Chapter 12 – Exponents and Powers
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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 = am − 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
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