Bhagya – Page 287 – PSEB Solutions

PSEB 7th Class Science Notes Chapter 9 Soil

November 23, 2021 by Bhagya

This PSEB 7th Class Science Notes Chapter 9 Soil will help you in revision during exams.

PSEB 7th Class Science Notes Chapter 9 Soil

→ The top layer of the earth in which crops and plants can grow is called soil.

→ Soil is made up of broken rocks, organic matter, animals, plants, and microorganisms.

→ There are different layers of soil, which can be seen in the soil profile.

→ Soil is made up of both organic and inorganic components.

→ The dead and rotten leaves of plants or the bodies of plants, insects, or dead animals buried in the soil, animal dung, etc. combine to form organic matter called humus.

→ Soil that contains a mixture of organic and inorganic substances is very useful for crops.

→ Depending on the size of the particles, the soil is clayey, sandy, rocky, and loamy.

→ Depending on the chemical nature of the soil, the soil may be acidic, basic, or neutral.

→ Acid soils have a pH of 1 to 6.

→ Alkaline soils have a pH of 8 to 14.

→ Neutral soil has a pH of 7.

→ Ph paper is used to determine the nature of the soil.

→ Black soil contains iron salts and is good for growing cotton.

→ Soil containing sulphur is good for growing onions.

→ Different types of soil are required to grow different types of crops.

→ It takes many years for the formation of the top layer of soil.

→ Removal of topsoil due to floods, winds, storms, and mining is called Erosion.

→ By digging the soil, the animals with their feet loosen the soil and the soil, which has been loosened, gets eroded quickly by wind and water.

→ By planting trees, building check dams, planting grass on the sides of farmland, and building along sides of rivers and canals soil erosion can be prevented.

→ Soil: A mixture of rock/horizontal particles and humus is called soil.

→ Soil Profile: A vertical section through different layers of soil is called the soil profile.

→ Humus: The dead and decaying organisms present in soil are called humus.

→ Soil Moisture: Soil retains water in it, which is called soil moisture.

→ Soil erosion: The removal of the top layer of soil by water, wind, or ice is called Soil erosion.

→ Weathering: It is a method in which soil is formed by the breakdown of rocks by the action of wind, water, and climate.

PSEB 7th Class Maths Solutions Chapter 8 Comparing Quantities Ex 8.2

November 23, 2021 by Bhagya

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 8 Comparing Quantities Ex 8.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.2

1. Convert the following fractions into percents

(i). \(\frac {1}{8}\)
Solution:
\(\frac {1}{8}\) = \(\frac {1}{8}\) × 100
= \(\frac {25}{2}\)
= 12.5
Thus, \(\frac {1}{8}\) = 12.5%

(ii). \(\frac {49}{50}\)
Solution:
\(\frac {49}{50}\) = \(\frac {49}{50}\) × 100
= 98
Thus, \(\frac {49}{50}\) = 98%

(iii). \(\frac {5}{4}\)
Solution:
\(\frac {5}{4}\) = \(\frac {5}{4}\) × 100
= 125
Thus, \(\frac {5}{4}\) = 125%

(iv). 1\(\frac {3}{8}\)
Solution:
1\(\frac {3}{8}\) = \(\frac {11}{8}\) × 100
= \(\frac {275}{2}\)
= 137\(\frac {1}{2}\)
Thus, 1\(\frac {3}{8}\) = 137\(\frac {1}{2}\)%

2. Convert the following percents into fractions in simplest form

(i). 25%
Solution:
25% = \(\frac {25}{100}\)
= \(\frac {1}{4}\)

(ii). 150%
Solution:
150% = \(\frac {150}{100}\)
= \(\frac {3}{2}\)

(iii). 7\(\frac {1}{2}\)
Solution:
7\(\frac {1}{2}\) = \(\frac {15}{2}\) × \(\frac {1}{100}\)
= \(\frac {3}{40}\)

3. (i) Anita secured 324 marks out of 400 marks. Find the percentage of marks secured by Anita.
(ii) Out of 32 students, 8 are absent from the class. What is the percentage of students who are absent ?
(iii) There are 120 voters, 90 out of them voted. What percent did not vote ?
Solution:
(i) Anita scored 324 marks out of 400 marks.
∴ Percentage of marks secured by Anita = \(\left(\frac{324}{400} \times 100\right) \%\) = 50%

(ii) Out of 32 students 8 students are absent
∴ Percentage of students who are absent = \(\frac {8}{32}\) × 100% = 25%

(iii) Total voters = 120
Voters who voted = 90
The voters who did not vote = 120 – 90
= 30
Percentage voters who did not vote = \(\frac {30}{120}\) × 100%
= 25%

4. Estimate the part of figure which is shaded and hence find the percentage of the part which is shaded.
PSEB 7th Class Maths Solutions Chapter 8 Comparing Quantities Ex 8.2 1
Solution:
(i) Shaded part = \(\frac {2}{4}\) = \(\frac {1}{2}\)
Percentage of shaded part = \(\left(\frac{1}{2} \times 100\right) \%\)
= 50%

(ii) Shaded part = \(\frac {2}{6}\) = \(\frac {1}{3}\)
Percentage of shaded part = \(\left(\frac{1}{3} \times 100\right) \%\)
= 33\(\frac {1}{3}\)%

(iii) Shaded part = \(\frac {5}{8}\)
Percentage of shaded part = \(\left(\frac{5}{8} \times 100\right) \%\)
= \(\frac {125}{2}\)%
= 62.5%

5. Convert the following percentages into ratios in simplest form :

(i). 14%
Solution:
14% = 14 × \(\frac {1}{100}\)
= \(\frac {7}{50}\)
= 7 : 50

(ii). 1\(\frac {3}{4}\)%
Solution:
1\(\frac {3}{4}\)% = \(\frac {7}{4}\) × \(\frac {1}{100}\)
= \(\frac {7}{400}\)
= 7 : 400

(iii). 33\(\frac {1}{3}\)%
Solution:
33\(\frac {1}{3}\)% = \(\frac {100}{3}\) × \(\frac {1}{100}\)
= \(\frac {1}{3}\)
= 1 : 3

6. Express the following ratios as percentages :

(i). 5 : 4
Solution:
5:4 = \(\frac {5}{4}\) × 100
= 125%

(ii). 1 : 1
Solution:
1 : 1 = \(\frac {1}{1}\) × 100
= 100%

(iii). 2 : 3
Solution:
2 : 3 = \(\frac {2}{3}\) × 100
= \(\frac {200}{3}\)%
= 66\(\frac {2}{3}\)%

(iv). 9 : 16
Solution:
9 : 16 = \(\frac {9}{16}\) × 100
= \(\frac {225}{4}\)%
= 56\(\frac {1}{4}\)%

7. Chalk contains calcium, carbon and sand in the ratio 12 : 3 : 10. Find the percentage of carbon in the chalk.
Solution:
Calcium : Carbon : sand
= 12 : 3 : 10
Total of ratios = 12 + 3 + 10
= 25
Percentage of carbon is chalk
= \(\frac {3}{25}\) × 100
= 12%

8. Convert each part of the following ratios into percentage :

(i). 3 : 1
Solution:
Total of ratios = 3 + 1 = 4
Percentage of first part = \(\frac {3}{4}\) × 100
= 75%
Percentage of second part = \(\frac {1}{4}\) × 100
= 25%

(ii). 1 : 4
Solution:
Total of ratios = 1 + 4 = 5
Percentage of first part = \(\frac {1}{5}\) × 100
= 20%
Percentage of second part = \(\frac {4}{5}\) × 100
= 80%

(iii). 4 : 5 : 6.
Solution:
Total of ratios = 4 + 5 + 6 = 15
Percentage of first part = \(\frac {4}{15}\) × 100
= \(\frac {8}{3}\)%
= 26\(\frac {2}{3}\)%
Percentage of second part = \(\frac {5}{15}\) × 100
= \(\frac {100}{3}\)%
= 33\(\frac {1}{3}\)%
Percentage of third part = \(\frac {6}{15}\) × 100
= 40%

9. Convert the following percentages to decimals :

(i). 28%
Solution:
28% = \(\frac {28}{100}\)
0.28

(ii). 3%
Solution:
3% = \(\frac {3}{100}\)
= 0.03

(iii). 37\(\frac {1}{2}\)%
Solution:
37\(\frac {1}{2}\)%
= \(\frac {75}{2}\) × \(\frac {1}{100}\)=
= \(\frac {3.75}{100}\)
= 0.375

10. Convert the following decimals to percentage :

(i). 0.65
Solution:
0.65 = (0.65 × 100)%
= \(\left(\frac{65}{100} \times 100\right) \%\)
= 65%

(ii). 0.9
Solution:
0.9 = (0.9 × 100)%
= \(\left(\frac{9}{10} \times 100\right) \%\)
= 90%

(iii). 2.1
Solution:
2.1 = (2.1 × 100)%
= \(\left(\frac{21}{10} \times 100\right) \%\)
= 210%

11. (i) If 65% of students in a class have a bicycle, then what percent of the students do not have a bicycle ?
(ii) We have a basket full of apples, oranges and mangoes. If 50% are apples, 30% are oranges, then what percent are mangoes ?
Solution:
(i) Percentage of student having a bicycle = 65%
Percentage of students do not have a bicycle = (100 – 65)%
= 35%

(ii) Percentage of apples = 50%
Percentage of oranges = 30%
Percentage Of mangoes = (100 – (50% + 30%)
= (100 – 80)%
= 20%

12. The population of a city decreased from 25,000 to 24,500. Find the percentage decrease.
Solution:
Original population = 25000
Decreased population = 24,500
Decrease in population = (25000 – 24500)
= 500
Percentage decrease = \(\frac{\text { Decrease in population }}{\text { Original population }} \times 100 \%\)
= \(\frac {500}{25000}\) × 100%
= 2%

13. Arun bought a plot for ₹ 3,50,000. The next year, the price went upto ₹ 3,70,000. What was the percentage of price increase ?
Solution:
Original price of plot = ₹ 3,50,000
The increased price of plot = ₹ 3,70,000
Increase in price = ₹ 3,70,000 – ₹ 3,50,000
= ₹ 20,000.
Percentage of price increase
= \(\left(\frac{\text { Increase in price }}{\text { Original price }} \times 100\right)\)
= \(\frac {20,000}{350000}\) × 100%
= \(\frac {40}{7}\)%
= 5\(\frac {5}{7}\)

14. Find :

(i). 15% of 250
Solution:
15% of 250 = \(\frac {15}{100}\) × 250
= \(\frac {375}{10}\)
= 37.5

(ii). 25% of 120 litres
Solution:
25% of 120 litres = \(\frac {25}{100}\) × 20 litres
= 30 liters.

(iii). 4% of 12.5
Solution:
4% of 12.5 = \(\frac {4}{100}\) × \(\frac {125}{10}\)
= \(\frac {5}{10}\)
= 0.5

(iv). 12% of ₹ 250.
Solution:
12% of ₹ 250 = ₹\(\frac {12}{100}\) × 250
= ₹300

15. Multiple Choice Questions :

Question (i).
The ratio 2 : 3 expressed as percentage is
(a) 40%
(b) 60%
(c) 66\(\frac {2}{3}\)%
(d) 33\(\frac {1}{3}\)%
Answer:
(c) 66\(\frac {2}{3}\)%

Question (ii).
If 30% of x is 72, then x is equal to
(a) 120
(b) 240
(c) 360
(d) 480
Answer:
(b) 240

Question (iii).
0.025 when expressed as a percent is
(a) 250%
(b) 25%
(c) 4%
(d) 2.5%
Answer:
(d) 2.5%

Question (iv).
In a class, 45% of students are girls. If there are 22 boys in the class, then the total number of students in the class is
(a) 30
(b) 36
(c) 40
(d) 44
Answer:
(c) 40

Question (v).
What percent of \(\frac {1}{7}\) is \(\frac {2}{35}\) ?
(a) 20%
(b) 25%
(c) 30%
(d) 40%
Answer:
(d) 40%

PSEB 7th Class Science Notes Chapter 8 Winds, Storms and Cyclones

November 23, 2021 by Bhagya

This PSEB 7th Class Science Notes Chapter 8 Winds, Storms and Cyclones will help you in revision during exams.

PSEB 7th Class Science Notes Chapter 8 Winds, Storms and Cyclones

→ The air around us exerts pressure.

→ Moving air is called wind.

→ Very strong wind lowers the pressure.

→ Air expands on heating and contracts on cooling.

→ Hot air is lighter than cold air.

→ The wind moves from high-pressure areas to low-pressure areas.

→ Wind speed is measured with an Anemometer.

→ The direction of wind speed is measured by the wind vane.

→ Wind currents (movement) are caused by the uneven heating of the earth.

→ Monsoon winds are filled with moisture (water vapours) and bring rain.

→ Cyclones are destructive.

→ A cyclone crossed the coast of Orissa on October 18, 1999.

→ Cyclones have higher wind speeds.

→ A cyclone is a very strong whirlwind that revolves around very low-pressure areas.

→ A hurricane is a storm with strong winds blowing through, in a funnel-shaped cloud.

→ The loud noise produced during lightning is called thunder.

→ Heavy rain with strong winds is called a storm.

→ Hurricanes in the United States and typhoons in Japan are cyclones.

→ Tornadoes are dark cone-like clouds that form between the earth’s crust and the sky.

→ All kinds of natural disasters like cyclones, tornadoes, etc. destroy trees, wires, and communication systems.

→ Special policies are adopted during disasters.

→ A cyclone warning is given 48 hours in advance with the help of satellite and radar.

→ And self-help is the best help. So it would be helpful to plan your safety in advance and take precautionary measures before any cyclone actually strikes.

→ Wind: Fast-moving air is called wind.

→ Monsoon winds: The winds that come from the sea and carry water are called monsoon winds.

→ Tornado: Dark coloured cone-like clouds whose cone structure is from sky to earth is called Tornado.

→ Cyclone: A violent wind that moves in a circle causing a storm is called a cyclone.

PSEB 9th Class Maths Solutions Chapter 2 Polynomials Ex 2.3

November 23, 2021 by Bhagya

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 2 Polynomials Ex 2.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.3

Question 1.
Find the remainder when x³ + 3x² + 3x + 1 is divided by
Answer:
The remainder theorem states that when polynomial p (x) of degree greater than or equal to 1 is divided by linear polynomial x – a, the remainder is p (a).
Here. p(x) = x³ + 3x² + 3x + 1

(i) x + 1
Answer:
Divisor g (x) = x + 1.
Comparing x + 1 with zero, we get x = – 1.
Then, remainder
= p(- 1)
= (- 1)³ + 3(- 1)² + 3(- 1) + 1
= – 1 + 3 – 3 + 1
= 0

(ii) x – \(\frac{1}{2}\)
Answer:
Divisor g (x) = x – \(\frac{1}{2}\)
x – \(\frac{1}{2}\) = 0 gives x = \(\frac{1}{2}\)
Then, remainder
= p\(\left(\frac{1}{2}\right)\)
= \(\left(\frac{1}{2}\right)^{3}+3\left(\frac{1}{2}\right)^{2}+3\left(\frac{1}{2}\right)+1\)
= \(\frac{1}{8}+\frac{3}{4}+\frac{3}{2}+1\)
= \(\frac{27}{8}\)

(iii) x
Answer:
Divisor g (x) = x.
x = 0 gives x = 0.
Then, remainder = p (0)
= (0)³ + 3(0)² + 3(0) + 1
= 0 + 0 + 0 + 1
= 1

(iv) x + π
Answer:
Divisor g (x) = x + π.
x + π = 0 gives x = – π.
Then, remainder
= p(- π)
= (- π)³ + 3(- π)² + 3(- π) + 1
= – π³ + 3π² – 3π + 1

(v) 5 + 2x
Answer:
Divisor g(x) = 5 + 2x.
5 + 2x = 0 gives x = – \(\frac{5}{2}\)
Then, remainder
= P\(\left(-\frac{5}{2}\right)\)
= \(\left(-\frac{5}{2}\right)^{3}+3\left(-\frac{5}{2}\right)^{2}+3\left(-\frac{5}{2}\right)+1\)
= \(-\frac{125}{8}+\frac{75}{4}-\frac{15}{2}+1\)
= \(\frac{-125+150-60+8}{8}\)
= \(-\frac{27}{8}\)

Question 2.
Find the remainder when x³ – ax² + 6x – a is divided by x – a.
Answer:
Here, p (x) = x³ – ax² + 6x – a and divisor
g (x) = x – a.
x – a = 0 gives x = a.
Then, remainder = p (a)
= (a)³ – a(a)² + 6(a) – a
= a³ – a³ + 6a – a
= 5a

Question 3.
Check whether 7 + 3x is a factor of 3x³ + 7x.
Answer:
Here, p (x) = 3x³ + 7x and divisor g (x) = 7 + 3x.
7 + 3x = 0 gives x = –\(\frac{7}{3}\).
Then, remainder = p\(\left(-\frac{7}{3}\right)\)
= \(3\left(-\frac{7}{3}\right)^{3}+7\left(-\frac{7}{3}\right)\)
= \(-\frac{343}{9}-\frac{49}{3}\)
= \(\frac{-343-147}{9}\)
= – \(\frac{490}{9}\) ≠ 0
Since the remainder is not zero when
p (x) = 3x³ + 7x is divided by 7 + 3x, it is clear that 7 + 3x is not a factor of 3x³ + 7x.

PSEB 7th Class Science Notes Chapter 7 Weather, Climate and Adaptations of Animals to Climate

November 23, 2021 by Bhagya

This PSEB 7th Class Science Notes Chapter 7 Weather, Climate and Adaptations of Animals to Climate will help you in revision during exams.

PSEB 7th Class Science Notes Chapter 7 Weather, Climate and Adaptations of Animals to Climate

→ The weather of a place changes day by day and week by week.

→ The weather depends on temperature, humidity, and rainfall.

→ Humidity is a measure of water vapour in the air.

→ Indian Meteorological Department of weather forecasting daily collects statistical data of heat, wind speed at various places and makes weather predictions.

→ Atmospheric conditions in terms of temperature, humidity, rainfall, wind speed, etc. of a place are called the weather of that place.

→ The weather can change in an instant.

→ Factors on which the weather depends, are called elements of weather.

→ Special high-low thermometers are used to measure the temperature.

→ The highest temperature of the day is usually in the afternoon and the lowest is normal usually in the morning.

→ All changes in the weather are caused by the sun.

→ In winters, the length of the day is shorter and the night is earlier.

→ The length of the weather of a place is based on the data collected at that place is called the climate of that place.

→ The climate of different places is different. It changes from hot and dry to hot and humid.

→ Energy reflected and absorbed by the earth’s surface, ocean and atmosphere play important roles in determining weather at any place.

→ Climate has a great impact on living things.

→ Animals are adapted to the conditions in which they live.

→ The Polar Regions are located near the poles, such as the North Pole and the South Pole.

→ Canada, Greenland, Iceland, Norway, Sweden, Finland and Alaska, and Siberian areas of Russia in America, are the Polar Regions

→ Tropical rainforests are found in India, Malaysia, Indonesia, Brazil, the Republic of Congo, Kenya, Uganda, and Nigeria. The Polar Regions have a cold climate.

→ Rainfall is measured by an instrument called the rain gauge.

→ Penguins and polar bears live in Polar Regions.

→ The Polar Regions are covered with white ice.

→ The white hairs on the polar bear’s body help protect it and catch prey.

→ Penguins are well-known animals found in Polar Regions. It is also white and merges well with the background.

→ In addition to polar bears and penguins, many other animals are found in Polar Regions.

→ Many fishes can live in cold water.

→ The climate of the Subtropical Regions is generally warmer, as these areas are closer to the equator.

→ Temperatures in these regions vary from 15°C to 40°C.

→ In areas near the equator, the length of night and day are approximately equal throughout the year.

→ Weather: Everyday changes in the atmosphere in terms of temperature, humidity, rainfall, wind speed, etc., is called the weather of that place.

→ Climate: The average weather pattern taken over a long time, say 25 years is called the climate of the place.

→ Adaptation: The special characteristics of plants and organisms, that is, the nature that enables them to live in a habitat, is called adaptation.

→ Migration: Moving from one place to another to avoid harsh climatic conditions by birds and animals is called migration.

PSEB 7th Class Science Notes Chapter 6 Physical and Chemical Changes

November 23, 2021 by Bhagya

This PSEB 7th Class Science Notes Chapter 6 Physical and Chemical Changes will help you in revision during exams.

PSEB 7th Class Science Notes Chapter 6 Physical and Chemical Changes

→ Change is the tendency of life. There are many changes in our daily life.

→ There are two types of changes:

Physical changes
Chemical changes

→ There is always a reason for the change.

→ Some changes can be controlled and some others cannot be controlled.

→ No new matter is formed in physical change.

→ Chemical changes usually cannot be reversed.

→ The properties of new substances produced in a chemical change are completely different (new).

→ Changes can be classified based on their similarities.

→ The properties of a substance such as size, measure, color, state are called its physical properties.

→ The change that takes place in the physical properties of a substance is called a physical change.

→ The magnesium strip (ribbon) is bums with bright white light.

→ When Carbon dioxide is passed through lime water, it becomes milky.

→ Chemical change produces sound, light, heat, smell, gas, color, and so on.

→ Burning is a chemical change in which there is always an outflow of heat.

→ There is a layer of Ozone in the atmosphere.

→ For the occurrence of rust both oxygen and water are required.

→ In the galvanization process, a layer of zinc is deposited on the iron.

→ Iron can be saved from rust by applying paint.

→ Large crystals can be obtained from a saturated solution of a substance by crystallisation method.

→ Physical changes: Changes in which only the physical properties of matter change and no new matter is created are called physical changes. Example: Salt solution in water.

→ Chemical changes: Changes that involve the formation of new substances with new properties are called chemical changes. Example: Burning of coal.

→ Rust: The process in which iron gets covered with a layer of brownish substance in the presence of moist air is called rust.

→ Galvanization: The process of depositing zinc on iron to protect it from corrosion is called Galvanization.

→ Crystallisation: The process of obtaining large size crystals of a soluble substance is called crystallisation.

PSEB 9th Class Maths Solutions Chapter 2 Polynomials Ex 2.2

November 23, 2021 by Bhagya

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 2 Polynomials Ex 2.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.2

Question 1.
Find the value of the polynomial 5x – 4x² + 3 at (i) x = 0, (ii) x = – 1 and (iii) x = 2.
Answer:
Here, p (x) 5x – 4x² + 3
(i) The value of polynomial p (x) at x = 0 is given by
p(0) = 5(0) – 4(0)² + 3
= 3

(ii) The value of polynomial p (x) at x = – 1 is given by
p(- 1) = 5(- 1) – 4 (- 1)² + 3
= – 5 – 4 + 3
= – 6

(iii) The value of polynomial p (x) at x = 2 is given by
p(2) = 5(2) – 4(2)² + 3
= 10 – 16 + 3
= – 3

Question 2.
Find p(0), p(1) and p(2) for each of the following polynomials:
(i) p (y) = y² – y + 1
Answer:
p(y) = y² – y + 1
∴ p(0) = (0)² – (0) + 1 = 1
∴ P(1) = (1)² – (1) + 1 = 1 – 1 + 1 = 1
∴ p(2) = (2)² – (2) + 1 = 4 – 2 + 1 = 3

(ii) p (t) = 2 + t + 2t² – t³
Answer:
p(t) = 2 + t + 2t² – t³
∴ p(0) = 2 + 0 + 2(0)² – (0)³ = 2
∴ p (1) = 2 + (1) + 2 (1)² – (1)³
= 2 + 1 + 2 – 1
= 4
∴ p(2) = 2 + (2) + 2(2)² – (2)³
= 2 + 2 + 8 – 8 = 4

(iii) p(x) = x³
Answer:
p(x) = x³
∴ p (0) = (0)³ = 0
∴ p ( 1) = (1)³ = 1
∴ p (2) = (2)³ = 8

(iv) p(x) = (x – 1) (x + 1)
Answer:
p(x) = (x – 1) (x + 1)
∴ p(0) = (0 – 1) (0 + 1) = (- 1) × 1 = – 1
∴ p(1) = (1 – 1) (1 + 1) = 0 × 2 = 0
∴ p(2) = (2 – 1)(2 + 1) = 1 × 3 = 3

Question 3.
Verify whether the following are zeros of the polynomial, indicated against them:
(i) p(x) = 3x + 1, x = – \(\frac{1}{3}\)
Answer:
Here, p (x) = 3x + 1
Then, p\(\left(-\frac{1}{3}\right)\) = 3\(\left(-\frac{1}{3}\right)\) + 1 = – 1 + 1 = 0
Hence, – \(\frac{1}{3}\) is a zero of polynomial
p(x) = 3x + 1

(ii) p(x) = 5x – π, x = \(\frac{4}{5}\)
Answer:
Here, p(x) = 5x – π,
Then, p\(\left(\frac{4}{5}\right)\) = 5\(\left(\frac{4}{5}\right)\) – π = 4 – π ≠ 0
Hence, \(\frac{4}{5}\) is not a zero of polynomial
p(x) = 5x – π

(iii) p(x) = x² – 1, x = 1, – 1
Answer:
Here, p(x) = x² – 1
Then, p(1) = (1)² – 1 = 1 – 1 = 0 and
p(- 1) = (- 1)² – 1 = 1 – 1 = 0.
Hence, 1 and – 1 both are zeroes of polynomial p(x) = x² – 1.

(iv) p(x) = (x + 1) (x – 2), x = – 1, 2
Answer:
Here, p (x) = (x + 1) (x – 2)
Then, p(- 1) = (- 1 + 1) (- 1 – 2) = 0 × (-3)= 0
and p (2) = (2 + 1) (2 – 2) = 3 × O = O.
Hence, – 1 and 2 both are zeros of polynomial p(x) = (x + 1) (x – 2).

(v) p(x) = x², x = 0
Answer:
Here, p(x) = x²
Then, p(0) = (0)² = 0
Hence, 0 is a zero of polynomial p (x) = x².

(vi) p(x) = lx + m, x = –\(\frac{n}{l}\)
Answer:
Here, p (x) = lx + m
Then, p \(\left(-\frac{m}{l}\right)\) = l\(\left(-\frac{m}{l}\right)\) + m = – m + m = 0
Hence, \(-\frac{m}{l}\) is a zero of polynomial
p(x) = lx + m.

(vii) p(x) = 3x² – 1, x = – \(\frac{1}{\sqrt{3}}\), \(\frac{2}{\sqrt{3}}\)
Answer:
Here, p(x) = 3x² – 1
Then, p\(\left(-\frac{1}{\sqrt{3}}\right)\) = 3\(\left(-\frac{1}{\sqrt{3}}\right)^{2}\) – 1
= 3\(\left(\frac{1}{3}\right)\) – 1 = 1 – 1 = 0
and p\(\left(\frac{2}{\sqrt{3}}\right)\) = 3\(\left(\frac{2}{\sqrt{3}}\right)^{2}\) – 1
= 3\(\left(\frac{4}{3}\right)\) – 1 = 4 – 1 = 3 ≠ 0
Hence, –\(\frac{1}{\sqrt{3}}\) is a zero of polynomial p (x) = 3x² – 1, but \(\frac{2}{\sqrt{3}}\) is not a zero of polynomial p(x) = 3x² – 1.

(viii) p (x) = 2x + 1, x = \(\frac{1}{2}\)
Answer:
Here, p(x) = 2x + 1
Then, p\(\left(\frac{1}{2}\right)\) = 2\(\left(\frac{1}{2}\right)\) + 1 = 1 + 1 = 2 ≠ 0
Hence, \(\frac{1}{2}\) is not a zero of polynomial
p(x) = 2x + 1.

Question 4.
Find the zero of the polynomial in each of the following cases:
(i) p(x) = x + 5
Answer:
To find the zero of polynomial p (x) = x + 5,
we solve the equation p (x) = 0.
∴ x + 5 = 0
∴ x = – 5
Thus, – 5 is the zero of polynomial
p(x) = x + 5.

(ii) p(x) = x – 5
Answer:
To find the zero of polynomial p(x) = x – 5,
we solve the equation p (x) = 0.
∴ x – 5 = 0
∴ x = 5
Thus, 5 is the zero of polynomial
p(x) = x – 5.

(iii) p (x) = 2x + 5
Answer:
To find the zero of polynomial p(x) = 2x + 5,
we solve the equation p (x) = 0.
∴ 2x + 5 = 0
∴ 2x = – 5
Thus, –\(\frac{5}{2}\) is the zero of polynomial
p(x) = 2x + 5.

(iv) p (x) = 3x – 2
Answer:
To find the zero of polynomial p (x) = 3x – 2,
we solve the equation p (x) = 0.
∴ 3x – 2 = 0
∴ 3x = 2
Thus, \(\frac{2}{3}\) is the zero of polynomial
p(x) = 3x – 2.

(v) p(x) = 3x
Answer:
To find the zero of polynomial p (x) = 3x.
we solve the equation p (x) = 0.
∴ 3x = 0
∴ x = 0
Thus, 0 is the zero of polynomial P(x) = 3x.

(vi) p(x) = ax, a ≠ 0
Answer:
To find the zero of polynomial p (x) = ax,
a ≠ 0, we solve the equation p (x) = 0.
∴ ax = 0
∴ x = 0 (∵ a ≠ 0)
Thus, 0 is the zero of polynomial
p(x) = ax, a ≠ 0.

(vii) p(x) = cx + d, c ≠ 0, C, d are real numbers.
Answer:
To find the zero of polynomial
p(x) = cx + d, c ≠ 0, c, d are real numbers, we solve the equation p (x) = 0.
∴ cx + d = 0
∴ cx = – d
∴ x = – \(\frac{d}{c}\)
Thus, – \(\frac{d}{c}\) is the zero of polynomial p (x) = cx + d, c ≠ 0, c, d are real numbers.

PSEB 7th Class Science Notes Chapter 5 Acids, Bases and Salts

November 23, 2021 by Bhagya

This PSEB 7th Class Science Notes Chapter 5 Acids, Bases and Salts will help you in revision during exams.

PSEB 7th Class Science Notes Chapter 5 Acids, Bases and Salts

→ We eat many foods with different tastes in our daily life.

→ Some foods taste bitter, sour, sweet, and salty.

→ The sour taste of the substance is due to the acid present in them.

→ The word acid is derived from the Latin word ‘acere’, which means sour.

→ Substances that have a bitter taste and are soapy to touch, are called the base.

→ Indicators are substances that give different colours to substances or solutions of acidic and basic nature.

→ They are used to test the acidic or alkaline nature of substances.

→ Turmeric, litmus, and China rose petals (Gudhal) are natural indicators.

→ Neutral solutions do not change the color of litmus red or blue because they are neither acidic nor basic.

→ Phenolphthalein is a synthetic indicator that is prepared in the laboratory.

→ Some acids are strong and some are weak.

→ The reaction between an acid and a base is called Neutralisation.

→ Antacids are used to treat indigestion.

→ Solution of baking soda (Sodium hydrogen carbonate) rubbed to treat the effects of insect bites.

→ The basicity of the soil is eliminated by the use of organic matter.

→ The acidity of the soil is treated with bases like a quick (calcium oxide) or slaked lime (Calcium hydroxide).

→ Waste products from factories should be released into the water only after treatment with basic substances.

→ Acids: Substances that have a sour taste and which react with blue litmus solution to make its color red are called acids.

→ Bases (Alkali): Substances that have a bitter taste and which react with red litmus solution to make it blue are called bases.

→ Neutralisation: The reaction between an acid and an alkali is called neutralisation.

→ Neutral solution: A solution that is neither acidic in nature nor alkaline in nature is called a neutral solution or solution which does not change the color of the indicator is called a neutral solution.

→ Indicators: Indicators are substances that show different colours by reacting with different acids, salts, and bases.

PSEB 7th Class Science Notes Chapter 4 Heat

November 23, 2021 by Bhagya

This PSEB 7th Class Science Notes Chapter 4 Heat will help you in revision during exams.

PSEB 7th Class Science Notes Chapter 4 Heat

→ The method of determining whether an object is hot or cold by touching it is not reliable.

→ Temperature is the degree of hotness or coldness of an object.

→ The hot or cold state of an object, i.e. the temperature is measured with an instrument called a thermometer.

→ The temperature of a human or living organism is measured with a medical thermometer (or clinical thermometer)

→ A scale is present on a medical thermometer. This scale is either in Celsius [°C] or Fahrenheit [°F] or both.

→ Medical thermometers have a narrow, homogeneous tube of glass with a bulb at the bottom.

→ There is a kink near the bulb of a medical thermometer that prevents the mercury level from falling as a result of its weight.

→ Should clean with antiseptic solution before and after use of a medical thermometer.

→ The scale range of medical thermometers is from 35°C to 42°C.

→ Before using a medical thermometer, make sure that the mercury level is below 35° C.

→ If not, hold the thermometer firmly and shake it to bring a level of mercury below 35°C.

→ The normal temperature of a healthy person is 37°C or 98.4°F.

→ There are other thermometers to measure the temperature of objects.

→ One of these is the Lab thermometer. The lab thermometer ranges from -10°C to 110° C.

→ The temperature of the object should be measured with a lab thermometer when the mercury level in the thermometer becomes stable.

→ The method by which heat is transferred from the hotter end of an object to the cold end by the particles of material of the object is called conduction. Solids are heated by the conduction method.

→ Materials that transmit heat are called conductors.

→ Items made of iron, silver, copper, aluminium are heat conductors.

→ Substances that do not conduct heat well are called insulators or non-conductors, e.g. Wood, plastic, and rubber.

→ Air is not a good conductor of heat.

→ The method of heat transfer in which heat is transmitted by the motion of hot molecules of a substance is called convection.

→ Heat is transmitted in liquids and gases by the convection method,

→ In coastal areas, the wind that blows from the sea to the coast during the day is called Sea breeze.

→ In coastal areas, the air that blows from the coast to the sea at night is called land breeze.

→ Radiation is the method of transfer of heat due to the emission of radiations by hot objects without a medium.

→ Dark coloured fabrics absorb heat more than light coloured fabrics.

→ So in winter, we wear dark colours and in summer we wear light coloured clothes.

→ Woolen garments keep us warm in winter because of the air trapped in the wool fibres which is a bad conductor of heat.

→ Heat: It is an agent which produces in us the sensation of warmth. This is a kind of energy.

→ Temperature: Degree of hotness or coldness measured on a definite scale.

→ Thermometer: An instrument used to measure the temperature of an object.

→ Celsius Scale: The Celsius scale is a scale for measuring temperature. Sometimes it is also called as the centigrade scale.

→ Resistor: A substance through which heat cannot be transmitted properly is considered to be resistor or resistant to heat.

→ Conduction: This is the method of heat transfer in which heat passes from the hot end of the object to the cold end through the molecules of the material, but the molecules of the object remain fixed in their place.

→ Convection(Liquid or Gas): This is the method of heat transfer in which heat is transferred from the source of heat to the cold part due to the motion of hot molecules and the cold molecules come down to take their place. This method is used for heating liquids and gases.

→ Radiation: This is the method of heat transfer in which heat is transmitted from a hot source or body directly to a cool body without affecting the medium.

→ Sea breeze: Due to the heat of the sun during the day, the soil of the land gets heated very quickly when water molecules of the ocean are not so hot. Therefore, the air near the ground gets warmer and rises up due to being lighter.

→ To replace it, cool air begins to flow from the ocean, causing air currents to flow. The wind that blows from the ocean to the land is called the sea breeze.

→ Land breeze: Due to high heat absorption capacity, water cools later than land, due to which cold air starts flowing towards seawater which is called land breeze.

→ Fahrenheit Scale: The Fahrenheit scale is a scale designed to measure temperature.

PSEB 7th Class Science Notes Chapter 3 Fibre to Fabric

November 23, 2021 by Bhagya

This PSEB 7th Class Science Notes Chapter 3 Fibre to Fabric will help you in revision during exams.

PSEB 7th Class Science Notes Chapter 3 Fibre to Fabric

→ Fibre of wool and silk is derived from animals in nature.

→ The fibre of wool is obtained from sheep, goats, or yaks.

→ The thin skin of a sheep has two types of fibres:

Coarse beard hair
Fine soft hair under the hair close to the skin.

→ Some breeds of sheep possess only fine under hair, their parents are specially chosen to give birth to sheep which have only soft under hair, such selection of parents for the specific characters is called ‘selective breeding’.

→ There are different types of wool, such as sheep’s wool, Angora, and cashmere wool.

→ Sheep are reared to obtain wool.

→ Sorters disease is spread by the bacterium Anthrax.

→ Raising silkworms to get silk is called Sericulture.

→ Female silkworms lay hundreds of eggs.

→ The larvae that hatch out from eggs are called caterpillars.

→ Caterpillars grow in size and become pupae (adult).

→ Then pupa swings its head from side to side in the form of a figure of eight(8).

→ During this movement, caterpillar secret fibre is made up of a protein that hardens on exposure to air and becomes silk fibre.

→ Caterpillar also covers itself completely with these silk fibers and this covering is known as Cocoon.

→ The most common silk moth is a mulberry silk moth. Other types of silk are Tassar silk, Mooga silk, and Kosa silk, etc.

→ The process by which cocoons are boiled or given steam to extract silk fibres is called Reeling of silk.

→ Reeling is done with special machines.

→ Silk thread is used to weave silk fabric.

→ Tuft: A cluster of sheep’s body hair.

→ Reeling: The process of extracting silk fibres by boiling or steaming silk cocoons.

→ Scouring: The sheared skin with hair is washed thoroughly to remove grease, dust, and sweat, which is called scouring.

→ Cocoon: A layer of silk fibres that covers the caterpillar is called a cocoon.

→ Sericulture (rearing silkworms): Rearing of silkworms to get silk, is called Sericulture.

→ Shearing or shaving: The process of removing a thin layer of sheep’s hair and skin from a sheep’s body is called shearing.

→ Throwing: Raw silk is spun to make raw silk stronger (thick). This process is called throwing. This protects the fibres from breaking down.

→ Combing: The process of removing small swollen fibres called burrs is called combing.

→ Dyeing or colouring: Sheep wool can be dyed in different colours as the natural color is only black, brown or white.

→ Selective Breeding: Two sheep with special required traits are selected as parents for breeding and reproducing a sheep of required traits. This process is said selective reproduction.