MG SCHOOL SCIENCE

Wednesday, 11 November 2026

Ydgh

 

MG SCHOOL SCIENCE

CLASS 7 SCIENCE

MEASUREMENT OF TIME AND MOTION

Complete Chapter Notes | Definitions | Formulas | Solved Examples

Learn • Understand • Explore • Achieve

📘 CLASS 7
🔬 SCIENCE
⭐ CHAPTER NOTES

📑 CHAPTER CONTENTS

1. Introduction

Time and motion are important concepts in our everyday life. We use clocks to measure time and different instruments to measure the speed and distance travelled by objects.

From a running athlete to a moving train, everything around us can be studied through the concepts of time, distance and motion.

Learning Objectives
  • Understand different methods of measuring time.
  • Learn the working of a simple pendulum.
  • Understand speed and its mathematical formula.
  • Differentiate uniform and non-uniform motion.
  • Solve numerical problems based on speed.

2. Measurement of Time

Since ancient times, people have observed natural events to understand the passage of time.

Ancient Methods of Measuring Time

1. Sundial

A sundial uses the shadow of an object produced by sunlight to indicate the time of day.

2. Water Clock

A water clock measures time through the controlled flow of water from one container to another.

3. Hourglass

An hourglass measures time using sand flowing through a narrow opening from the upper chamber to the lower chamber.

4. Candle Clock

A candle clock measures the passage of time by observing how much of a marked candle burns.

Interesting Fact: The Samrat Yantra at Jantar Mantar, Jaipur, is a giant sundial used for astronomical observations and time measurement.

Development of Clocks

  • Ancient people used natural cycles and simple devices.
  • Mechanical clocks developed using gears and weights.
  • Pendulum clocks improved timekeeping accuracy.
  • Modern clocks use quartz crystals and atomic transitions.

3. The Simple Pendulum

Simple Pendulum: A simple pendulum consists of a small heavy object called a bob suspended from a fixed support by a light string.
Diagram of a Simple Pendulum
O A B String Mean position Extreme Extreme

A – Left extreme | O – Mean position | B – Right extreme

Important Terms

  • Bob: The small heavy object suspended by a string.
  • Mean position: The central resting position of the bob.
  • Extreme positions: The farthest positions on either side.
  • Oscillation: One complete to-and-fro movement.
  • Time period: Time taken to complete one oscillation.
One Complete Oscillation: Movement from one extreme position to the other extreme position and back to the starting extreme position.

Experiment: Finding the Time Period

  1. Take a string and tie a small heavy bob to one end.
  2. Fix the other end to a rigid support.
  3. Displace the bob slightly and release it.
  4. Measure the time taken for 10 oscillations.
  5. Divide the total time by 10.
Time Period = Total Time / Number of Oscillations
The time period of a simple pendulum depends on its length. For small oscillations, it does not depend on the mass of the bob.

4. SI Units of Time

The SI unit of time is the second.

UnitSymbolConversion
SecondsStandard unit
Minutemin1 min = 60 s
Hourh1 h = 60 min
Dayd1 day = 24 h
1 hour = 60 × 60 = 3600 seconds

5. Slow or Fast and Speed

When two objects cover different distances in the same time, the object covering more distance is moving faster.

Speed: Speed is the distance travelled by an object per unit time.
Speed = Total Distance / Total Time

Units of Speed

  • Metre per second (m/s)
  • Kilometre per hour (km/h)

Measuring Instruments

  • Speedometer: Measures the speed of a vehicle.
  • Odometer: Measures the total distance travelled by a vehicle.
Vehicle Measuring Instruments

Speedometer → Speed

Odometer → Distance travelled

6. Relationship Between Speed, Distance and Time

Speed = Distance / Time
Distance = Speed × Time
Time = Distance / Speed

Speed Formula Triangle

Distance Speed Time

Cover the quantity you want to calculate to find its formula.

Solved Example 1

Question: A cyclist travels 5 km in 1 hour. Find the speed.

Given:

Distance = 5 km
Time = 1 hour

Formula:

Speed = Distance / Time

Calculation:

Speed = 5 / 1 = 5 km/h

Answer: 5 km/h.

Solved Example 2

Question: A car moves at 60 km/h for 3 hours. Find the distance.

Distance = Speed × Time

= 60 × 3

= 180 km

Answer: 180 km.

Solved Example 3

Question: A train travels 240 km at 80 km/h. Calculate the time.

Time = Distance / Speed

= 240 / 80

= 3 hours

Answer: 3 hours.

Unit Conversion

To convert km/h into m/s, multiply by 5/18.

To convert m/s into km/h, multiply by 18/5.
Example: Convert 72 km/h into m/s. 72 × 5/18 = 20 m/s. Answer: 20 m/s.

7. Uniform and Non-uniform Linear Motion

Linear Motion

Motion of an object along a straight line is called linear motion.

Uniform Linear Motion

An object is said to be in uniform linear motion when it travels equal distances in equal intervals of time along a straight line.

Uniform Motion
Time IntervalDistance
1st minute10 m
2nd minute10 m
3rd minute10 m

Non-uniform Linear Motion

An object is said to be in non-uniform linear motion when it travels unequal distances in equal intervals of time along a straight line.

Non-uniform Motion
Time IntervalDistance
1st minute10 m
2nd minute15 m
3rd minute8 m
Uniform MotionNon-uniform Motion
Equal distances in equal time intervals.Unequal distances in equal time intervals.
Constant speed.Changing speed.
Example: Ideal constant-speed motion on a straight track.Example: A vehicle changing speed on a road.

8. Important Definitions

1. Sundial: An instrument that uses the Sun's shadow to indicate time.
2. Water Clock: A device that measures time through the flow of water.
3. Hourglass: A device that measures time using flowing sand.
4. Pendulum: A suspended object that swings to and fro.
5. Oscillation: One complete to-and-fro movement.
6. Time Period: Time taken to complete one oscillation.
7. Speed: Distance travelled per unit time.
8. Average Speed: Total distance travelled divided by total time taken.
9. Linear Motion: Motion along a straight line.
10. Uniform Motion: Equal distances travelled in equal time intervals.
11. Non-uniform Motion: Unequal distances travelled in equal time intervals.
12. Speedometer: Instrument used to measure the speed of a vehicle.
13. Odometer: Instrument used to measure the distance travelled by a vehicle.

9. Important Questions and Answers

Q1. What is the SI unit of time?
The SI unit of time is the second (s).
Q2. What is a simple pendulum?
A simple pendulum consists of a small heavy bob suspended from a fixed support by a light string.
Q3. Define time period.
The time taken by a pendulum to complete one oscillation is called its time period.
Q4. What is speed?
Speed is the distance travelled by an object per unit time.
Q5. Write the formula for speed.
Speed = Total distance travelled / Total time taken.
Q6. What is the difference between a speedometer and an odometer?
A speedometer measures the speed of a vehicle, whereas an odometer measures the total distance travelled.
Q7. What is uniform linear motion?
Motion along a straight line in which equal distances are covered in equal intervals of time is called uniform linear motion.
Q8. What is non-uniform linear motion?
Motion along a straight line in which unequal distances are covered in equal intervals of time is called non-uniform linear motion.
Q9. What is average speed?
Average speed is the ratio of total distance travelled to the total time taken.
Q10. Name four ancient devices used to measure time.
Sundial, water clock, hourglass and candle clock.

10. Interactive MCQ Practice Test

Choose the correct answer and click Submit Test.

1. What is the SI unit of time?
2. Which instrument uses the shadow of the Sun?
3. What is the formula for speed?
4. Which instrument measures vehicle speed?
5. Equal distances in equal intervals of time indicate:
6. How many seconds are there in one hour?
7. What is the SI unit of speed?
8. Which device uses flowing sand?

11. Quick Revision

Remember These Important Points
  • Second is the SI unit of time.
  • Speed = Distance / Time.
  • Distance = Speed × Time.
  • Time = Distance / Speed.
  • Speedometer measures speed.
  • Odometer measures distance.
  • Uniform motion covers equal distances in equal time intervals.
  • Non-uniform motion covers unequal distances in equal time intervals.
  • One complete to-and-fro movement is one oscillation.

Monday, 9 November 2026

Unit Test | Our Environment

MG SCHOOL SCIENCE

CLASS 10 – SCIENCE

ADVANCED UNIT TEST – VERSION 4

OUR ENVIRONMENT

Maximum Marks: 25 Time: 50 Minutes Competency-Based Assessment
Student Name
Class / Section
Roll Number
Date

General Instructions

  1. All questions are compulsory.
  2. Choose the correct answer for each MCQ.
  3. Write answers in clear scientific language.
  4. Show all steps in numerical calculations.
  5. Draw neat diagrams wherever necessary.

SECTION A – Multiple Choice Questions 5 × 1 = 5

1. A farmer sprays a persistent pesticide on crops. The pesticide enters a food chain. Which organism is likely to accumulate the highest concentration?

2. If producers contain 50,000 J of energy, approximately how much energy reaches the secondary consumer?

3. Which of the following would most directly reduce the availability of nutrients for reuse in an ecosystem?

4. Which gas is formed when ultraviolet radiation acts on oxygen molecules in the upper atmosphere?

5. Which practice best follows the principle of sustainable waste management?

SECTION B – One Mark Questions 4 × 1 = 4

6. What are abiotic components of an ecosystem? Give one example.

7. What is the first trophic level in a grazing food chain?

8. Name the process by which green plants prepare food using sunlight.

9. Mention one harmful effect of excessive ultraviolet radiation on human health.

SECTION C – Two Marks Questions 3 × 2 = 6

10. Explain why decomposers are called the natural recyclers of an ecosystem.

11. Why can non-biodegradable substances remain in the environment for a long time? Explain one consequence.

12. State two differences between the flow of energy and the cycling of nutrients in an ecosystem.

SECTION D – Three Marks Questions 2 × 3 = 6

13. Study the following food chain:

Grass → Grasshopper → Lizard → Hawk

(a) Identify the secondary consumer. (1)

(b) Identify the top consumer. (1)

(c) If the grass contains 60,000 J of energy, calculate the approximate energy available to the lizard. (1)

14. Explain the formation, importance and depletion of the ozone layer in the stratosphere.

SECTION E – Four Marks Question 1 × 4 = 4

15. Case Study – Ecosystem and Pollution

A freshwater lake supports algae, insect larvae, small fish and large fish. A nearby factory releases untreated waste containing persistent toxic chemicals into the lake. These chemicals enter the food chain and are not easily broken down. Over time, the concentration of the chemicals becomes higher in organisms at successive trophic levels.

(a) Identify the producers in the lake ecosystem. (1)

(b) Name the phenomenon responsible for the increasing chemical concentration. (1)

(c) Which group is likely to face the greatest chemical concentration: insect larvae, small fish or large fish? (1)

(d) Suggest one practical measure to prevent this pollution. (1)

COMPLETE KEY ANSWERS – OUR ENVIRONMENT

SECTION A – MCQ Answers

  1. (d) Eagles feeding at the top trophic level.
  2. (b) 500 J.
  3. (b) Removal or destruction of decomposers.
  4. (a) Ozone.
  5. (c) Segregating waste and reusing or recycling suitable materials.

SECTION B – One Mark Answers

  1. Abiotic components are the non-living physical and chemical factors of an ecosystem, such as air, water, soil or sunlight.
  2. Producers, such as green plants.
  3. Photosynthesis.
  4. Excessive UV radiation can cause skin cancer or eye damage such as cataracts.

SECTION C – Two Marks Answers

  1. Decomposers such as bacteria and fungi break down dead plants, animals and organic wastes. They release simpler substances and return nutrients to the environment, where producers can reuse them.
  2. Many non-biodegradable substances are not easily broken down by microorganisms. They can accumulate in soil and water, harm organisms and may enter food chains.
  3. Energy flow: It is unidirectional and energy is progressively dissipated as heat. Nutrient cycling: Nutrients are reused and circulate between organisms and the physical environment.

SECTION D – Three Marks Answers

  1. (a) Secondary consumer: Lizard.
    (b) Top consumer: Hawk.
    (c) Energy calculation:
    Grass = 60,000 J
    Grasshopper = 6,000 J
    Lizard = 600 J
    Answer: 600 J.
  2. Formation: Ultraviolet radiation acts on oxygen molecules (O₂), splitting some into oxygen atoms. These atoms combine with O₂ to form ozone (O₃).
    Importance: The ozone layer absorbs much of the harmful ultraviolet radiation from the Sun.
    Depletion: CFCs and certain other ozone-depleting substances release reactive chlorine or bromine compounds in the stratosphere, which destroy ozone.

SECTION E – Four Marks Answer

  1. (a) Algae (including microscopic phytoplankton). (1 mark)
    (b) Biological magnification. (1 mark)
    (c) Large fish. (1 mark)
    (d) Treat industrial wastewater before discharge and monitor effluent quality. (1 mark)

TEACHER'S MARKING SCHEME

SectionEvaluationMarks
AEach correct MCQ5
BCorrect facts and definitions4
CTwo relevant scientific points6
DCorrect explanation and calculation6
EFour correct case-study answers4
TOTAL25

Important revision points:

  • Abiotic and biotic components together form an ecosystem.
  • Producers occupy the first trophic level in a grazing food chain.
  • Only about 10% of energy is transferred to the next trophic level.
  • Decomposers help recycle nutrients.
  • Persistent non-biodegradable chemicals may undergo biological magnification.
  • The ozone layer protects life by absorbing much of the Sun's harmful UV radiation.

Assessment note: MCQs are automatically scored out of 5. The remaining 20 marks require teacher evaluation.

Unit Test | Our Environment | Class 10 Science

MG SCHOOL SCIENCE

CLASS 10 – SCIENCE

ADVANCED UNIT TEST – VERSION 3

OUR ENVIRONMENT

Maximum Marks: 25 Time: 50 Minutes HOTS & Competency Based
Student Name
Class / Section
Roll Number
Date

General Instructions

  1. All questions are compulsory.
  2. Read the questions carefully before answering.
  3. Show all steps in numerical calculations.
  4. Use suitable scientific terms in your answers.
  5. Draw diagrams wherever necessary.

SECTION A – Multiple Choice Questions 5 × 1 = 5

1. In a food chain, if the population of snakes decreases considerably, which change may occur initially?

2. A producer contains 1,00,000 J of energy. According to the 10% law, how much energy reaches the tertiary consumer in the chain producer → primary consumer → secondary consumer → tertiary consumer?

3. Which statement about decomposers is correct?

4. The ozone layer is important because it absorbs much of the harmful:

5. Which action is the most suitable way to reduce non-biodegradable waste?

SECTION B – One Mark Questions 4 × 1 = 4

6. What is a food web?

7. Name any two decomposing microorganisms.

8. What is the main source of energy for most ecosystems?

9. What is meant by ozone layer depletion?

SECTION C – Two Marks Questions 3 × 2 = 6

10. Why is the flow of energy in an ecosystem unidirectional? Explain briefly.

11. Explain two harmful effects of excessive use of non-biodegradable pesticides.

12. Why are food webs generally more stable than simple food chains?

SECTION D – Three Marks Questions 2 × 3 = 6

13. Study the food chain below:

Grass → Deer → Tiger

(a) Identify the producer. (1)

(b) Identify the primary consumer. (1)

(c) If grass stores 80,000 J of energy, calculate the approximate energy available to the tiger using the 10% law. (1)

14. Explain how human activities can disturb the balance of an ecosystem. Give three examples.

SECTION E – Four Marks Question 1 × 4 = 4

15. Read the following case study and answer the questions:

Case Study – A Polluted Lake
A lake receives agricultural runoff containing a persistent pesticide. The pesticide enters the aquatic food chain through microscopic producers. It passes to zooplankton, small fish and large fish. The chemical is not easily broken down and its concentration increases at higher trophic levels.

(a) Identify the producers in this aquatic food chain. (1)

(b) Name the phenomenon described in the case study. (1)

(c) Which organism is expected to have the highest pesticide concentration: zooplankton, small fish or large fish? (1)

(d) Suggest one measure to prevent this type of pollution. (1)

COMPLETE ANSWER KEY – OUR ENVIRONMENT

SECTION A – MCQ Answers

  1. (a) Frog population may increase.
  2. (c) 100 J.
  3. (b) They recycle nutrients by breaking down dead organic matter.
  4. (b) Ultraviolet radiation.
  5. (c) Reusing and recycling suitable materials.

SECTION B – One Mark Answers

  1. A food web is a network of interconnected food chains in an ecosystem.
  2. Bacteria and fungi.
  3. The Sun.
  4. Ozone layer depletion is the thinning or reduction in the concentration of ozone in the stratosphere.

SECTION C – Two Marks Answers

  1. Energy enters most ecosystems from the Sun and is captured by producers. It passes to consumers through food chains. Much energy is used in life processes and dissipated as heat, so it does not return to the previous trophic level.
  2. Persistent pesticides can accumulate in organisms and become more concentrated at higher trophic levels through biological magnification. They can harm wildlife and may cause health problems in humans exposed through contaminated food.
  3. Food webs provide alternative food sources and feeding pathways. If one organism becomes scarce, some consumers may use other food sources, helping the ecosystem maintain balance.

SECTION D – Three Marks Answers

  1. (a) Producer: Grass.
    (b) Primary consumer: Deer.
    (c) Energy calculation:
    Grass = 80,000 J
    Deer = 10% of 80,000 = 8,000 J
    Tiger = 10% of 8,000 = 800 J.
  2. Three examples of human activities that disturb ecosystems:
    • Deforestation: Destroys habitats and reduces biodiversity.
    • Pollution: Contaminates air, water and soil and harms organisms.
    • Excessive use of pesticides: Can harm non-target organisms and cause biological magnification.
    Other relevant examples, such as overhunting and destruction of wetlands, may also be accepted.

SECTION E – Four Marks Answer

  1. (a) Microscopic producers such as phytoplankton or algae. (1 mark)
    (b) Biological magnification. (1 mark)
    (c) Large fish. (1 mark)
    (d) Reduce pesticide runoff by using integrated pest management, safer alternatives and responsible application of agricultural chemicals. (1 mark)

TEACHER'S MARKING SCHEME

SectionEvaluationMarks
AEach correct MCQ5
BCorrect scientific facts4
CTwo relevant points per answer6
DCorrect explanation and numerical calculation6
EFour correct case-study answers4
TOTAL25

Important concepts for revision:

  • Energy flow in an ecosystem is unidirectional.
  • Only approximately 10% of energy passes to the next trophic level.
  • Decomposers recycle nutrients in the ecosystem.
  • Biological magnification increases the concentration of certain persistent chemicals at higher trophic levels.
  • The ozone layer protects living organisms from much of the Sun's harmful ultraviolet radiation.

Assessment note: Only the 5 MCQs are automatically evaluated. The remaining 20 marks require teacher assessment.

Unit Test | Our Environment | Class 10 Science

MG SCHOOL SCIENCE

CLASS 10 – SCIENCE

ADVANCED UNIT TEST – VERSION 2

OUR ENVIRONMENT

Maximum Marks: 25 Time: 50 Minutes HOTS & Competency Based
Student Name
Class / Section
Roll Number
Date

General Instructions

  1. All 15 questions are compulsory.
  2. Write answers clearly and use scientific terms.
  3. Show calculations for numerical questions.
  4. Draw neat diagrams wherever required.
  5. Each section carries the marks indicated in its heading.

SECTION A – Multiple Choice Questions 5 × 1 = 5

1. A food chain contains grass, deer and tiger. If the deer population suddenly decreases, which change is most likely to occur initially?

2. A producer has 40,000 J of energy. According to the 10% law, how much energy is available to the secondary consumer in a simple food chain?

3. Which of the following is the most appropriate reason for the limited length of food chains?

4. Which environmental problem is directly associated with increased ultraviolet radiation reaching Earth's surface?

5. A pesticide enters a pond and passes through plankton, small fish and large fish. Which organism is likely to have the highest pesticide concentration?

SECTION B – One Mark Questions 4 × 1 = 4

6. What are producers in an ecosystem?

7. Name the organisms that feed on dead and decaying organic matter and break it down.

8. What is the approximate percentage of energy transferred from one trophic level to the next according to the 10% law?

9. Name one international agreement aimed at protecting the ozone layer.

SECTION C – Two Marks Questions 3 × 2 = 6

10. Explain why decomposers are essential for maintaining the balance of an ecosystem.

11. Why are plastics considered harmful to the environment? Give two reasons.

12. Distinguish between a food chain and a food web.

SECTION D – Three Marks Questions 2 × 3 = 6

13. Study the following food chain:

Grass → Rabbit → Snake → Eagle

(a) Identify the primary consumer. (1)

(b) Identify the tertiary consumer. (1)

(c) If grass stores 30,000 J of energy, calculate the energy available to the snake using the 10% law. (1)

14. Explain biological magnification. Describe how it can affect human beings through contaminated food.

SECTION E – Four Marks Question 1 × 4 = 4

15. Read the case study and answer the questions:

Case Study:
A lake ecosystem contains algae, zooplankton, small fish and large fish. Due to agricultural runoff, a non-biodegradable pesticide enters the lake. The pesticide accumulates in organisms and its concentration increases at successive trophic levels.

(a) Identify the producer in this ecosystem. (1)

(b) Name the process by which pesticide concentration increases at higher trophic levels. (1)

(c) Which organism is likely to have the highest pesticide concentration: zooplankton, small fish or large fish? (1)

(d) Suggest one effective measure to reduce this type of pollution. (1)

COMPLETE ANSWER KEY – OUR ENVIRONMENT

SECTION A – MCQ Answers

  1. (a) Grass population may increase.
  2. (b) 400 J. Producer = 40,000 J; primary consumer = 4,000 J; secondary consumer = 400 J.
  3. (b) Energy decreases at successive trophic levels.
  4. (a) Ozone layer depletion.
  5. (c) Large fish.

SECTION B – One Mark Answers

  1. Producers are organisms, such as green plants and algae, that prepare their own food, usually by photosynthesis.
  2. Decomposers, such as bacteria and fungi.
  3. Approximately 10%.
  4. Montreal Protocol (1987).

SECTION C – Two Marks Answers

  1. Decomposers break down dead plants, animals and organic wastes. They release simpler substances and return nutrients to the environment, making them available for reuse by producers.
  2. Plastics are generally non-biodegradable and remain in the environment for long periods. They can pollute soil and water, harm animals through ingestion or entanglement, and break into microplastics.
  3. Food chain: A linear sequence showing the transfer of food and energy from one organism to another. Food web: A network of interconnected food chains showing multiple feeding relationships in an ecosystem.

SECTION D – Three Marks Answers

  1. (a) Primary consumer: Rabbit.
    (b) Tertiary consumer: Eagle.
    (c) Energy calculation:
    Grass = 30,000 J
    Rabbit = 3,000 J
    Snake = 300 J
    Answer: 300 J.
  2. Biological magnification is the progressive increase in the concentration of certain non-biodegradable harmful chemicals at successive trophic levels in a food chain. Contaminated organisms are eaten by predators, causing the chemicals to accumulate. Humans may be exposed by eating contaminated fish, meat or other food, which can cause harmful health effects depending on the chemical and exposure.

SECTION E – Four Marks Answer

  1. (a) Algae.
    (b) Biological magnification.
    (c) Large fish, as it is at the highest trophic level among the listed organisms.
    (d) Reduce pesticide runoff by using safer pest-management methods, applying pesticides responsibly, and preventing agricultural chemicals from entering water bodies.

TEACHER'S MARKING SCHEME

SectionEvaluationMarks
AOne mark for each correct MCQ5
BCorrect definitions and facts4
CTwo relevant scientific points per answer6
DCorrect explanation and numerical calculation6
EFour correct case-study answers4
Total25

Important scientific concepts:

  • Energy flow through an ecosystem is unidirectional.
  • Approximately 10% of energy is transferred to the next trophic level.
  • Biological magnification affects certain persistent, non-biodegradable chemicals.
  • The Montreal Protocol is an international agreement to protect the ozone layer by controlling ozone-depleting substances.

Assessment note: The MCQ section is automatically evaluated out of 5 marks. The remaining 20 marks require teacher evaluation.

Unit Test | Our Environment | Class 10 Science

MG SCHOOL SCIENCE

CLASS 10 – SCIENCE

ADVANCED UNIT TEST

OUR ENVIRONMENT

Maximum Marks: 25 Time: 50 Minutes Application-Based Questions
Student Name
Class / Section
Roll Number
Date

General Instructions

  1. All questions are compulsory.
  2. Read each question carefully.
  3. Draw neat and labelled diagrams wherever necessary.
  4. Write scientific terms and examples correctly.
  5. Use the answer key only after completing the test.

SECTION A – Multiple Choice Questions 5 × 1 = 5

1. Which of the following represents the correct direction of energy flow in an ecosystem?

2. If producers in a food chain store 20,000 J of energy, approximately how much energy is transferred to primary consumers according to the 10% law?

3. Which organisms play the most important role in recycling nutrients by breaking down dead organic matter?

4. Which substance is mainly responsible for the depletion of the ozone layer?

5. The concentration of harmful non-biodegradable chemicals increases at successive trophic levels. This phenomenon is called:

SECTION B – One Mark Questions 4 × 1 = 4

6. What is an ecosystem?

7. Define trophic level.

8. Name the ozone-depleting chemicals commonly used in older refrigerators and aerosol products.

9. Give one example of a biodegradable substance.

SECTION C – Two Marks Questions 3 × 2 = 6

10. Differentiate between biodegradable and non-biodegradable substances. Give one example of each.

11. Why is the flow of energy in an ecosystem considered unidirectional?

12. Explain why the number of organisms generally decreases at higher trophic levels.

SECTION D – Three Marks Questions 2 × 3 = 6

13. Consider the food chain:

Grass → Grasshopper → Frog → Snake

(a) Identify the producer.

(b) Identify the secondary consumer.

(c) If producers contain 10,000 J of energy, calculate the approximate energy available to the frog using the 10% law.

14. What is biological magnification? Explain why top-level consumers are more severely affected by non-biodegradable pesticides.

SECTION E – Four Marks Question 1 × 4 = 4

15. Study the following food chain and answer the questions:

🌿 Grass → 🦗 Grasshopper → 🐸 Frog → 🐍 Snake → 🦅 Eagle

(a) Identify the producer and the top consumer. (1 mark)

(b) Write the trophic level of the frog. (1 mark)

(c) If the grass stores 50,000 J of energy, calculate the energy available to the frog using the 10% law. (1 mark)

(d) Why is less energy available at higher trophic levels? (1 mark)

ANSWER KEY – OUR ENVIRONMENT

SECTION A – MCQ Answers

  1. (b) Sun → Producers → Consumers
  2. (c) 2,000 J
  3. (c) Decomposers
  4. (b) Chlorofluorocarbons (CFCs)
  5. (b) Biological magnification

SECTION B – One Mark Answers

  1. An ecosystem is a functional unit of nature in which living organisms interact with one another and with their physical environment.
  2. A trophic level is each feeding position or step occupied by an organism in a food chain.
  3. Chlorofluorocarbons (CFCs).
  4. Vegetable peels, paper, leaves or food waste.

SECTION C – Two Marks Answers

  1. Biodegradable substances: Substances that can be broken down by microorganisms into simpler substances, e.g. vegetable peels. Non-biodegradable substances: Substances that are not easily broken down by natural biological processes, e.g. plastic or many synthetic pesticides.
  2. Energy flows from the Sun to producers and then to consumers. At each trophic level, much energy is used in life processes and lost as heat. It is not returned to the previous trophic level, so energy flow is unidirectional.
  3. Only a small fraction of energy is transferred to the next trophic level. Consequently, less energy is available to support organisms at higher levels, and food chains generally have fewer organisms at their top levels.

SECTION D – Three Marks Answers

  1. (a) Producer: Grass.
    (b) Secondary consumer: Frog.
    (c) Energy calculation:
    Grass = 10,000 J
    Grasshopper = 10% of 10,000 = 1,000 J
    Frog = 10% of 1,000 = 100 J.
  2. Biological magnification is the increase in concentration of harmful non-biodegradable chemicals at successive trophic levels in a food chain. Such chemicals are not easily broken down or removed from the body. They accumulate in organisms and become more concentrated as predators consume many contaminated prey. Therefore, top-level consumers often have the highest concentration.

SECTION E – Four Marks Answer

  1. (a) Producer: Grass. Top consumer: Eagle. (1 mark)
    (b) Frog is at the third trophic level (producer = first, grasshopper = second, frog = third). (1 mark)
    (c) Energy transfer:
    Grass = 50,000 J
    Grasshopper = 5,000 J
    Frog = 500 J. (1 mark)
    (d) Only about 10% of energy is transferred to the next trophic level; much of the energy is used in life processes and dissipated as heat. (1 mark)

Teacher's Marking Scheme

SectionEvaluationMarks
AEach correct MCQ5
BCorrect definitions and examples4
CTwo relevant scientific points per answer6
DCorrect explanation and calculations6
EFour correct sub-answers4
Total25

Important concept: According to the 10% law, approximately 10% of the energy available at one trophic level is transferred to the next trophic level. This is an ecological approximation, not an exact value in every ecosystem.

Teacher's note: The MCQ section is automatically evaluated out of 5 marks. The remaining 20 marks require manual assessment using the answer key.

Unit Test | Heredity | Class 10 Science

MG SCHOOL SCIENCE

CLASS 10 – SCIENCE

ADVANCED UNIT TEST

HEREDITY

Maximum Marks: 25 Time: 50 Minutes Higher Order Thinking
Student Name
Class / Section
Roll Number
Date

General Instructions

  1. All questions are compulsory.
  2. Read each question carefully before answering.
  3. Draw neat diagrams or genetic crosses wherever required.
  4. Write scientific terms correctly.
  5. Use the answer key only after completing the test.

SECTION A – Multiple Choice Questions 5 × 1 = 5

1. A pure tall pea plant (TT) is crossed with a pure dwarf plant (tt). What will be the phenotype of the F₁ generation?

2. Which molecule carries hereditary information in most living organisms?

3. In a human female, the sex chromosome combination is:

4. A trait that appears in the F₁ generation when two contrasting pure traits are crossed is generally called:

5. Which of the following is an acquired characteristic?

SECTION B – One Mark Questions 4 × 1 = 4

6. Define heredity.

7. What is an allele?

8. State the number of chromosomes in a normal human gamete.

9. Name the scientist known as the Father of Genetics.

SECTION C – Two Marks Questions 3 × 2 = 6

10. Differentiate between inherited and acquired traits. Give one example of each.

11. Why do offspring produced by sexual reproduction show variations?

12. Explain why the sex of a human child is not determined by the mother.

SECTION D – Three Marks Questions 2 × 3 = 6

13. Explain Mendel's experiment by crossing pure tall (TT) and pure dwarf (tt) pea plants. Give the F₁ result and explain why all plants are tall.

14. Explain the role of DNA copying in reproduction and how small changes in DNA can produce variations.

SECTION E – Four Marks Question 1 × 4 = 4

15. In pea plants, tallness (T) is dominant over dwarfness (t). Two heterozygous tall plants are crossed (Tt × Tt).

(a) Write the gametes produced by each parent.

(b) Complete the Punnett square.

(c) Write the genotypic ratio.

(d) Write the phenotypic ratio.

GametesTt
T
t

ANSWER KEY – HEREDITY

Section A – MCQ Answers

  1. (b) All tall
  2. (c) DNA
  3. (b) XX
  4. (c) Dominant
  5. (c) A scar developed after an injury

Section B – One Mark Answers

  1. Heredity is the transmission of characteristics from parents to offspring through genetic material.
  2. Alleles are alternative forms of the same gene found at the same position on homologous chromosomes.
  3. 23 chromosomes.
  4. Gregor Johann Mendel.

Section C – Two Marks Answers

  1. Inherited traits: Passed from parents to offspring through genes, e.g. natural eye colour. Acquired traits: Developed during an individual's lifetime, e.g. a scar or developed muscles.
  2. Sexual reproduction involves genetic material from two parents. During gamete formation and fertilisation, different combinations of genetic material are formed, producing variations among offspring.
  3. The mother has XX chromosomes and all her ova carry an X chromosome. The father has XY chromosomes and produces X-bearing and Y-bearing sperm. Therefore, the type of sperm that fertilises the ovum determines whether the child is XX or XY.

Section D – Three Marks Answers

  1. Mendel crossed a pure tall plant (TT) with a pure dwarf plant (tt).
    CrossTT
    tTtTt
    tTtTt
    All F₁ plants are tall (Tt). The allele T is dominant and expresses itself in the presence of the recessive allele t.
  2. DNA copying transfers hereditary information from parent cells to offspring. The copying process is highly accurate but may involve small changes. These changes can create variations in offspring. Some variations may help organisms survive better in changing environments.

Section E – Four Marks Answer

  1. Cross: Tt × Tt
    Gametes: T and t from each parent.
    GametesTt
    TTTTt
    tTttt
    Genotypic ratio: 1 TT : 2 Tt : 1 tt
    Phenotypic ratio: 3 Tall : 1 Dwarf.

Teacher's Evaluation Scheme

  • Section A: 1 mark for each correct option.
  • Section B: 1 mark for each correct definition or fact.
  • Section C: 2 marks for two relevant scientific points.
  • Section D: 3 marks for correct explanation and key concepts.
  • Section E: 1 mark each for correct gametes, Punnett square, genotypic ratio and phenotypic ratio.

Scientific note: In humans, the father’s sperm determines whether the fertilised egg has XX or XY sex chromosomes. The probability is approximately equal for either outcome.

Assessment note: The MCQ section is automatically evaluated out of 5. The other 20 marks require teacher evaluation.