Download CBSE Class 10 Science Assignments
Access comprehensive school assignments for Practical Worksheets using the CBSE Class 10 Science Physics Light Eye Electricity Magnetic Effects and Environment Assignment with Answers Set 03. Designed to align with the 2026-27 CBSE academic guidelines, these practice sets help Class 10 Science students reinforce core concepts and improve their problem-solving accuracy.
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Access the complete assignment PDF for Class 10 Science below. Regular practice with these targeted academic tasks builds familiarity with standard question patterns and helps secure higher marks in final school evaluations.
Light: Reflection and Refraction
1. Explain the laws of reflection and refraction.
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Answer: Laws of reflection: (1) The angle of incidence equals the angle of reflection. (2) The incident ray, the reflected ray and the normal at the point of incidence all lie in the same plane.
Laws of refraction: (1) The incident ray, the refracted ray and the normal lie in the same plane. (2) Snell's law: for light passing from one medium to another, sin i / sin r = constant, called the refractive index of the second medium with respect to the first.
Teacher's Note: Use the key term in your answer.
2. Define refractive index, centre of curvature, focal length, optical centre, pole, principal axis and aperture.
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Answer: Refractive index: the ratio of the speed of light in vacuum to its speed in the medium, n = c/v.
Centre of curvature: the centre of the sphere of which the mirror or lens surface is a part.
Focal length: the distance between the pole (or optical centre) and the principal focus.
Optical centre: the central point of a lens, through which a ray passes without deviation.
Pole: the centre of the reflecting surface of a spherical mirror.
Principal axis: the straight line through the pole and the centre of curvature of a mirror (or through the optical centre of a lens and its centres of curvature).
Aperture: the diameter of the reflecting surface of a spherical mirror.
Teacher's Note: Give an example if the question allows.
3. (a) Draw ray diagrams for a concave mirror: object beyond C, between P and F, and at infinity.
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Answer: Beyond C: the image is formed between F and C; it is real, inverted and smaller than the object.
Between P and F: the image is behind the mirror; it is virtual, erect and enlarged.
At infinity: the image is formed at F; it is real, inverted and very small, like a point.
Teacher's Note:
3. (b) Draw ray diagrams for a convex lens: object beyond 2F₁, between F₁ and 2F₁, and at F₁.
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Answer: Beyond 2F₁: the image is between F₂ and 2F₂; real, inverted and diminished.
Between F₁ and 2F₁: the image is beyond 2F₂; real, inverted and enlarged.
At F₁: the refracted rays are parallel, so the image is at infinity; it is real, inverted and highly enlarged.
Teacher's Note:
4. At what distance from a concave lens of focal length 20 cm should a 6 cm tall object be placed so that it forms an image at 15 cm from the lens? Find the size of the image.
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Answer: f = −20 cm, v = −15 cm. From 1/v − 1/u = 1/f:
1/u = 1/v − 1/f = −1/15 + 1/20 = −1/60, so u = −60 cm.
Magnification m = v/u = (−15)/(−60) = 0.25. Height of image = 0.25 × 6 = 1.5 cm.
The object is 60 cm from the lens, and the image is virtual, erect and 1.5 cm tall.
Teacher's Note: Learn this point and write it in your own words.
5. A 4 cm object is placed 15 cm in front of a concave mirror of focal length 10 cm. Where should a screen be placed? Find the height of the image.
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Answer: u = −15 cm, f = −10 cm, h = 4 cm. From 1/v + 1/u = 1/f: 1/v = −1/10 + 1/15 = −1/30, so v = −30 cm.
m = −v/u = −(−30)/(−15) = −2. Height of image = −2 × 4 = −8 cm.
The screen should be placed 30 cm in front of the mirror. The image is real, inverted and 8 cm tall.
Teacher's Note: Use the key term in your answer.
The Human Eye and the Colourful World
1. Explain the three defects of vision with causes and correction.
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Answer:
| Defect | What happens | Cause | Correction |
|---|---|---|---|
| Myopia (near-sightedness) | Distant objects look blurred | Eyeball too long, or lens too curved: image forms in front of the retina | Concave lens |
| Hypermetropia (far-sightedness) | Near objects look blurred | Eyeball too short, or lens too flat: image forms behind the retina | Convex lens |
| Presbyopia | Cannot see near or far objects clearly | Weak ciliary muscles and hard lens with age | Bifocal lens (concave and convex) |
Teacher's Note: Give an example if the question allows.
2. Define astigmatism, presbyopia, dispersion, scattering, accommodation and atmospheric refraction.
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Answer: Astigmatism: a defect in which the cornea is not perfectly spherical, so lines in one direction look blurred; corrected by a cylindrical lens.
Presbyopia: loss of the eye's power of accommodation with age.
Dispersion: splitting of white light into its seven colours when it passes through a prism.
Scattering: the spreading of light in all directions by small particles in its path.
Accommodation: the ability of the eye lens to change its focal length to see near and far objects.
Atmospheric refraction: the bending of light by the layers of air of different density in the atmosphere.
Teacher's Note: Write a short, clear answer.
3. Give reasons.
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Answer: (a) Twinkling of stars: starlight is refracted by air layers of changing density, so the apparent position of the star keeps changing and its brightness flickers.
(b) Rainbow: sunlight is dispersed by raindrops, which act as small prisms; it is refracted, internally reflected and refracted again, and we see the colours spread in a band.
(c) Blue colour of sky: air molecules scatter short wavelengths (blue) much more than long wavelengths (red), so the blue light reaches our eyes from all directions.
(d) Early sunrise and delayed sunset: because of atmospheric refraction, we can see the Sun about 2 minutes before it actually rises and 2 minutes after it sets.
Teacher's Note: Remember this for your exam.
4. The near point of a hypermetropic eye is 60 cm. Find the nature and power of the lens needed to read a book at 25 cm.
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Answer: u = −25 cm, v = −60 cm. 1/f = 1/v − 1/u = −1/60 + 1/25 = 7/300, so f = 300/7 ≈ 42.9 cm = 0.43 m.
Power P = 1/f = +2.33 D. A convex lens of power +2.33 D is needed.
Teacher's Note: Learn this point and write it in your own words.
5. Give the functions of the iris, cornea, pupil and retina.
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Answer: Cornea: the transparent front layer, where most of the refraction of light takes place.
Iris: the coloured muscle that controls the size of the pupil.
Pupil: the opening through which light enters the eye; it controls the amount of light.
Retina: the screen at the back of the eye where the image forms; its rods and cones change the light into electrical signals sent to the brain.
Teacher's Note: Use the key term in your answer.
Electricity
1. Define resistance, resistivity and Ohm's law.
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Answer: Resistance: the property of a conductor that opposes the flow of current. R = V/I; its SI unit is the ohm (Ω).
Resistivity: the resistance of a conductor of unit length and unit area of cross-section; ρ = RA/l, with the unit Ω m.
Ohm's law: at constant temperature the current through a conductor is directly proportional to the potential difference across it, V = IR.
Teacher's Note: Give an example if the question allows.
2. What are the advantages of connecting appliances in parallel?
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Answer: Each appliance gets the full voltage. Each works independently, so if one fails, the others continue to work. Each can be switched on and off separately. The total resistance is lower, so appliances of different current rating can be used.
Teacher's Note: Write a short, clear answer.
3. Explain: (a) Why is tungsten used for lamp filaments? (b) Why are alloys used in heating devices?
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Answer: (a) Tungsten has a very high melting point (about 3380 °C) and high resistivity, so it glows white-hot without melting.
(b) Alloys have higher resistivity than pure metals and do not oxidise (burn) easily at high temperature, so they last long.
Teacher's Note: Remember this for your exam.
4. A resistance wire of 4 Ω is doubled on itself. Find the new resistance.
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Answer: When the wire is folded in two, its length becomes half and the area of cross-section becomes double. R = ρl/A, so the new resistance R′ = ρ(l/2)/(2A) = R/4 = 1 Ω.
Teacher's Note: Learn this point and write it in your own words.
5. Write Joule's law of heating.
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Answer: The heat produced in a resistor is directly proportional to the square of the current, to the resistance and to the time for which the current flows: H = I²Rt.
Teacher's Note: Use the key term in your answer.
6. List the factors on which the resistance and the resistivity of a conductor depend.
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Answer: Resistance depends on the length of the conductor (directly), the area of cross-section (inversely), the material and the temperature. Resistivity depends only on the nature of the material and the temperature, not on the shape or size.
Teacher's Note: Give an example if the question allows.
Magnetic Effects of Electric Current
1. State Fleming's right-hand rule, Fleming's left-hand rule and Maxwell's right-hand rule.
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Answer: Fleming's left-hand rule: stretch the thumb, forefinger and middle finger of the left hand at right angles to each other. If the forefinger points along the magnetic field and the middle finger along the current, the thumb shows the direction of force (motion) on the conductor.
Fleming's right-hand rule: with the right hand, if the forefinger points along the magnetic field and the thumb along the motion of the conductor, the middle finger shows the direction of the induced current.
Maxwell's right-hand thumb rule: imagine holding the conductor in the right hand with the thumb along the current; the curled fingers show the direction of the magnetic field lines.
Teacher's Note: Write a short, clear answer.
2. Give the principle of an electric motor and a generator.
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Answer: Electric motor: a current-carrying conductor placed in a magnetic field experiences a force; it changes electrical energy into mechanical energy. Generator: when a conductor moves in a magnetic field, a current is induced (electromagnetic induction); it changes mechanical energy into electrical energy.
Teacher's Note: Remember this for your exam.
3. Differentiate between AC and DC, and between a bar magnet and a temporary magnet.
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Answer:
| AC | DC |
|---|---|
| Direction reverses periodically (50 Hz in India) | Flows in one direction only |
| Can be sent over long distances with less loss | Cannot be sent far economically |
| From generators and mains | From cells and batteries |
| Bar (permanent) magnet | Temporary magnet (electromagnet) |
|---|---|
| Keeps its magnetism for a long time | Is a magnet only while current flows |
| Fixed strength | Strength can be changed by changing the current or the turns |
| Poles are fixed | Poles can be reversed by reversing the current |
Teacher's Note: Learn this point and write it in your own words.
4. Define solenoid, electromagnet, commutator, slip rings, overloading and short-circuiting.
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Answer: Solenoid: a coil of many circular turns of insulated copper wire wound closely in the shape of a cylinder.
Electromagnet: a soft iron core placed inside a solenoid, which becomes a strong magnet when current flows.
Commutator: a split ring that reverses the direction of current in the coil of a DC motor every half turn.
Slip rings: two complete rings used in an AC generator to connect the coil with the brushes without reversing the current.
Overloading: too many appliances connected to one circuit, so that the current exceeds its safe limit.
Short-circuiting: the live and neutral wires touch each other, so that resistance becomes very small and the current very large.
Teacher's Note: Use the key term in your answer.
5. Explain the functions of (a) the armature, (b) the brushes, (c) the split ring of an electric motor.
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Answer: (a) Armature: the coil of insulated wire wound on a soft iron core, which rotates in the magnetic field.
(b) Brushes: carbon strips that touch the split ring and pass the current from the battery to the coil.
(c) Split ring (commutator): reverses the direction of current in the coil after every half rotation, so that the coil keeps rotating in the same direction.
Teacher's Note: Give an example if the question allows.
Our Environment
1. Define food chain, food web and biological magnification.
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Answer: Food chain: a chain of organisms in which each eats the one below it, and the energy passes along, e.g. grass → deer → lion.
Food web: many food chains linked together in an ecosystem.
Biological magnification: the increase in the concentration of harmful chemicals, such as pesticides, at each higher level of a food chain.
Teacher's Note: Write a short, clear answer.
2. In the food chain Plant → Deer → Lion, 100 J of energy is available to the lion. How much energy was available to the producer?
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Answer: Only 10% of energy passes to the next trophic level. The deer had 100 × 10 = 1000 J, and the plant had 1000 × 10 = 10,000 J. So 10,000 J was available to the producer.
Teacher's Note: Remember this for your exam.
3. What is ozone? What are the causes and effects of ozone depletion? How is ozone formed?
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Answer: Ozone (O₃) is a molecule of three oxygen atoms. It forms in the upper atmosphere when UV rays split oxygen molecules (O₂) into atoms, which combine with other oxygen molecules: O₂ → O + O; O + O₂ → O₃.
Cause of depletion: chlorofluorocarbons (CFCs) used in refrigerators, aerosols and foams.
Effects: more UV rays reach the earth, which can cause skin cancer, cataracts and damage to crops and aquatic life.
Teacher's Note: Learn this point and write it in your own words.
4. Natural water bodies are not cleaned regularly, but an aquarium needs regular cleaning. Why?
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Answer: A natural water body is a self-sustaining ecosystem with producers, consumers and decomposers in balance, and the wastes are recycled. An aquarium is not a complete ecosystem, with few organisms and no proper decomposers, so wastes build up and it needs regular cleaning.
Teacher's Note: Use the key term in your answer.
5. What are decomposers? What would happen if they were completely removed?
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Answer: Decomposers, such as bacteria and fungi, break down dead plants and animals and their waste into simple substances that return to the soil. Without them dead matter would pile up, nutrients would not be recycled to the producers and life would end in the ecosystem.
Teacher's Note: Give an example if the question allows.
Free study material for Science
Download Practice Assignments: Class 10 Science Practical Worksheets
Class 10 Science Practical Worksheets Printable Assignments
Access structured practice assignments for Practical Worksheets designed in alignment with the latest CBSE curriculum for Class 10 Science. These printable sets cover objective and descriptive problem types to support thorough revision.
Why Practice Class 10 Science Assignments?
- Syllabus Compliance: Sets reflect current CBSE evaluation criteria and official marking frameworks.
- Multi-Format Practice: Includes varied problem types designed to deepen comprehension across all sub-topics.
- Time Management: Routine practice optimizes pacing to finish school examinations comfortably within schedule.
Effective Strategy for Class 10 Science Assignments
- Textbook Review: Always study the core NCERT book for Class 10 Science prior to beginning the assignment.
- Independent Attempt: Solve Practical Worksheets questions on your own initially before cross-checking with expert solutions.
- Error Tracking: Record challenging concepts in a dedicated notebook and practice online MCQ tests for revision.
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You can download free PDF assignments for Class 10 Science Practical Worksheets from StudiesToday.com. These practice sheets have been updated for the 2026-27 session covering all concepts from latest NCERT textbook.
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