Class 10 Science Model Paper 2026 (Based on New Pattern)
⚠️ Important Update for 2025-26 Batch: Based on the latest CBSE Sample Paper, the structure has changed significantly! The paper is now strictly divided by subject:
- Section A: Biology (Q1 - Q16)
- Section B: Chemistry (Q17 - Q29)
- Section C: Physics (Q30 - Q39)
This 80 Marks Model Paper is designed with high-probability competency-based questions. Practice this to score 95+ in your Boards!
Section A: Biology (Marks)
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Reason (R): Ventricles have to pump blood into various organs at high pressure.
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Secretion: Bile Juice (secreted by the liver).
Significance: Fats are present in the intestine in the form of large globules which makes it difficult for enzymes to act on them. Bile salts break them down into smaller globules (emulsification) increasing the efficiency of enzyme action (lipase).
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Definition: In humans, blood passes through the heart twice in one complete cycle of the body. This is called double circulation.
It involves two pathways:
- Pulmonary Circulation: Deoxygenated blood moves from Right Ventricle $\rightarrow$ Lungs $\rightarrow$ Left Atrium (Oxygenated).
- Systemic Circulation: Oxygenated blood moves from Left Ventricle $\rightarrow$ Body Parts $\rightarrow$ Right Atrium (Deoxygenated).
(Note for student: You must draw a box diagram showing Heart with 4 chambers, Lungs at top, and Body organs at bottom connected by vessels).
(i) Write the genotype of the $F_1$ progeny.
(ii) If the $F_1$ plants are self-pollinated, what will be the phenotypic ratio of the $F_2$ generation?
(iii) If a total of 160 seeds are collected in the $F_2$ generation, approximately how many are expected to be short with green seeds?
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(i) Genotype of F1: $TtYy$ (Heterozygous Tall and Heterozygous Yellow).
(ii) Phenotypic Ratio of F2: 9:3:3:1
(9 Tall Yellow : 3 Tall Green : 3 Short Yellow : 1 Short Green).
(iii) Calculation:
Short and Green seeds correspond to the recessive trait ($ttyy$).
Ratio share = 1 out of 16.
Number of seeds = $\frac{1}{16} \times 160 = 10$.
Answer: 10 seeds.
Case Study: During a marathon, an athlete experiences severe muscle cramps in his legs after running for 30 minutes. He knows that his breathing rate increased significantly during the run, yet his muscles ran out of oxygen.
Attempt either subpart A or B.
A. Identify the 3-carbon molecule formed in the muscle cells that causes the cramps. Write the chemical equation for this pathway.
B. The figure below represents the breakdown of glucose in different pathways. (Imagine a flowchart showing Glucose $\rightarrow$ Pyruvate $\rightarrow$ Three paths).
- Name the specific part of the cell where the first step (Glucose to Pyruvate) takes place.
- Compare the amount of energy released in aerobic respiration vs anaerobic respiration in yeast.
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Answer for A:
The 3-carbon molecule is Lactic Acid.
Equation: $Glucose \xrightarrow{Cytoplasm} Pyruvate \xrightarrow{Muscle (Lack of O_2)} Lactic Acid + Energy$.
Answer for B:
(i) Cytoplasm.
(ii) Aerobic respiration releases a large amount of energy (38 ATP), whereas Anaerobic respiration in yeast (fermentation) releases a very small amount of energy (2 ATP).
A. "The sex of a newborn child is a matter of chance and none of the parents are responsible for it."
(i) Justify this statement with the help of a flow chart showing determination of sex in humans.
(ii) Why is the ratio of male to female offspring in a population roughly 1:1?
- The part where fertilization takes place.
- The part where implantation of the embryo occurs.
(iii) What happens to the lining of the uterus if the egg is NOT fertilized?
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Answer for A:
(i) A male produces two types of sperms (X and Y) in equal numbers. A female produces only one type of egg (X).
- If X-sperm fertilizes egg (X) $\rightarrow$ XX (Girl)
- If Y-sperm fertilizes egg (X) $\rightarrow$ XY (Boy)
Since a male produces 50% X and 50% Y sperms, it is a matter of chance.
(ii) The probability of an X sperm or a Y sperm fusing with the egg is equal, hence the 1:1 ratio.
Answer for B:
(i) Labels: 1. Oviduct (Fallopian Tube), 2. Uterus.
(ii) Placenta: A disc-embedded tissue in the uterine wall. Function: It allows exchange of nutrients (glucose, oxygen) from mother to foetus and removes waste from foetus to mother.
(iii) The thick and spongy lining of the uterus breaks down and comes out through the vagina as blood and mucus. This is called Menstruation.
Section B: Chemistry
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Reason (R): A large amount of energy is required to break the strong inter-ionic attraction.
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A. State the colour change observed during heating.
B. Name the gases evolved and the type of chemical reaction.
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A. Colour Change: The green colour of Ferrous Sulphate crystals ($FeSO_4 \cdot 7H_2O$) changes to white (loss of water of crystallization) and then to reddish-brown (formation of Ferric Oxide, $Fe_2O_3$).
B. Gases & Type:
Gases: Sulphur Dioxide ($SO_2$) and Sulphur Trioxide ($SO_3$) (Smell of burning sulphur).
Type: Thermal Decomposition Reaction.
A. A metal 'X' is found in the form of its sulphide ore. It is low in the activity series.
(i) Name the process used to extract this metal.
(ii) Write the chemical equations involved.
(iii) What happens when 'X' corrodes?
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Answer for A:
(i) Roasting (Heating in excess air).
(ii) Example Copper (Cu) or Mercury (Hg). Let's take Copper:
$2Cu_2S + 3O_2 \xrightarrow{heat} 2Cu_2O + 2SO_2$
$2Cu_2O + Cu_2S \xrightarrow{heat} 6Cu + SO_2$
(iii) Copper reacts with moist $CO_2$ in air to form a green coating of basic copper carbonate.
Answer for B:
Soaps are sodium salts of long chain fatty acids. The molecule has two ends: Hydrophilic (ionic head) and Hydrophobic (carbon tail). The hydrophobic tail attaches to dirt (oil) and hydrophilic head to water, forming a structure called Micelle. This pulls the dirt out.
Hard Water: Soap reacts with Calcium and Magnesium ions in hard water to form an insoluble precipitate called 'Scum', wasting the soap.
(i) Draw a schematic diagram of the setup he should use. Label the Anode, Cathode, and Electrolyte.
(ii) Write the reaction taking place at the Cathode.
(iii) Where do the insoluble impurities settle?
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(i) Setup details:
- Anode: Impure Copper block (Connected to positive terminal).
- Cathode: Thin strip of Pure Copper (Connected to negative terminal).
- Electrolyte: Acidified Copper Sulphate solution ($CuSO_4 + H_2SO_4$).
(ii) Reaction at Cathode:
$Cu^{2+}(aq) + 2e^- \rightarrow Cu(s)$
(Pure copper deposits here).
(iii) Insoluble impurities settle at the bottom of the anode as Anode Mud.
Case Study: The pH of a solution is a measure of its hydrogen ion concentration. Plants require a specific pH range for their healthy growth. If the soil becomes too acidic or too basic, the yield is affected.
A farmer tested the soil of his field and found the pH to be 4.5. He consulted an agricultural scientist for a remedy.
Answer the following questions:
A. Which chemical substance should the farmer add to the soil to adjust the pH? Why?
B. Rainwater often has a pH of 5.6. If this rain flows into a river, how does it affect aquatic life?
C. Explain how the pH change in the mouth leads to tooth decay. What is the role of toothpaste in preventing it?
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Answer A: He should add Quick Lime (Calcium Oxide) or Slaked Lime (Calcium Hydroxide) or Chalk (Calcium Carbonate).
Reason: The soil is acidic (pH 4.5). These substances are basic in nature and will neutralize the excess acid.
Answer B: When acid rain (pH < 5.6) flows into rivers, it lowers the pH of the river water. This makes it difficult for aquatic life (fish, etc.) to survive as the water becomes acidic.
Answer C (OR Part):
- Bacteria in the mouth produce acid by degrading sugar.
- If pH falls below 5.5, the tooth enamel (Calcium Hydroxyapatite) starts corroding.
- Toothpaste: It is basic in nature. It neutralizes the excess acid in the mouth and prevents decay.
(a) Identify 'X'.
(b) What is its chemical formula?
(c) State the reaction determining its production from sodium chloride.
(d) What happens when it is heated during cooking? Write the equation.
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(a) Baking Soda.
(b) Sodium Hydrogen Carbonate ($NaHCO_3$).
(c) Solvay Process:
$NaCl + H_2O + CO_2 + NH_3 \rightarrow NH_4Cl + NaHCO_3$
(d) On heating, it decomposes to give sodium carbonate, water and carbon dioxide (which makes dough rise).
$2NaHCO_3 \xrightarrow{Heat} Na_2CO_3 + H_2O + CO_2$
Section C: Physics
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Reason (R): It always forms a virtual, erect, and diminished image, covering a wider field of view.
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(i) Identify the type of mirror used.
(ii) What is the focal length of the mirror?
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(i) Type: Concave Mirror (Because only concave mirrors can form real images on a screen/wall).
(ii) Focal Length: 15 cm.
Reason: Rays from a distant object (infinity) converge at the focus. So, the distance between the mirror and the screen (image) is the focal length.
A. In the circuit diagram given below, two resistors of $4 \Omega$ and $8 \Omega$ are connected in parallel to a 12V battery.
(i) Calculate the total resistance of the circuit.
(ii) Calculate the total current drawn from the battery.
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Answer A:
(i) $\frac{1}{R_p} = \frac{1}{4} + \frac{1}{8} = \frac{2+1}{8} = \frac{3}{8}$. So, $R_p = \frac{8}{3} = 2.67 \Omega$.
(ii) Current $I = \frac{V}{R} = \frac{12}{8/3} = \frac{36}{8} = 4.5 A$.
Answer B:
Power $P = 880 W$, $V = 220 V$.
Current $I = \frac{P}{V} = \frac{880}{220} = 4 A$.
Resistance $R = \frac{V}{I} = \frac{220}{4} = 55 \Omega$.
(ii) List two possible causes for this defect.
(iii) Which lens is used for its correction?
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(i) Hypermetropia (Far-sightedness).
(ii) Causes:
1. The focal length of the eye lens is too long.
2. The eyeball has become too small.
(iii) Correction: Convex Lens (Converging lens).
(ii) A wire of resistance $R$ is cut into five equal parts. These parts are then connected in parallel. If the equivalent resistance of this combination is $R'$, find the ratio $R/R'$.
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(i) Disadvantages of Series:
1. If one appliance fails, the circuit breaks and none work.
2. Different appliances need different currents, but in series, the current is the same.
3. Total resistance increases, reducing the current.
(ii) Numerical:
Resistance of one part = $R/5$.
In parallel: $\frac{1}{R'} = \frac{1}{R/5} + \frac{1}{R/5} + \frac{1}{R/5} + \frac{1}{R/5} + \frac{1}{R/5}$
$\frac{1}{R'} = \frac{5}{R} \times 5 = \frac{25}{R}$
$R' = \frac{R}{25}$
Ratio $\frac{R}{R'} = \frac{R}{R/25} = 25$
Answer: 25:1
(ii) State two ways to increase the strength of the magnetic field produced by a solenoid.
(iii) What acts as the North Pole in a current-carrying solenoid?
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(i) (Student should draw lines similar to a bar magnet: straight inside, curved outside).
(ii) Ways to increase strength:
1. Increasing the current flowing through it.
2. Increasing the number of turns in the coil.
(Bonus: Inserting a soft iron core).
(iii) The face of the coil where the current flows in an Anti-clockwise direction acts as the North Pole.
Case Study: Slide Projector
A slide projector uses a convex lens to project a magnified image of a small slide onto a large screen. The slide is placed between $F$ and $2F$ of the lens to get a real, inverted, and magnified image on the screen.
Answer the following questions:
A. What is the nature of the image formed on the screen?
B. If the slide is placed at the focus ($F$) of the projection lens, where will the image be formed?
Attempt either subpart C or D.
C. A slide projector has a lens of focal length 20 cm. The slide is placed 30 cm from the lens. Calculate the distance of the screen from the lens.
D. If the magnification produced by the projector lens is -10, and the object (slide) is 3 cm tall, what is the height of the image on the screen?
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A. Real, Inverted, and Magnified.
B. At Infinity (Highly magnified).
Answer C:
$f = +20 cm$ (Convex), $u = -30 cm$.
Lens Formula: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
$\frac{1}{v} - \frac{1}{-30} = \frac{1}{20}$
$\frac{1}{v} = \frac{1}{20} - \frac{1}{30} = \frac{3-2}{60} = \frac{1}{60}$
$v = +60 cm$. Screen distance is 60 cm.
Answer D (OR):
$m = \frac{h_i}{h_o}$
$-10 = \frac{h_i}{3}$
$h_i = -30 cm$.
Height of image is 30 cm (Negative sign indicates it is inverted).
A. An object 4.0 cm in size is placed at 25.0 cm in front of a concave mirror of focal length 15.0 cm.
(i) At what distance from the mirror should a screen be placed in order to obtain a sharp image?
(ii) Find the nature and size of the image.
(iii) Draw a ray diagram for this case.
(i) Identify the defect of vision.
(ii) Draw a ray diagram to illustrate this defect.
(iii) Calculate the power of the lens required to correct this defect.
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Solution for A (Mirror Formula):
$u = -25 cm$, $f = -15 cm$, $h_o = 4 cm$.
$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$
$\frac{1}{v} - \frac{1}{25} = -\frac{1}{15}$
$\frac{1}{v} = \frac{1}{25} - \frac{1}{15} = \frac{3-5}{75} = \frac{-2}{75}$
$v = -37.5 cm$. (Screen placed 37.5 cm in front of mirror).
Magnification: $m = -\frac{v}{u} = -(\frac{-37.5}{-25}) = -1.5$
$h_i = m \times h_o = -1.5 \times 4 = -6 cm$.
Nature: Real, Inverted, Enlarged.
Solution for B (Eye Defect):
(i) Myopia (Near-sightedness) because the far point is 5m (not infinity).
(ii) Diagram: Image forms in front of the retina.
(iii) $f = -Far Point = -5 m$.
Power $P = \frac{1}{f (in meters)}$
$P = \frac{1}{-5} = -0.2 Diopter$.