class 10 science question paper solution sample paper 2025-26

Directorate of Education, GNCT of Delhi 

Mid Term Examination Practice Paper (Session: 2025-26) 

Class:X; Subject: SCIENCE (086) 

Maximum Marks: 80 Duration: 3 hours 

General Instructions: 

1. This question paper consists of 39 questions in 3 sections. Section A is Biology, Section B is Chemistry, and Section C is Physics. 

2. All sections are compulsory. However internal choice is provided in some questions. A student is expected to attempt only one of these questions.

Section A 

1. Which process in plants is responsible for the transport of food? 
A. Photosynthesis 

B. Respiration 

C. Transpiration 

D. Translocation 

Answer: D. Translocation

Explanation:
Translocation is the process by which food (mainly sugars like glucose) is transported through the plant. This occurs via the phloem, moving from the leaves (where food is made during photosynthesis) to other parts of the plant, including roots, stems, and fruits.

2. A person accidentally touches a hot object and withdraws his hand immediately. This is an example of: 

A. Involuntary action 

B. Hormonal action 

C. Voluntary action 

D. Reflex action 

Answer: D. Reflex action

Explanation:
A reflex action is an involuntary and automatic response to a stimulus. When you touch something hot, the sensory receptors in your skin send a signal to your spinal cord, which immediately sends a signal to the muscles to withdraw your hand. This happens very quickly, without needing the brain to process it consciously.

3. Which of the following will have the least amount of energy in a food chain? 

A. Producers 

B. Secondary consumers 

C. Primary consumers 

D. Tertiary consumers

Answer: D. Tertiary consumers

Explanation:
In a food chain, energy decreases as you move up from producers to primary consumers, secondary consumers, and tertiary consumers. Producers (plants) have the most energy since they capture energy from the sun. As energy moves through the food chain (from producers to primary consumers to secondary consumers and then tertiary consumers), about 90% of energy is lost at each trophic level due to metabolic processes like movement, growth, and heat. So, tertiary consumers, being at the top of the food chain, have the least amount of energy.

 

4. The stomatal pore opens when 
A. The guard cells swell as water flows into them 

B. The guard cells shrink as water flows outside 

C. The guard cells swell as water flows outside 

D. The guard cells shrink as water flows into them 

Answer: A. The guard cells swell as water flows into them

Explanation:
Stomata are tiny pores found on the surface of leaves, and their opening and closing are controlled by guard cells. When the guard cells absorb water (via osmosis), they swell and curve, causing the stomatal pore to open. This allows for the exchange of gases (like CO₂ entering for photosynthesis and O₂ leaving). When the guard cells lose water, they shrink, causing the stomatal pore to close.

5. The gustatory receptors and olfactory receptors in humans are present in 

A. skin and tongue respectively 

B. nasal cavity and eyes respectively 

C. tongue and nasal cavity respectively 

D.tongue and skin respectively 

Answer: C. Tongue and nasal cavity respectively

Explanation:
Gustatory receptors are responsible for the sense of taste, and they are located on the tongue. These receptors detect different tastes such as sweet, salty, sour, bitter, and umami. Olfactory receptors, which are responsible for the sense of smell, are located in the nasal cavity. Both senses work together to help us perceive the flavor of food.

6. Which portion of the brain is responsible for the precision of voluntary actions and maintaining the posture and balance of the body. 

A. Medulla 

B. Hypothalamus 

C. Cerebrum 

D. Cerebellum 

Answer: D. Cerebellum

Explanation:
The cerebellum is located at the back of the brain and is responsible for coordinating voluntary movements, maintaining posture, and balancing the body. It helps ensure that movements are smooth and precise. It also plays a role in motor learning and the timing of muscle movements.

7. Which are the correct statements related with ecosystem?

(i) Ecosystem consists of biotic and abiotic components. 

(ii) Garden and crop field are human-made ecosystems. 

(iii) Biotic components of ecosystem comprise of physical factors like wind, soil etc. 

(iv) Forest, ponds and lakes are artificial ecosystem. 

(v) The biotic and abiotic components of ecosystem interact with each other. 

A. (i), (ii), (v) 

B.(ii), (iv), (v) 

C. (i), (iii), (iv) 

D. (i), (iv), (v) 

Answer: A. (i), (ii), (v)

Explanation:

• (i) Ecosystem consists of biotic and abiotic components – An ecosystem includes both living (biotic) and non-living (abiotic) components. The biotic components include plants, animals, and microorganisms, while abiotic components include things like water, soil, temperature, and sunlight.

• (ii) Garden and crop field are human-made ecosystems – These are examples of artificial or man-made ecosystems, as they are created and maintained by humans for purposes like food production.

• (v) The biotic and abiotic components of ecosystem interact with each other – In an ecosystem, biotic and abiotic components are interconnected. For example, plants (biotic) require sunlight (abiotic) to grow, and animals depend on plants for food.

The other options are incorrect because:

• (iii) Biotic components of ecosystem comprise of physical factors like wind, soil, etc. – This is incorrect; wind and soil are abiotic components, not biotic.

• (iv) Forest, ponds, and lakes are artificial ecosystems – This is incorrect. Forests, ponds, and lakes are natural ecosystems, not artificial ones.

8. The following question consists of two statements- Assertion (A) and Reason (R). 

Answer these questions by selecting appropriate option given below: 

A. Both A and R are true, and R is the correct explanation of A. 

B. Both A and R are true, and R is not the correct explanation of A. 

C. A is true but R is false. 

D. A is false but R is true. 

Assertion (A):Ozone at higher levels of atmosphere is beneficial for organisms. 

Reason (R):Ozone shields the surface of earth from ultraviolet radiations from sun. 

Answer: A. Both A and R are true, and R is the correct explanation of A.

Explanation:

• The ozone layer is located in the stratosphere and plays a crucial role in absorbing and blocking most of the sun's harmful ultraviolet (UV) radiation. This radiation can cause skin cancer, cataracts, and damage to plants and marine life.

• Assertion (A) is true because ozone at higher altitudes indeed provides a protective shield for living organisms by filtering out harmful UV radiation.

• Reason (R) correctly explains why ozone is beneficial—it protects organisms from UV radiation, making the reason an accurate explanation for the assertion.

9. The following question consists of two statements- Assertion (A) and Reason (R). 

Answer these questions by selecting appropriate option given below: 

A. Both A and R are true, and R is the correct explanation of A. 

B. Both A and R are true, and R is not the correct explanation of A. 

C. A is true but R is false. 

D. A is false but R is true. 

Assertion (A):Energy requirements of the autotrophic organisms are fulfilled by photosynthesis. 

Reason (R):Carbon dioxide is oxidised to form carbohydrates. 

Answer: C. A is true but R is false.

Explanation:

• Assertion (A) is true because autotrophic organisms (like plants, algae, and some bacteria) produce their own food through photosynthesis. This process captures sunlight energy to convert carbon dioxide and water into glucose and oxygen.

• Reason (R) is incorrect because in photosynthesis, carbon dioxide is reduced (not oxidized) to form carbohydrates (like glucose). In oxidation, electrons are lost, while in reduction, electrons are gained, which is what happens to carbon dioxide in photosynthesis.

10. How the positive geotropism is different from negative geotropism? Explain with the help of suitable examples. 

Explanation:

• Positive geotropism refers to the growth of plant parts towards gravity. For example, roots exhibit positive geotropism because they grow downward into the soil, seeking water and nutrients.

• Negative geotropism refers to the growth of plant parts away from gravity. For example, stems exhibit negative geotropism as they grow upward, away from the ground, towards the light to maximize photosynthesis.

11. Attempt either option A or B. 

A. Describe how Amoeba obtains its food and digests it. Which type of nutrition is exhibited by it? 

Answer:

• Amoeba is a single-celled organism that obtains its food through a process called phagocytosis.

  • The amoeba surrounds a food particle (like bacteria) with its cell membrane, forming a food vacuole.

  • Lysosomes containing digestive enzymes fuse with the food vacuole, breaking down the food into simpler substances.

  • The absorbed nutrients are used for energy and growth, and waste materials are expelled through exocytosis.

• Nutrition type: Amoeba exhibits holozoic nutrition, where food is ingested, digested, and absorbed in solid form.

 OR 

B. Name any two digestive enzymes present in pancreatic juices and mention their important function. 

Answer:

• Two digestive enzymes in pancreatic juices:

  1. Amylase: It breaks down starch into simpler sugars (like maltose) in the small intestine.

  2. Trypsin: It breaks down proteins into smaller peptides and amino acids in the small intestine.

12. “The flow of energy in unidirectional in a food chain”.Justify the statement with the help of a terrestrial food chain comprising four trophic levels. 

Explanation:
In a food chain, energy flows in one direction, from producers to consumers:

1. Producers (plants): Plants convert solar energy into chemical energy through photosynthesis.

2. Primary consumers (herbivores): Herbivores consume plants, taking in the stored energy.

3. Secondary consumers (carnivores): Carnivores feed on herbivores, transferring energy up the food chain.

4. Tertiary consumers (top predators): These animals consume other carnivores and are at the top of the food chain.

At each level, only about 10% of the energy is transferred to the next level, with the rest being lost as heat, used for metabolism, or excreted. Thus, energy decreases at each trophic level and flows in one direction—from producers to consumers.

13. What are plant hormones? Name the plant hormone responsible for the following: 

(i)inhibition of growth 

(ii) growth of stem 

(iii) promotion of cell division 

(iv) growth of tendril around a support  

Explanation:

• Plant hormones are chemicals that regulate various physiological processes in plants, such as growth, development, and responses to environmental stimuli.

Hormones responsible for:

1. Inhibition of growth: Abscisic acid (ABA) – This hormone plays a role in slowing down growth, especially during stress conditions.

2. Growth of stem: Auxins – These hormones promote stem elongation and help the plant grow towards light (phototropism).

3. Promotion of cell division: Cytokinins – These hormones promote cell division and are often involved in growth and development.

4. Growth of tendril around a support: Auxins – Auxins cause the tendril to coil around a support, helping the plant to climb.

14. A. Describe two metabolic pathways through which glucose is broken down in the human body in different situations to release energy, and specify the end products of each pathway. 

Answer:

1. Glycolysis:

   • In the cytoplasm, glucose is broken down into pyruvate (or pyruvic acid), releasing a small amount of energy. This occurs in both aerobic and anaerobic conditions.

   • End products: 2 molecules of pyruvate, 2 ATP, and 2 NADH.

2. Aerobic Respiration (when oxygen is available):

   • Pyruvate enters the mitochondria and is further oxidized in the citric acid cycle and electron transport chain.

   • End products: 36 ATP, carbon dioxide (CO₂), and water.

3. Anaerobic Respiration (Fermentation) (when oxygen is not available):

   • In the absence of oxygen, pyruvate undergoes lactic acid fermentation (in muscles) or alcoholic fermentation (in yeast).

   • End products: In lactic acid fermentation, lactic acid and 2 ATP are produced. In alcoholic fermentation, ethanol and CO₂ are produced.

B. How the mode of transport of oxygen is different from the transport of carbon dioxide in our body? 

Answer:

• Oxygen transport:

  • Oxygen is carried by hemoglobin in red blood cells. A small amount of oxygen is dissolved in plasma, but most is bound to hemoglobin to form oxyhemoglobin.

• Carbon dioxide transport:

  • About 70% of carbon dioxide is transported as bicarbonate ions (HCO₃⁻) in the plasma, after reacting with water and carbonic acid. A small amount is carried by hemoglobin (as carbaminohemoglobin) and dissolved in plasma.

15. During science class, the students learned the significance of feedback mechanisms in endocrine system. Attempt either subpart A or B. 

A. What is the importance of feedback mechanism of hormonal action in animals 

Answer:
Feedback mechanisms in the endocrine system play a crucial role in maintaining homeostasis (the stable internal environment) within the body. These mechanisms help regulate the levels of hormones, ensuring that they remain balanced and do not fluctuate too much. There are two types of feedback mechanisms:

1. Negative Feedback: This is the most common type of feedback in the endocrine system. It works to reverse a change, keeping the system stable. For example, if the blood sugar level rises, the pancreas releases insulin to lower the blood sugar level. When the sugar level returns to normal, insulin secretion decreases.

2. Positive Feedback: This amplifies a change rather than reversing it. An example is the release of oxytocin during childbirth. As contractions occur, oxytocin is released, which increases the intensity of contractions, speeding up the process until the baby is born.

Without feedback mechanisms, the body would not be able to regulate vital processes like metabolism, growth, or reproductive functions, leading to disorders or imbalances.

OR 

B. How the hormonal mechanism of control and coordination in animals are different from nervous control? 

Answer:
Hormonal control and nervous control are two key systems that regulate and coordinate body functions, but they work in different ways:

1. Hormonal Control:

   • Medium of transmission: Hormones are released by glands into the bloodstream and travel to distant target organs or tissues.

   • Speed of response: Hormonal responses are slower, and their effects may last for a longer time. For example, growth hormones control the development of bones and tissues over weeks or months.

   • Control mechanism: The endocrine system releases hormones into the blood based on signals from the nervous system or feedback mechanisms, regulating processes like metabolism, growth, and reproduction.

   • Examples: Hormones such as insulin, thyroxine, and adrenaline regulate metabolic functions, growth, and stress responses, respectively.

2. Nervous Control:

   • Medium of transmission: Electrical signals (nerve impulses) travel along neurons to specific tissues and organs.

   • Speed of response: Nervous responses are instantaneous (milliseconds). For example, touching something hot triggers an immediate reflex action to withdraw the hand.

   • Control mechanism: The nervous system consists of the central nervous system (CNS) and the peripheral nervous system (PNS), transmitting rapid signals to control movements, thoughts, and reflexes.

   • Examples: Nervous control is responsible for muscle movements, sensory responses, and reflex actions.

C. Name the hormone released by pancreas when the sugar levels in blood rises. 

The hormone released by the pancreas when blood sugar levels rise is insulin. Insulin helps lower blood sugar by promoting the uptake of glucose into cells for energy or storage in the liver as glycogen.

D. Name a hormone secreted by adrenal gland and state its important function. 

One of the key hormones secreted by the adrenal glands is adrenaline (also called epinephrine).

Important function:
Adrenaline is released during the fight or flight response. It prepares the body to respond to stress by increasing heart rate, dilating the airways, increasing blood flow to muscles, and releasing energy stores. This helps the body react quickly to stressful or emergency situations.

16. Attempt either option A or B. 

A. (i) Describe the pathway of oxygenated blood circulation in the human body. 

The oxygenated blood follows this pathway in the human circulatory system:

1. Lungs: Oxygenated blood is first pumped to the lungs, where it picks up oxygen and releases carbon dioxide.

2. Left Atrium: Oxygen-rich blood returns to the heart via the pulmonary veins into the left atrium.

3. Left Ventricle: From the left atrium, the oxygenated blood flows through the mitral valve into the left ventricle.

4. Aorta: The left ventricle pumps oxygenated blood into the aorta, the main artery of the body.

5. Systemic Circulation: The blood is distributed through the aorta into various arteries, reaching all parts of the body. Oxygen is delivered to tissues, and carbon dioxide is collected as waste.

(ii) Describe two important functions of lymph in humans. 

Answer:

1. Immune Function: Lymph contains white blood cells (specifically lymphocytes) that play a vital role in the immune system by fighting off infections and foreign pathogens. Lymph nodes act as filtering stations, where lymph is screened for pathogens and viruses.

2. Transport of Fats: Lymph absorbs fats from the small intestine after digestion and transports them in the form of chylomicrons to the bloodstream, aiding in the distribution of nutrients throughout the body.

OR 

B. Give reasons for the following: 

(i) Veins are provided with valves whereas arteries lack it. 

Answer:
Veins are under low pressure, and valves prevent the backflow of blood as it moves towards the heart, especially from the legs and arms. Arteries, on the other hand, carry blood under high pressure from the heart, so there is no need for valves to prevent backflow.

(ii) The septum separates right side and left side of heart. 

Answer:
The septum divides the heart into two halves: the right side pumps deoxygenated blood to the lungs for oxygenation, and the left side pumps oxygenated blood to the rest of the body. This separation ensures that oxygen-rich and oxygen-poor blood do not mix, maintaining the efficiency of circulation.

(iii) Fishes have two chambered heart. 

Answer:
Fishes have a two-chambered heart (one atrium and one ventricle), which is sufficient for their single circulatory system. Blood flows in a single loop: from the heart to the gills (where it is oxygenated) and then to the body. This is more energy-efficient for the lower metabolic demands of fish.

(iv) Wall of arteries are thick and elastic. 

Answer:
Arteries carry blood at high pressure from the heart. Their thick, elastic walls help them withstand this pressure and maintain the flow of blood. The elasticity also helps arteries expand and contract with each heartbeat, ensuring smooth blood flow.

(v) A blood clot develops at the site of injury after some time. 

Answer:
When a blood vessel is injured, the body activates platelets and proteins (such as fibrinogen) in the blood to form a clot. This clot forms a barrier to prevent excessive blood loss and protects against infection. The clotting cascade is triggered, leading to the formation of fibrin, which solidifies the blood at the wound site.

Section B 

17. A student in laboratory uses a universal indicator and note that solution shows the pH value of 3. What does it indicate? 

A. The solution is neutral 

B. The solution is strongly basic 

C. The solution is strongly acidic 

D. The solution is weakly basic. 

Answer: C. The solution is strongly acidic

Explanation:
A pH value of 3 indicates a strongly acidic solution. The pH scale ranges from 0 to 14, with values less than 7 indicating acidity. A pH value of 3 is much closer to 0, meaning it is strongly acidic.

18. On heating lead nitrate in a test tube, emission of brown fumes are of: 

A. Lead oxide 

B. Nitrogen dioxide 

C. Oxygen 

D. Carbon dioxide 

Answer: B. Nitrogen dioxide

Explanation:
When lead nitrate (Pb(NO₃)₂) is heated, it decomposes to form lead oxide (PbO), nitrogen dioxide (NO₂), and oxygen (O₂). The brown fumes are nitrogen dioxide, which is a reddish-brown gas that is formed in this reaction.

The reaction is:
$2Pb(N{O}_3{)}_2\to 2PbO+4N{O}_2+O_2$

19. Basic salts are formed when 

A. Strong acid reacts with weak base 

B. Strong acid reacts with strong base 

C. Weak acid reacts with weak base 

D. Weak acid reacts with strong base  

​Answer: D. Weak acid reacts with strong base

Explanation:
When a weak acid reacts with a strong base, the resulting salt is basic in nature. This happens because the conjugate base of the weak acid can accept protons, making the salt basic. An example is the reaction between acetic acid (weak acid) and sodium hydroxide (strong base), which forms sodium acetate (basic salt).

20. Which of these metals does not reacts either with cold or hot water but reacts with steam? 

A. Potassium and sodium 

B. Iron and aluminium 

C. Silver and gold 

D. Lead and copper 

Answer: B. Iron and aluminium

Explanation:
Iron (Fe) and aluminium (Al) do not react with cold or hot water, but they do react with steam at high temperatures to form their respective oxides and release hydrogen gas. For example, when iron reacts with steam, it forms iron oxide and hydrogen gas:
$3Fe + 4H_2O \rightarrow Fe_3O_4 + 4H_2$

21. Metal hydrogen carbonates reacts with acids to produce 

A. Corresponding salt, hydrogen gas and water 

B. Metal carbonates and carbon dioxide gas 

C. Corresponding salt, carbon dioxide and water 

D. Metal carbonates and oxygen gas 

Answer: C. Corresponding salt, carbon dioxide, and water

Explanation:
When a metal hydrogen carbonate (e.g., sodium bicarbonate NaHCO₃) reacts with an acid, it produces a salt, carbon dioxide (CO₂), and water. For example, when sodium bicarbonate reacts with hydrochloric acid, the reaction is:
$NaHCO_3 + HCl \rightarrow NaCl + CO_2 + H_2O$

22. The following question consists of two statements- Assertion (A) and Reason (R). 

Answer these questions by selecting appropriate option given below: 

A. Both A and R are true, and R is the correct explanation of A. 

B. Both A and R are true, and R is not the correct explanation of A. 

C. A is true but R is false. 

 D.A is false but R is true 

Assertion (A):A yellow precipitate is formed when solutions of lead(II) nitrate and potassium iodide are mixed. 

Reason (R):Lead(II) iodide is insoluble in water and form precipitates. 

Assertion (A): A yellow precipitate is formed when solutions of lead(II) nitrate and potassium iodide are mixed.
Reason (R): Lead(II) iodide is insoluble in water and forms precipitates.

Answer: A. Both A and R are true, and R is the correct explanation of A.

Explanation:

• When lead(II) nitrate (Pb(NO₃)₂) and potassium iodide (KI) solutions are mixed, they react to form lead(II) iodide (PbI₂), which is insoluble in water and forms a yellow precipitate.

• The reason for the yellow precipitate is that lead(II) iodide is insoluble in water, explaining why it forms a precipitate.

The reaction is:
$Pb(NO₃)_2 + 2KI \rightarrow PbI_2 \,(\text{yellow precipitate}) + 2KNO_3$

23. Write balanced chemical equation for the reaction involved in the preparation of bleaching powder. State any one use also.

Balanced chemical equation for the preparation of bleaching powder:

$Ca(OH)_2 + Cl_2 \rightarrow Ca(OCl)_2 + H_2O$

Use:
One common use of bleaching powder (calcium hypochlorite, Ca(OCl)₂ is for disinfecting water in water treatment plants to kill harmful bacteria and other pathogens.

24. A student placed a copper wire in iron sulphate solution and observed no colour change even after an hour. Explain the reason behind this observation. 

Answer:
Copper does not displace iron from iron sulfate (FeSO₄) solution because copper (Cu) is less reactive than iron (Fe). According to the reactivity series, a more reactive metal can displace a less reactive metal from its compound, but since copper is less reactive than iron, it does not replace iron from iron sulfate. Hence, there is no reaction, and no color change occurs.

25. A student in the science lab diluted a solution of hydrochloric acid with water under supervision of science teacher. 

A. How will it affect the hydrogen ion concentration and pH? 

Answer:
Diluting hydrochloric acid (HCl) with water will decrease the hydrogen ion (H⁺) concentration because the acid becomes more dilute. This will increase the pH of the solution, making it less acidic (the pH moves closer to 7).

B. Why does the acid become less corrosive upon dilution? 

Answer:
When the concentration of the acid decreases, there are fewer hydrogen ions (H⁺) available to react with surfaces. Since hydrogen ions are responsible for the corrosive nature of acids, a lower concentration makes the acid less corrosive.

C. Explain the safe procedure that must be followed by the student while diluting a strong concentrated acid ? 

Answer:
When diluting a strong acid like hydrochloric acid, the student must:

1. Always add acid to water, never the other way around, because adding water to acid can cause a violent exothermic reaction, which may splash the acid.

2. Use proper protective gear such as gloves, goggles, and a lab coat.

3. Dilute the acid slowly while stirring, to avoid splashing or sudden temperature increases.

4. Perform the dilution in a well-ventilated area or under a fume hood to avoid inhaling fumes.

26. Attempt either option A or B. A. 

(i) Differentiate between roasting and calcination. 

Answer:

Roasting	Calcination
Involves heating ores in the presence of oxygen (air).	Involves heating ores in the absence of oxygen (air).
Typically used for sulfide ores.	Typically used for carbonate ores.
Example: Roasting of zinc sulfide (ZnS) to form zinc oxide (ZnO).	Example: Calcination of limestone (CaCO₃) to form lime (CaO) and carbon dioxide (CO₂).

(ii) Draw a flow chart representing various steps involved in the extraction of metal of low reactivity from its ore. 

Answer:

1. Mining of Ore

2. Crushing & Grinding

3. Concentration (e.g., Froth flotation or magnetic separation)

4. Smelting (for ores with oxides)

5. Reduction (using carbon or hydrogen)

6. Purification (if necessary)

OR 

B. A technician uses the thermit reaction to repair a broken steel joint in a machine. 

(i) Explain how this reaction helps in joining metal parts. 

Answer:
The thermit reaction is used to generate intense heat that can melt metals. In this reaction, aluminum powder reacts with iron(III) oxide to produce molten iron and aluminum oxide. The heat generated is sufficient to melt the iron, allowing metal parts to be joined together.

The reaction is:
$Fe_2O_3 + 2Al \rightarrow 2Fe + Al_2O_3$

This molten iron is used to join or repair metal parts, especially in railways

(ii) Write equation for the chemical reaction involved. 

The thermit reaction is:
$Fe_2O_3 + 2Al \rightarrow 2Fe + Al_2O_3$

(iii) Why is aluminium preferred over zinc or copper in this case? 

Answer:
Aluminum is preferred over zinc or copper in the thermit reaction because it is more reactive than iron and can easily reduce iron oxide to molten iron. Zinc and copper are less reactive than aluminum, so they cannot produce the same high temperatures necessary for the reaction.

27. In an experiment, black copper(II) oxide (CuO) is heated while hydrogen gas is passed over it. 

A reddish-brown solid (copper) forms, and water vapours condense on the cooler part of the test tube. Attempt either subpart A or B. 

A. Write balanced chemical equation for the reaction. 

A. Write the balanced chemical equation for the reaction.

Answer:
The balanced chemical equation for the reduction of copper(II) oxide by hydrogen is:
$CuO + H_2 \rightarrow Cu + H_2O$

OR 

B. Why is this reaction considered as redox reaction? 

Answer:

This reaction is a redox reaction because it involves both oxidation and reduction processes:

• Copper(II) oxide (CuO) is reduced to copper (Cu) by hydrogen gas (H₂). Reduction is the gain of electrons, so Cu²⁺ (in CuO) gains electrons to form Cu.

• Hydrogen (H₂) is oxidized to form water (H₂O). Oxidation is the loss of electrons, so H₂ loses electrons to form H₂O.

The oxidation half-reaction:
$H_2 \rightarrow 2H^+ + 2e^-$
The reduction half-reaction:
$CuO + 2e^- \rightarrow Cu + O^{2-}$

C. Identify the substance acting as oxidising agent and reducing agent. 

Reaction:

$$\ce{CuO + H2 -> Cu + H2O}$$

• Hydrogen (H₂) reduces copper(II) oxide (CuO) to copper (Cu).

• At the same time, hydrogen is oxidised to water (H₂O).

Answer:

• Oxidising agent: Copper(II) oxide (CuO)
  (It accepts electrons and gets reduced to copper)

• Reducing agent: Hydrogen (H₂)
  (It donates electrons and gets oxidised to water)

D. Identify the substances being oxidised and reduced during this reaction. 

Substance oxidised: Hydrogen (H₂)

(It loses electrons to form H₂O → oxidation)

Substance reduced: Copper(II) oxide (CuO)
(The Cu²⁺ ion in CuO gains electrons to become metallic Cu → reduction)

28. Attempt either option A or B. A. 

(i)Explain the formation of Lithium oxide (Li2O) with the help of electron dot structure. 

⚛ Formation of Lithium Oxide (Li₂O) – Electron Dot Structure

Step 1: Valence electrons

• Lithium (Li) has 1 valence electron (Group 1).

• Oxygen (O) has 6 valence electrons (Group 16), and it needs 2 electrons to complete its octet.

---

✅ Electron Dot Structure:

$$\ce{Li \cdot + \cdot O \cdot \cdot \cdot \cdot \cdot + \cdot Li \rightarrow Li^+ \ [\cdot \cdot O \cdot \cdot ]^{2-} \ Li^+}$$

Or, neatly:

Two lithium atoms each donate 1 electron to oxygen:

$$\ce{2Li \cdot + \cdot \cdot O \cdot \cdot \cdot + \cdot Li \rightarrow 2Li^+ + O^{2-} \rightarrow Li_2O}$$

---

🧪 Balanced Chemical Equation:

$$\boxed{\ce{4Li + O2 -> 2Li2O}}$$

Lithium reacts with oxygen to form lithium oxide.

---

✍️ Explanation:

• Each Li atom donates 1 electron → becomes Li⁺ ion.

• Oxygen gains 2 electrons (1 from each Li) → becomes O²⁻ ion.

• Ionic bond forms between 2Li⁺ and 1O²⁻, resulting in Li₂O.

(ii) Why ionic compounds conduct electricity in molten or aqueous state but not in solid state?

In solid state, ions are fixed in a lattice and cannot move → no conductivity.

In molten/aqueous state, ions are free to move → they carry electric current

 

OR 

B. (i) Why is aluminium oxide considered an amphoteric oxide? Explain with examples.

• Amphoteric oxides react with both acids and bases.

• Aluminium oxide (Al₂O₃) reacts with:

  • Acids like HCl to form salt and water

  • Bases like NaOH to form aluminates

Examples:

• Al₂O₃ + 6HCl → 2AlCl₃ + 3H₂O

• Al₂O₃ + 2NaOH → 2NaAlO₂ + H₂O

 

(ii) Why hydrogen gas is not evolved when a metal reacts with nitric acid? 

Nitric acid is a strong oxidising agent.

Instead of releasing H₂ gas, it oxidises the metal and gets reduced itself (to NO, NO₂, etc.).

Hence, hydrogen gas is not evolved.

(iii) Define alloy. Name alloy and its constituents commonly used during welding of electrical wires.

• Alloy: A homogeneous mixture of two or more metals, or a metal and non-metal, to improve properties.

Example used in welding electrical wires:

• Solder = Tin (Sn) + Lead (Pb)

Section C 

29. When a narrow beam of white light is passes through a glass prism, it splits into its component colours. This phenomenon is called 

A. Tyndall effect 

B. Dispersion of light

C. Total reflection of light 

D. Reflection of light 

B. Dispersion of light

Correct Answer: B. Dispersion of light

Explanation:

Dispersion is the phenomenon where white light splits into its seven component colours (VIBGYOR) due to different degrees of refraction for each colour as light passes through the prism.

30. A concave mirror forms an enlarged and virtual image when the object is placed: 

A. At infinity 

B. Beyond centre of curvature 

C. Between pole and focus 

D. At centre of curvature 

Correct Answer: C. Between pole and focus

Explanation:
When the object is placed between the pole and focus of a concave mirror, the mirror forms a virtual, erect, and enlarged image behind the mirror.

31. Which colour of light is scattered the least in the atmosphere? 

A. Blue 

B. Red 

C. Violet 

D. Green 

Explanation:
Red light has the longest wavelength, so it is scattered the least by atmospheric particles. That’s why the sun appears red during sunrise and sunset.

32. A concave lens has a focal length of 50 cm. Its power will be 

A. +2 D 

B. -2 D 

C. +0.5 D 

D. -0.5 D 

33. The following question consists of two statements- Assertion (A) and Reason (R). Answer these questions by selecting appropriate option given below: 

A. Both A and R are true, and R is the correct explanation of A. 

B. Both A and R are true, and R is not the correct explanation of A. 

C. A is true but R is false. D. A is false but R is true 

Assertion (A): The ability of the eye to focus on both near and distant objects, by adjusting its focal length, is called the accommodation. 

Reason (R):Cornea in eyes helps in adjusting its focal length. 

Explanation:

• Accommodation is indeed the ability of the eye to change the focal length to focus on objects at various distances.

• But it's not the cornea that adjusts the focal length — it’s the ciliary muscles that change the shape of the lens to adjust focus.

34. Attempt either option A or B. 

 A.If Earth's atmosphere had a uniform density, how would it affect the twinkling of stars? Explain. 

Answer:
Twinkling of stars is caused by the refraction of light through the Earth's atmosphere, which has varying densities. When light from a star enters Earth's atmosphere, it passes through layers of different densities, bending multiple times in slightly different directions, making the star appear to shift and twinkle.

• If the atmosphere had uniform density, the light would travel in a straight line without continuous bending.

• Result: Twinkling would not occur, and stars would appear steady because there would be no variation in refraction.

OR 

B.Draw a ray diagram to show the rainbow formation and label the point: 

(i) where dispersion of light takes place. 

A rainbow is formed due to the combination of refraction, dispersion, and total internal reflection of sunlight in raindrops. Here's what happens:

Sunlight enters the raindrop → light is refracted and dispersed (white light splits into different colors).

The dispersed light reflects internally from the back surface of the drop (total internal reflection).

The light is refracted again as it exits the drop, and we see the colors spread out as a rainbow.

(ii) where total internal refraction occurs. 

(i) Dispersion of light: Occurs at the first surface where sunlight enters the raindrop.
→ Label this point as (i).

(ii) Total internal reflection: Happens at the inner back surface of the raindrop.
→ Label this point as (ii).

35. A. If the refractive index of a glass slab is increased, how does it affect the speed of light in the slab? 

B. Determine the speed of light in medium X if the refractive index of X with respect to vacuum is 1.90. (Speed of light in vacuum is 3 × 108 m/s). 

 Effect of Increasing Refractive Index on Speed of Light

The refractive index (n) of a medium is related to the speed of light in that medium by the formula:

$$n = \frac{c}{v}$$

Where:

• $n$ = refractive index

• $c$ = speed of light in vacuum ($3 \times 10^8 \, \text{m/s}$)

• $v$ = speed of light in the medium

✅ So, if the refractive index of a glass slab increases, the speed of light in the slab decreases.

---

B. Calculate Speed of Light in Medium X

Given:

• Refractive index $n = 1.90$

• Speed of light in vacuum $c = 3 \times 10^8 \, \text{m/s}$

Using the formula:

$$v = \frac{c}{n} = \frac{3 \times 10^8}{1.90} = 1.5789 \times 10^8 \, \text{m/s}$$

🔹 Answer: The speed of light in medium X is approximately

$$\boxed{1.58 \times 10^8 \, \text{m/s}}$$

36. Describe the important function of following parts of human eye: 

(i) Iris 

(ii) Ciliary muscle 

(iii) Retina 

---

(i) Iris

• Function: The iris controls the size of the pupil, thereby regulating the amount of light that enters the eye.

• It expands or contracts depending on the lighting conditions:

  • In bright light, the iris contracts the pupil to reduce light entry.

  • In dim light, it dilates the pupil to allow more light in.

---

(ii) Ciliary Muscle

• Function: The ciliary muscle controls the shape of the eye lens to focus light properly on the retina.

• It helps in accommodation, allowing the eye to focus on objects at different distances:

  • For near objects, the muscle contracts, making the lens thicker.

  • For distant objects, it relaxes, making the lens thinner.

---

(iii) Retina

• Function: The retina is the light-sensitive layer at the back of the eye that converts light into electrical signals.

• It contains photoreceptor cells:

  • Rods – detect light intensity (black and white vision; useful in dim light).

  • Cones – detect color and function in bright light.

• The signals from the retina are sent to the brain via the optic nerve, allowing us to perceive images.

---

37. An object is placed 40 cm from a convex lens and the image is formed 20 cm on the other side of the lens. Calculate the focal length of the lens, nature of image and magnification. 

---

Given:

• Object distance, ( u = -40 , \text{cm} ) (since the object is placed on the left side, we take it as negative in lens formula convention)

• Image distance, ( v = +20 , \text{cm} ) (image is on the other side of the lens, so positive for a convex lens)

---

Step 1: Find the focal length ( f ) using the lens formula:

[
\f1​=v1​−u1​

Substitute the values:

$$\frac{1}{f} = \frac{1}{20} - \frac{1}{-40} = \frac{1}{20} + \frac{1}{40} = \frac{2}{40} + \frac{1}{40} = \frac{3}{40}$$

So,

$$f = \frac{40}{3} = 13.33 \, \text{cm}$$

---

Step 2: Nature of the image

• Since ( v = +20 , \text{cm} ), the image is real (formed on the opposite side of the lens).

• The image distance is positive, so the image is inverted.

• The image is closer to the lens than the object, so it is smaller than the object.

---

Step 3: Calculate magnification ( m )

[
m = \frac{v}{u} = \frac{20}{-40} = -0.5
]

• The negative sign indicates the image is inverted.

• Magnification of 0.5 means the image is half the size of the object.

---

Final answers:

Quantity	Value
Focal length ( f )	+13.33 cm
Nature of image	Real, inverted, smaller
Magnification ( m )	-0.5

---

38. Nisha, after her class room discussion understands that in some refractive defects of vision, the image is formed either in front of or behind the retina. She wants to understand how the use of appropriate lenses can help to overcome these problems. 

A. In which defect of the vision, the image is formed behind the retina? Name the corrective lens used for correction. 

B. Enlist two causes that may result in the defect of vision where the image is formed in front of the retina. Attempt either subpart C or D. 

C. Draw ray diagram showing: 

 (i) a myopic eye and 

(ii) its correction using appropriate lens. 

---

A. Defect where the image is formed behind the retina

• Defect: Hypermetropia (Farsightedness)

• Explanation: In hypermetropia, the eye is too short or the lens is too weak, so the image forms behind the retina.

• Corrective lens: Convex lens (converging lens)

  • This lens helps to converge the light rays before they enter the eye so that the image can be formed correctly on the retina.

---

B. Defect where the image is formed in front of the retina

• Defect: Myopia (Nearsightedness)

• Causes:

  1. Elongated eyeball (eyeball is longer than normal).

  2. Excessively curved lens or strong lens power.

• Due to this, light rays converge before reaching the retina.

---

C. Ray diagram for myopia and its correction

I’ll describe it here, and if you want, I can create the diagram for you:

---

(i) Ray diagram showing a myopic eye

• Parallel rays from a distant object converge in front of the retina because the eyeball is too long.

• The image formed is blurry on the retina.

(ii) Ray diagram showing correction using concave lens

• The concave lens diverges the incoming light rays before they enter the eye.

• This makes the rays appear to come from a nearer point, allowing the myopic eye to focus the image correctly on the retina.

OR 

D. In elderly people, the power of eyes to see nearby and distant objects diminishes. 

(i) Name the defect of vision that is likely to develop in such situations and which lens is used for correction of this defect

(ii) Explain the reason for this defect. 

---

(i) Name of the defect and corrective lens

• Defect: Presbyopia

• Corrective lens: Convex lens (converging lens)

---

(ii) Reason for this defect

• As people age, the lens of the eye becomes less flexible and the ciliary muscles weaken.

• This reduces the eye’s ability to accommodate, meaning it cannot easily change the shape of the lens to focus on nearby objects.

• As a result, the near point (the closest distance at which one can see clearly) moves farther away, making it difficult to see nearby objects clearly.

• Convex lenses are used to help focus the image of nearby objects onto the retina.

OR

39. Attempt either option A or B. A. An object of 5cm in height is placed at 10 cm in front of concave mirror of focal length 15 cm. 

(i) Find the position, nature and size of image using mirror formula. 

(ii) Why concave mirror is preferred in searchlight and headlights of vehicles?

Given:

• Object height, ( h_o = 5 \text{ cm} )

• Object distance, ( u = -10 \text{ cm} ) (object is in front of the mirror, so negative by sign convention)

• Focal length, ( f = -15 \text{ cm} ) (for concave mirror, focal length is negative)

---

(i) Find the position, nature, and size of the image using mirror formula.

Mirror formula:

[
\frac{1}{f} = \frac{1}{v} + \frac{1}{u}
]
Where:

• ( v ) = image distance (to be found)

• ( u ) = object distance

• ( f ) = focal length

Substituting the values:

$$\frac{1}{-15} = \frac{1}{v} + \frac{1}{-10}$$ $$\frac{1}{v} = \frac{1}{-15} + \frac{1}{10} = -\frac{1}{15} + \frac{1}{10} = \frac{-2 + 3}{30} = \frac{1}{30}$$

v1​=−151​+101​=−151​+101​=30−2+3​=301​

So,

$$v = 30 \text{ cm}$$

The positive value of ( v ) indicates that the image is real and formed on the same side as the reflected light.

---

Magnification ( m ):

m=ho​/hi​​=−u/v​

Substituting:

$$m = -\frac{30}{-10} = 3$$

So,

$$h_i = m \times h_o = 3 \times 5 = 15 \text{ cm}$$

The image is 15 cm tall, so it is magnified and since magnification is positive, the image is erect.

---

Summary:

• Position of image ( v ) = +30 cm (in front of the mirror)

• Nature of image: Real and erect (positive magnification)

• Size of image: 15 cm (magnified 3 times)

---

(ii) Why is a concave mirror preferred in searchlight and headlights of vehicles?

Concave mirrors are preferred because:

• They converge light rays coming from the source at the focus into a parallel beam, producing a strong, focused, and intense beam of light.

• This allows the light to travel longer distances, improving visibility for vehicles.

• The concave mirror can magnify the image of the bulb filament, making the beam brighter.

 

 OR 

B. (i) Draw a labelled ray diagram for the image formation by a convex mirror when the object is placed between infinity and the pole of the mirror and describe the position, size and nature of image formed. (ii) List any two applications of convex mirror. 

Ray Diagram for Image Formation by a Convex Mirror

Setup:

• Object placed anywhere between infinity and the pole of the convex mirror.

• Convex mirror has focal length $f > 0$.

• Object distance $u < 0$ (object is always in front of the mirror, so negative by sign convention).

---

How to Draw the Ray Diagram:

1. Draw the convex mirror (curved outward surface) with the pole $P$, focal point $F$, and center of curvature $C$ behind the mirror.

2. Place the object in front of the mirror between infinity and the pole $P$.

3. Draw two rays from the top of the object:

   • Ray 1: Parallel to the principal axis → After reflection, it appears to come from the focal point $F$ behind the mirror.

   • Ray 2: Directed towards the pole $P$ → Reflects obeying the law of reflection (angle of incidence = angle of reflection).

4. The reflected rays diverge, but when extended backward (behind the mirror), they appear to meet at a point $I$.

5. The image forms at $I$, behind the mirror.

---

Labelled diagram:

(I’ll describe what the labels should be—if you want, I can generate an image too.)

• Mirror surface labeled "Convex Mirror"

• Pole $P$

• Focal point $F$ behind the mirror

• Object $O$ in front of the mirror

• Rays showing incident and reflected paths

• Image $I$ behind the mirror, smaller than object

---

Position, Size, and Nature of Image formed by Convex Mirror:

• Position: Behind the mirror, between the pole $P$ and the focal point $F$.

• Size: Smaller (diminished) compared to the object.

• Nature: Virtual, erect, and diminished.

---

(ii) Two Applications of Convex Mirrors:

1. Rear-view mirrors of vehicles: Convex mirrors give a wider field of view, allowing drivers to see more area behind and on the sides.

2. Security and surveillance: Used in shops, banks, and stores to monitor large areas due to their wide-angle view.