Power Play Class 8 Extra Questions Maths Chapter 2

Get the simplified Class 8 Maths Extra Questions Chapter 2 Power Play Class 8 Extra Questions and Answers with complete explanation.

Class 8 Power Play Extra Questions

Class 8 Maths Chapter 2 Power Play Extra Questions

Class 8 Maths Chapter 2 Extra Questions – Power Play Extra Questions Class 8

Very Short Answer Type Questions

Question 1.
Write the expression for the thickness of a paper after n folds if its initial thickness is t cm.
Answer:
Given, the initial thickness = t
So, the thickness of a paper after n folds = t × 2ⁿ.

Question 2.
Find the value of [2^-1 × 3^-1]^-1.
Answer:
We have, [2^-1 × 3^-1]^-1

Question 3.
Write the value of (13)^-13 + (13)¹³.
Answer:
We have, (13)^-13 + (13)¹³ = (13)^-13-13 = (13)^-26
[aⁿ ÷ a^m = a^n-m]
= \(\frac{1}{(13)^{26}}\) [∵ a^-n = \(\frac{1}{a^n}\)]

Question 4.
Find the standard form of 32500000000.
Answer:
The standard form of 32500000000 is 325 × 10¹⁰.

Question 5.
Write 0.0000056789 in the standard form.
Answer:
The standard form of 0.0000056789 is 5.6789 × 10^-6.

Question 6.
Write the usual form of 2.8 × 10^-10.
Answer:
The usual form of 2.8 × 10^-10 is 0.00000000028.

Question 7.
Expand the number 4050 in power of 10.
Answer:
We have, 4050 = 4 × 1000 + 0 × 100 + 5 × 10 + 0
= 4 × 10³ + 0 × 10² + 5 × 10¹ + 0 × 10⁰

Short Answer Type Questions

Question 1.
The cells of a bacteria double itself every hour. How many cells will there be after 12 h, if initially we start with 1 cell? Express the answer in powers.
Answer:
The cells of a bacteria double itself every hour i.e. the cells of a bacteria after 1 h = 2
[∵ if initially we start with 1 cell]
The cells of a bacteria after 2h = 2 × 2 = 2²
The cells of a bacteria after 3h = 2 × 2² = 2³
∴ The cells of a bacteria after 12 h = 2¹²

Question 2.
Express as a power of a rational number with negative exponent.
(i) \(\left[\left(-\frac{3}{2}\right)^{-2}\right]^{-3}\)
Answer:

(ii) (2⁵ ÷ 2⁸) × 2^-7
Answer:
(2⁵ ÷ 2⁸) × 2^-7 = (2⁵ × \(\frac{1}{2^8}\)) × 2^-7
= 2^5-8 × 2^-7
= 2⁸ × 2^-7
= 2^-3+(-7)
= 2^-3-7
= 2^-10

Question 3.
How many seconds are there approximately in 3.8 months, if one month has about 2.6 x 106 seconds?
Answer:
Given, the number of seconds in one-month = 2.6 × 10⁶
So, the number of seconds is 3.8 months
= 2.6 × 3.8 × 10⁶
= 9.88 × 10⁶

Question 4.
Express \(\frac{1.5 \times 10^6}{2.5 \times 10^4}\) in the standard form.
Answer:
We have, \(\frac{1.5 \times 10^6}{2.5 \times 10^4}=\frac{1.5 \times 10^2 \times 10^4}{2.5 \times 10^4}=\frac{150}{2.5}\) × 10^4-4
= \(\frac{150}{2.5}\) × 10⁰
= 60 × 1 = 60
= 6.0 × 10¹
= 6.0 × 10

Question 5.
An electron’s mass is approximately 9.1093826 × 10^-31 kg. What is this mass in grams?
Answer:
Given, an electron’s mass = 9.1093826 × 10^-31 kg
∵ 1 kg = 1000 g = 10 × 10 × 10 g = 10³ g
∴ An electron’s mass in grams
= 9.1093826 × 10^-31 × 10^-3 g
= 9.1093826 × 10^-31+3 g
= 9.1093826 × 10^-28 g

Long Answer Type Questions

Question 1.
The cells of a bacteria double in every 30 min.
A scientist begins with a single cell.
(i) How many cells will be there after (a) 10h? (b) 25 h?
Answer:
The cells of a bacteria double in every 30 min
i.e. number of cells of a bacteria after 30 min =2
[∵ a scientist begins with a single cell]

∴ Number of cells of a bacteria after 1 h
= 2 × 2 = 2² = 2^2×1

Number of cells of a bacteria after 1\(\frac{1}{2}\) h
= 2 × 2³ = 2⁴ = 2^{2×\(\frac{3}{2}\)}
and number of cells of a bacteria after 2 h = 2 × 2³ = ⁴ = 2^2×2

(a) Thus, the number of cells after 10 h = 2^2×10= 2²⁰
(b) The number of cells after 25 h =2^2×25 = 2⁵⁰

(ii) What type of value is depicted by the cells of bacteria?
Answer:
The value depicted by the cells of bacteria is that it double itself after 30 min or it grows itself in t h by 2^2×t.

Question 2.
If a = – 1 and b = 2, then find the value of the following.
(i) a^b + b^a
(ii) a^b – b^a
(iii) a^b x b^a
(iv) a^b + b^a
Answer:

Question 3.
Astronomy The table shows the mass of the planets, the Sun and the Moon in our solar system.

Celestial Body	Mass (kg)	Mass (kg) in Standard Notation
Sun	1990000000000000000000000000000
Mercury	330000000000000000000000
Venus	4870000000000000000000000
Earth	5970000000000000000000000
Mars	642000000000000000000000000000
Jupiter	1900000000000000000000000000
Saturn	568000000000000000000000000
Uranus	86800000000000000000000000
Neptune	102800000000000000000000000
Pluto	12700000000000000000000
Moon	73500000000000000000000

(i) Write the mass of each celestial body in standard notation.
Answer:
Mass of each celestial body in standard in notation is given below.

• Mass of Sun = 1.99 × 10³⁰ kg
• Mass of Mercury = 3.3 × 10²³ kg
• Mass of Venus = 4.870 × 10²⁴ kg
• Mass of Earth = 5.970 × 10²⁴ kg
• Mass of Mars = 6.42 × 10²⁹ kg
• Mass of Jupiter = 1.9 × 10²⁷ kg
• Mass of Saturn = 5.68 × 10²⁶ kg
• Mass of Uranus = 8.68 × 10²⁵ kg
• Mass of Neptune = 1.02 × 10²⁶ kg
• Mass of Pluto = 1.27 × 10²² kg
• Mass of Moon = 7.35 × 10²² kg

(ii) Arrange the planets and the Moon according to their mass, from least to greatest.
Answer:
On observing the masses of the all given planets and the Moon in the exponent form, we see that the order of planets and the Moon by mass from least to greatest is
Pluto < Moon < Mercury < Venus < Earth < Uranus < Neptune < Saturn < Jupiter < Mars

(iii) Which planet has same mass as the Earth (approx.)?
Answer:
∵ Mass of Earth = 5.970 × 10²⁴ kg and
mass of Venus = 4.870 × 10²⁴ kg
Thus, we can say that planet Venus has about same mass as the Earth.

Question 4.
The given table shows the crop production of a state in the year 2008 and 2009. Observe, the table given below and answer the given questions.

Crop	2008 Harvest (in hectare)	Increase/Decrease in 2009 (in hectare)
Bajra	1.4 × 103	-100
Jowar	1,7 × 106	– 440000
Rice	3.7 × 103	-100
Wheat	5.1 × 105	+190000

(i) For which crop(s) did the production decrease?
Answer:
Production decreased for bajra, jowar and rice,

(ii) Write the production of all the crops in 2009 in their standard form.
Answer:
Production of bajra in 2008 = 14 × 10³and decrease in production of bajra in 2009 = 100 = 0.1 × 10³
So, production of bajra in 2009
= 1.4 × 10³ – 0.1 × 10³
= (14 – 0.1) × 10³
= 1.3 × 10³

Production of jowar in 2008 = 1.7 × 10⁶ and
decrease in production of jowar in 2009 = 440000 = 0.44 × 10⁶

So, production of jowar in 2009
= 17 × 10⁶ – 0.44 × 10⁶ = (17 – 0.44) × 10⁶ = 126 × 10⁶
Production of rice in 2008 = 3.7 × 10³ and

decrease in production of rice in 2009 = 100 = 0.1 × 10³
So, production of rice in 2009
= 3.7 × 10³ – 0.1 × 10³
= (3.7 – 0.1) × 10³
= 3.6 × 10³

Production of wheat in 2008 = 5.1 × 10⁵
and increase in production of wheat in 2009 = 190000 = 19 × 10⁵

So, production of wheat in 2009
= 5.1 × 10⁵ + 19 × 10⁵
= (5.1 + 19) × 10⁵
= 7.0 × 10⁵

(iii) Assuming the same decrease in rice production each year as in 2009, how many acres will be harvested in 2015? Write in standard form.
Answer:
Decrease in production of rice in 2009 = 0.1 × 10³
The number of years from 2009 to 2015 is 6.
The total decrease = 6 × 0.1 × 10³ = 0.6 × 10³
So, production of rice in 2015
= 3.7 × 10³ – 0.6 × 10³
= (3.7 – 0.6) × 10³
= 3.1 × 10³ hec
Hence, 3.1 × 10³ hec will be harvested in 2015.

Skill Based Questions

Question 1.
Fill the boxes with the numbers having exponent so that each horizontal and vertical equations are correct. Use knowledge of powers and exponents.

Clues
(i) Evaluate the expressions using exponents.
(ii) Each row and column must form a valid product equation.
(iii) Use positive whole numbers only.
(iv) Each row and column gives the correct product i.e. AB = C.
Answer:

Question 2.
Exponent Puzzle Challenge!
Instruction In the grid below, some numbers are given in the form of powers (exponents). The product of each row and column is also given. Fill the boxes with appropriate numbers (written as powers of whole numbers) so that the product of each row and column is correct.

Your Task!
(i) Fill the boxes with numbers in exponential (power) form such that each row and column gives the correct product.
Answer:
Compute grid is as

(ii) Write each number in exponential form.
Answer:
Rows 1024 = 2¹⁰
625 = 5⁴
6561 = 3⁸
Columns 16200 = 2³ ×  3⁴ × 5²
72 = 2³ × 3² × 5⁰
3600 = 2⁴ × 3² × 5²

Case Study Based Question

Question 1.
The distance between planets varies depending on their position on their orbital path around Sun. The table given below shows the average distance between two planets.

Planets	Average distance between them
Venus to Earth	40000000 km
Earth to Mars	225000000 km

(i) What is the distance (in km) between Venus and Earth in exponential form?
Answer:
4 × 10⁷

(ii) How much further is Mars from Earth than Venus?
Answer:
1.85 × 10⁸

(iii) Finds the ratio of the distance between the Venus to the Earth and the distance between the Earth to Mars?
Answer:
8 : 45