15}}{10^{15}}=6.02\times 10^{15}$

(iv)0.00000000837

0.00000000837=0.00000000837×109109=8.37×109$0.00000000837 =0.00000000837 \times \frac{10^{9}}{10^{9}}=8.37\times 10^{-9}$

(v)31860000000

31860000000=31860000000×10101010=3.186×1010$31860000000 =31860000000 \times \frac{10^{10}}{10^{10}}=3.186\times 10^{10}$

Q.2. Convert the following numbers in usual form:

(i)3.02×106$3.02\times 10^{-6}$

(ii)4.5×104$4.5\times 10^{4}$

(iii)3×108$3\times 10^{-8}$

(iv)1.0001×109$1.0001\times 10^{9}$

(v) 5.8×1012$5.8\times 10^{12}$

(vi)3.61492×106$3.61492\times 10^{6}$

Solution:

(i)3.02×106$3.02\times 10^{-6}$

3.02×106=3.02106=0.00000302$3.02\times 10^{-6}=\frac{3.02}{10^{6}}=0.00000302$

(ii)4.5×104$4.5\times 10^{4}$

4.5×104=4.5×10000=45000$4.5\times 10^{4}=4.5\times 10000=45000$

(iii) 3×108$3\times 10^{-8}$

3×108=3108=0.00000003$3\times 10^{-8}=\frac{3}{10^{8}}=0.00000003$

(iv) 1.0001×109$1.0001 \times 10^{9}$

1.0001×109=1000100000$1.0001 \times 10^{9}=1000100000$

(v) 5.8×1012$5.8\times 10^{12}$

5.8×1012=5.8×1000000000000=5800000000000$5.8\times 10^{12}=5.8\times 1000000000000=5800000000000$

(vi) 3.61492×106$3.61492\times 10^{6}$

3.61492×106=3.61492×1000000=3614920$3.61492\times 10^{6}=3.61492 \times 1000000 =3614920$

Q.3. Conert the given numbers in the following statements in standard form:

(i) 1 micron is equal to 11000000$\frac{1}{1000000}$ m.

(ii)Charge of an electron is 0.000,000,000,000,000,000,28 coulomb.

(iii)Size of bacteria is 0.0000005 m.

(iv)Size of a plant cell is 0.00001275 m.

(v)Thickness if a thick paper is 0.07 mm.

Solution:

(i)1 micron =11000000=1106=1×106m$\frac{1}{1000000}=\frac{1}{10^{6}}=1\times 10^{-6}\: m$

(ii) Charge of an electron is 0.00000000000000000028 coulombs.

=0.00000000000000000028×10191019=2.8×1019coulomb$= 0.00000000000000000028\times \frac{10^{19}}{10^{19}}=2.8\times 10^{-19}\: coulomb$

(iii)Size of bacteria = 0.0000005 =510000000=5107=5×107m$\frac{5}{10000000}=\frac{5}{10^{7}}=5\times 10^{-7\: m}$

(iv)Size of a plant cell is 0.00001275 m =0.00001275×105105=1.275×105m$0.00001275\times \frac{10^{5}}{10^{5}}=1.275\times 10^{-5}\: m$

(v) Thickness of a thick paper = 0.07 mm =7100mm=7102=7×102mm$\frac{7}{100}\: mm=\frac{7}{10^{2}}=7\times 10^{-2}\: mm$

Q.4. In a stack there are 6 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:

Thickness of one book = 20 mm

Thickness of 6 books =20×6=120mm$20\times 6=120\: mm$

Thickness of one paper = 0.016 mm

Thickness of 5 papers =0.016×5=0.08mm$0.016\times 5=0.08\: mm$

Total thickness of a stack =120+0.08

=120.08 mm=120.08×102102=1.2008×102mm$120.08\times \frac{10^{2}}{10^{2}}=1.2008\times 10^{2}\: mm$