Chapter 12 – Exponents and Powers



Page No 197:

Question 1:

Evaluate
(i) 3−2 (ii) (−4)−2 (iii) 

Answer:

(i) 
(ii) 
(iii) 

Question 2:

Simplify and express the result in power notation with positive exponent.
(i)  (ii) 
(iii)  (iv) 
(v) 

Answer:

(i) (−4)5 ÷ (−4)8 = (−4)5 − 8 (am ÷ an = am − n)
= (− 4)−3
(ii) 
(iii) 
(iv) (3− 7 ÷ 3−10) × 3−5 = (3−7 − (−10)) × 3−5 (am ÷ an = a− n)
= 33 × 3−5
= 33 + (− 5) (am × an = am + n)
= 3−2
(v) 2−3 × (−7)−3 = 

Question 3:

Find the value of.
(i) (30 + 4−1) × 22 (ii) (2−1 × 4−1) ÷2−2
(iii)  (iv) (3−1 + 4−1 + 5−1)0
(v) 

Answer:

(i) 
(ii) (2−1 × 4−1) ÷ 2− 2 = [2−1 × {(2)2}− 1] ÷ 2− 2
= (2− 1 × 2− 2) ÷ 2− 2 
= 2−1+ (−2) ÷ 2−2 (am × an = am + n)
= 2−3 ÷ 2−2
= 2−3 − (−2) (am ÷ an = am − n)
= 2−3 + 2 = 2 −1
(iii) 
(iv) (3−1 + 4−1 + 5−1)0
= 1 (a0 = 1)
(v) 

Page No 198:

Question 4:

Evaluate (i)  (ii) 

Answer:

(i) 
(ii) 

Question 5:

Find the value of m for which 5m ÷5−3 = 55.

Answer:

5m ÷ 5−3 = 55
5m − (− 3) = 55 (am ÷ an = am − n)
5m + 3 = 55
Since the powers have same bases on both sides, their respective exponents must be equal.
m + 3 = 5
m = 5 − 3
m = 2

Question 6:

Evaluate (i)  (ii) 

Answer:

(i) 
(ii) 

Question 7:

Simplify. (i)  (ii) 

Answer:

(i) 
(ii) 

Page No 200:

Question 1:

Express the following numbers in standard form.
(i) 0.0000000000085 (ii) 0.00000000000942
(iii) 6020000000000000 (iv) 0.00000000837
(v) 31860000000

Answer:

(i) 0.0000000000085 = 8.5 × 10−12
(ii) 0.00000000000942 = 9.42 × 10−12
(iii) 6020000000000000 = 6.02 × 1015
(iv) 0.00000000837 = 8.37 × 10−9
(v) 31860000000 = 3.186 × 1010

Question 2:

Express the following numbers in 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

Answer:

(i) 3.02 × 10−6 = 0.00000302
(ii) 4.5 × 104 = 45000
(iii) 3 × 10−8 = 0.00000003
(iv) 1.0001 × 109 = 1000100000
(v) 5.8 × 1012 = 5800000000000
(vi) 3.61492 × 106 = 3614920

Question 3:

Express the number appearing in the following statements in standard form.
(i) 1 micron is equal to m.
(ii) Charge of an electron is 0.000, 000, 000, 000, 000, 000, 16 coulomb.
(iii) Size of a 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

Answer:

(i)  = 1 × 10−6
(ii) 0.000, 000, 000, 000, 000, 000, 16 = 1.6 × 10−19
(iii) 0.0000005 = 5 × 10−7
(iv) 0.00001275 = 1.275 × 10−5
(v) 0.07 = 7 × 10−2

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?

Answer:

Thickness of each book = 20 mm
Hence, thickness of 5 books = (5 × 20) mm = 100 mm
Thickness of each paper sheet = 0.016 mm
Hence, thickness of 5 paper sheets = (5 × 0.016) mm = 0.080 mm
Total thickness of the stack = Thickness of 5 books + Thickness of 5 paper sheets
= (100 + 0.080) mm
= 100.08 mm
= 1.0008 × 102 mm

Read this too

Post a Comment