Class 10 Science Model Paper 2026
Max Marks: 80 | Time: 3 Hours
Section A: Biology
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Reason (R): Variations enable a species to survive in drastically changing environmental conditions.
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Reason (R): DNA copying leads to the transmission of characters from parents to offspring.
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Reason (R): The energy captured by the autotrophs does not revert back to the solar input.
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- Gas Exchange: Massive amounts of gaseous exchange ($O_2$ and $CO_2$) takes place in leaves through these pores for photosynthesis and respiration.
- Transpiration: Large amount of water is lost in the form of vapour through stomata, which helps in cooling the plant and pulling water up.
A. How do Mendel's experiments show that traits may be dominant or recessive?
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Answer A: When Mendel crossed pure Tall (TT) and pure Short (tt) plants, all F1 progeny were Tall (Tt). This showed that the trait for Tallness suppressed the trait for Shortness. The trait that appeared (Tall) is Dominant, and the hidden one (Short) is Recessive.
Answer B: Organs that have the same basic structure and origin but perform different functions. Example: Forelimb of a frog and Forelimb of a human.
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Definition: The phenomenon of progressive increase in the concentration of non-biodegradable chemicals (like DDT/Pesticides) at each successive trophic level.
Levels: Yes, the concentration increases as we move up the trophic levels. Top carnivores (like humans or eagles) have the maximum concentration.
(ii) Why do desert plants take up Carbon Dioxide at night?
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(i) Three Events:
1. Absorption of light energy by chlorophyll.
2. Conversion of light energy to chemical energy and splitting of water molecules into hydrogen and oxygen.
3. Reduction of carbon dioxide to carbohydrates.
(ii) Desert Plants: To prevent water loss, their stomata are closed during the day. They take up $CO_2$ at night to prepare an intermediate, which is acted upon by energy absorbed during the day.
(i) Direction of blood flow.
(ii) Type of blood carried (except pulmonary).
(iii) Thickness of walls.
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| Feature | Arteries | Veins |
|---|---|---|
| Direction | Carry blood away from heart to body. | Carry blood towards the heart. |
| Blood Type | Oxygenated (Rich in $O_2$). | Deoxygenated (Rich in $CO_2$). |
| Walls | Thick and elastic (high pressure). | Thin, have valves to prevent backflow. |
Case Study: Mendel crossed a pure tall pea plant ($TT$) with a pure short pea plant ($tt$). He obtained only tall plants in the $F_1$ generation. When $F_1$ plants were self-pollinated, he obtained both tall and short plants in the $F_2$ generation.
Attempt the following:
A. Why did only tall plants appear in the $F_1$ generation?
B. What is the ratio of Tall to Short plants in the $F_2$ generation?
Attempt either subpart C or D.
C. Calculate the percentage of short plants in the $F_2$ generation.
D. If the $F_1$ plants ($Tt$) were crossed with a short plant ($tt$), what percentage of the offspring would be tall?
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Answer A: Because 'Tallness' is a dominant trait. In $Tt$ (F1), the dominant 'T' allele expresses itself, masking the recessive 't'.
Answer B: The Phenotypic ratio is 3:1 (3 Tall : 1 Short).
Answer C:
Short plants = 1 out of 4.
Percentage = $(1/4) \times 100 = 25\%$.
Answer D (OR):
Cross: $Tt \times tt$.
Offspring: $Tt$ (Tall), $tt$ (Short).
Ratio is 1:1.
Percentage of Tall = $50\%$.
A. (i) What is Vegetative Propagation? List two advantages.
(ii) Explain how Spore Formation takes place in Rhizopus with the help of a diagram.
(ii) Explain the process of seed formation after fertilization.
(iii) What is the function of sepals and petals?
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Answer A:
(i) Vegetative Propagation: New plants are obtained from parts of old plants (stem, root, leaf) without seeds.
Advantages: 1. Plants bear flowers/fruits earlier. 2. Genetically identical to parent (preserves good traits).
(ii) Rhizopus: It has thread-like structures called hyphae. Blob-on-a-stick structures are sporangia, which contain spores. When sporangia burst, spores are released and grow into new Rhizopus on moist surfaces.
Answer B:
(ii) After fertilization, the zygote divides to form an embryo within the ovule. The ovule develops a tough coat and converts into a Seed. The ovary ripens to form a Fruit.
(iii) Sepals: Protect flower in bud stage. Petals: Attract insects for pollination.
Section B: Chemistry
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Reason (R): Zinc is more reactive than hydrogen and displaces it from dilute acids.
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Observation: Brisk effervescence is observed due to the evolution of Carbon Dioxide gas ($CO_2$).
Equation:
$2CH_3COOH + Na_2CO_3 \rightarrow 2CH_3COONa + H_2O + CO_2$.
A. Give reasons for the following:
(i) Ionic compounds conduct electricity in molten state.
(ii) Metals are good conductors of heat.
(iii) Silver articles become black after some time when exposed to air.
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Answer A:
(i) In molten state, the rigid crystal structure breaks down, and ions become free to move and conduct electricity.
(ii) Metals have free electrons which carry heat energy from the hot end to the cold end.
(iii) Silver reacts with Sulphur in air to form black Silver Sulphide ($Ag_2S$).
Answer B:
Mg (2,8,2) loses 2 electrons to form $Mg^{2+}$.
Two Cl atoms (2,8,7) each gain 1 electron to form $2Cl^-$.
Result: $[Mg]^{2+} [Cl]^-_{2}$. Strong electrostatic attraction holds them.
(ii) List two reasons for the existence of a large number of carbon compounds.
(iii) Select saturated hydrocarbons from: $C_2H_6, C_2H_4, C_3H_4, C_5H_{12}$.
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(i) Carbon has 4 valence electrons. It cannot gain 4 electrons ($C^{4-}$) due to repulsion, nor lose 4 electrons ($C^{4+}$) due to high energy requirement. So, it shares electrons.
(ii) Reasons:
1. Catenation: Self-linking property to form long chains.
2. Tetravalency: Ability to bond with 4 other atoms.
(iii) Saturated (Alkanes $C_nH_{2n+2}$):
$C_2H_6$ (Ethane) and $C_5H_{12}$ (Pentane).
Case Study: Corrosion is the slow degradation of metals due to the action of atmospheric gases and moisture. Rusting of iron is the most common example of corrosion, which causes huge economic loss every year.
Attempt the following:
A. What are the two necessary conditions for rusting of iron?
B. Write the chemical formula of Rust.
Attempt either subpart C or D.
C. Explain how Galvanization protects iron from rusting.
D. Why do aluminium utensils not corrode easily even though aluminium is more reactive than iron?
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Answer A: 1. Presence of Air (Oxygen). 2. Presence of Moisture (Water).
Answer B: Hydrated Iron (III) Oxide: $Fe_2O_3 \cdot xH_2O$.
Answer C:
Galvanization involves coating iron with a layer of Zinc. Since Zinc is more reactive than Iron, it reacts with oxygen first (sacrificial protection), preventing iron from contacting air/moisture.
Answer D (OR):
Aluminium reacts with oxygen to form a thin, protective layer of Aluminium Oxide ($Al_2O_3$). This layer is stable and prevents further reaction of the metal underneath.
A. (i) Explain the Chlor-Alkali process with a chemical equation.
(ii) Name the three products formed and give one use of each.
(ii) Write the chemical name and formula of Bleaching Powder.
(iii) How is Bleaching Powder prepared?
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Answer A:
(i) When electricity is passed through brine (aq. NaCl), it decomposes to form Sodium Hydroxide.
$2NaCl(aq) + 2H_2O(l) \rightarrow 2NaOH(aq) + Cl_2(g) + H_2(g)$.
(ii) Products:
1. Chlorine: Water treatment.
2. Hydrogen: Rocket fuel.
3. NaOH: Making soaps/paper.
Answer B:
(i) Fixed number of water molecules chemically attached to a salt unit. Ex: Blue Vitriol ($CuSO_4 \cdot 5H_2O$), Washing Soda ($Na_2CO_3 \cdot 10H_2O$).
(ii) Calcium Oxychloride ($CaOCl_2$).
(iii) By the action of chlorine gas on dry slaked lime: $Ca(OH)_2 + Cl_2 \rightarrow CaOCl_2 + H_2O$.
Section C: Physics
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Reason (R): The speed of light increases as it enters a rarer medium.
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Planets are much closer to the Earth and are seen as extended sources of light (collection of many point sources). The total variation in the amount of light entering our eye from all the individual point sources averages out to zero, thereby nullifying the twinkling effect.
A. Two lamps rated 100 W, 220 V and 10 W, 220 V are connected in parallel to 220 V supply. Calculate the total current drawn from the line.
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Answer A:
Current $I_1 = P_1/V = 100/220 = 0.45 A$.
Current $I_2 = P_2/V = 10/220 = 0.045 A$.
Total Current = $0.45 + 0.045 \approx 0.5 A$.
Answer B:
Electric Power determines the rate of energy delivery.
SI Unit: Watt (W).
(ii) An electric bulb is rated 220V and 100W. When it is operated on 110V, the power consumed will be:
(a) 100W (b) 75W (c) 50W (d) 25W
(Show calculation).
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(i) Definition: The rate at which electric energy is consumed in an electric circuit. Commercial Unit: Kilowatt-hour (kWh) or 'Unit'.
(ii) Calculation:
First, find Resistance ($R$) of the bulb.
$P = V^2/R \Rightarrow R = V^2/P = (220)^2 / 100 = 484 \Omega$.
Now, Voltage ($V'$) = 110V.
New Power $P' = (V')^2 / R = (110)^2 / 484 = 12100 / 484 = 25 W$.
Answer: (d) 25W.
(ii) A positively charged particle (alpha particle) projected towards the west is deflected towards the north by a magnetic field. What is the direction of the magnetic field?
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(i) Fleming's Left-Hand Rule: Stretch the thumb, forefinger, and middle finger of the left hand mutually perpendicular. If the Forefinger points to Magnetic Field, and Middle finger points to Current, then Thumb points to the direction of Force (Motion).
(ii) Direction:
- Current (Alpha particle movement) = West.
- Force (Deflection) = North.
- Using Left-Hand Rule, the Magnetic Field points Upward.
(ii) A student uses a lens of focal length +20 cm. What is the power of the lens? What is its nature?
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(i) Differences:
- Real images can be obtained on a screen; Virtual images cannot.
- Real images are inverted; Virtual images are erect.
(ii) Power:
$f = +20 cm = +0.2 m$.
$P = 1/f(m) = 1/0.2 = +5 Diopter$.
Nature: Since power is positive, it is a Convex Lens (Converging).
Case Study: Solenoid
A coil of many circular turns of insulated copper wire wrapped closely in the shape of a cylinder is called a solenoid. The pattern of magnetic field lines around a current-carrying solenoid is similar to that of a bar magnet. This property is used to make electromagnets.
Attempt the following:
A. What is the difference between a Solenoid and a Bar Magnet regarding their magnetic strength?
B. Draw the magnetic field lines inside a solenoid.
Attempt either subpart C or D.
C. Explain how a soft iron core inside a solenoid acts as an electromagnet.
D. State two factors on which the strength of the magnetic field inside a solenoid depends.
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Answer A: A solenoid is a temporary magnet (strength can be changed, polarity can be reversed), whereas a bar magnet is a permanent magnet (fixed strength and polarity).
Answer B: (Draw parallel straight lines inside the coil).
Answer C:
Soft iron has high magnetic permeability. When current flows, the magnetic field of the solenoid induces strong magnetism in the soft iron core, making it a powerful magnet (Electromagnet).
Answer D (OR):
1. Number of turns per unit length.
2. Magnitude of current flowing.
(ii) State the laws of refraction.
(iii) An object 5 cm high is held 25 cm away from a converging lens of focal length 10 cm. Find the position, size and nature of the image formed.
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(ii) Laws of Refraction:
1. Incident ray, refracted ray, and normal all lie in the same plane.
2. Ratio of sine of angle of incidence to sine of angle of refraction is constant (Snell's Law: $\frac{\sin i}{\sin r} = n$).
(iii) Numerical:
Convex Lens ($f = +10, u = -25$).
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f} \Rightarrow \frac{1}{v} = \frac{1}{10} + \frac{1}{-25}$
$\frac{1}{v} = \frac{5-2}{50} = \frac{3}{50} \Rightarrow v = 16.67 cm$.
$m = v/u = 16.67 / -25 = -0.66$.
$h_i = m \times h_o = -0.66 \times 5 = -3.33 cm$.
Nature: Real, Inverted, Diminished.