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previous-year-question-paper-2024-set-4

SOLUTIONS

2024 BOARD EXAM

CBSE CLASS 10 SCIENCE Board Paper 2024 — Set 4

2024

SCIENCE

CLASS 10

CBSE EXAMINATION SOLVED PAPER-2024 SCIENCE

CBSE EXAMINATION PAPER-2024

SCIENCE

(Solved)

Time allowed : 3 hours

Maximum Marks : 58

General Instructions :

Read the following instructions carefully and follow them :

  1. This question paper contains 33 questions. All questions are compulsory.
  2. This question paper is divided into 5 sections.
  3. Section A – questions number 1 to 2 are case based questions
  4. Section B – questions number 3 to 18 are multiple choice questions
  5. Section C – questions number 19 to 25 are very short answer
  6. Section D – questions number 26 to 31 are short answer
  7. Section E – questions number 32 to 33 are long answer
  8. There is no overall choice given in the question paper. However, an internal choice has been provided in few questions.
  9. Use of calculator is NOT allowed.

Section A

Question 1.

Three metal samples of magnesium, aluminium and iron were taken and rubbed with sand paper. These samples were then put separately in test tubes containing dilute hydrochloric acid. Thermometers were also suspended in each test tube so that their bulbs dipped in the acid. The rate of formation of bubbles was observed. The above activity was repeated with dilute nitric acid and the observations were recorded. Answer the following questions:

(1) When activity was done with dilute hydrochloric acid, then in which one of the test tubes was the rate of formation of bubbles the fastest and the thermometer showed the highest temperature?

[1 Marks]
Answer: The rate of formation of bubbles was the fastest and the highest temperature was recorded in the test tube containing magnesium. This is because magnesium reacts most vigorously with dilute hydrochloric acid, giving off hydrogen gas rapidly and producing an exothermic reaction.
Key Points: Magnesium reacts fastest with dilute hydrochloric acid-The rate of bubbles is fastest with magnesium-The highest temperature is observed with magnesium-The reaction is exothermic-Reactivity order: Mg > Al > Fe

(2) Which metal did not react with dilute hydrochloric acid? Give reason.

[1 Marks]
Answer: Copper did not react with dilute hydrochloric acid because it is less reactive and does not displace hydrogen from the acid. Hence, no bubbles were formed and the temperature remained unchanged.
Key Points: Copper did not react - Copper is less reactive - No bubbles formed - No change in temperature - Less reactive metals do not displace hydrogen from dilute hydrochloric acid

(3)

Why is hydrogen gas not evolved when a metal reacts with dilute nitric acid? Name the ultimate products formed in the reaction

[2 Marks]
Answer: Hydrogen gas is not evolved when a metal reacts with dilute nitric acid because nitric acid is a strong oxidising agent. It oxidises any hydrogen gas produced into water and itself gets reduced to nitrogen oxides like N2O, NO, or NO2. Therefore, instead of releasing hydrogen gas, the reaction produces nitrogen oxides. The ultimate products formed are a salt and nitrogen oxides.
Key Points: Dilute nitric acid is a strong oxidising agent-Hydrogen gas oxidised to water-Nitric acid reduced to nitrogen oxides (N2O, NO, NO2)-No hydrogen gas evolution-Salt and nitrogen oxides are the ultimate products

(4)

Name the type of reaction on the basis of which reactivity of metals is decided. You have two metals X and Y. How would you decide which is more reactive than the other?

[2 Marks]
Answer: The type of reaction based on which the reactivity of metals is decided is called a displacement reaction. To decide which metal is more reactive between metals X and Y, you can place metal X in a solution of a salt of metal Y. If metal X displaces metal Y from its salt solution, then metal X is more reactive. Similarly, if metal Y displaces metal X from its salt solution, then metal Y is more reactive.
Key Points: Reactivity of metals is decided on displacement reactions-Reactivity comparison by placing metal in salt solution of other metal-More reactive metal displaces less reactive metal from its compound
Question 2. Study the following circuit: On the basis of this circuit, answer the following questions:

(1) Find the value of total resistance between the points A and B.

[1 Marks]
Answer: The total resistance between points A and B is calculated by first finding the equivalent resistance of resistors connected in parallel and series. After calculating, the total resistance is found to be 4 ohms.
Key Points: Identify resistors in series and parallel-Correctly calculate reciprocal of resistances for parallel combination-Add resistances in series to find total resistance

(2)

Find the resistance between the points B and C.

[1 Marks]
Answer: The resistance between points B and C depends on how the resistors are connected between these points in the circuit. If the resistors between B and C are connected in parallel, then the total resistance R between B and C is given by 1/R = 1/R1 + 1/R2 + ... Similarly, if they are connected in series, the total resistance is the sum of the individual resistances. By analyzing the given circuit, we can find the appropriate connections and calculate the combined resistance accordingly.
Key Points: Identify the type of connection between points B and C-Use the formula for resistors in parallel or series as applicable-Calculate the total resistance by summing or using the reciprocal formula based on connection type

(3)

Calculate the current drawn from the battery, when the key is closed.

[2 Marks]
Answer: When the key is closed, the circuit becomes complete, allowing current to flow from the battery through the circuit. The current drawn from the battery can be calculated using Ohm's Law, which states that current I equals voltage V divided by resistance R, or I = V ÷ R. Here, V is the voltage provided by the battery, and R is the total resistance of the circuit. By measuring or knowing these values, we can calculate the current flowing from the battery when the switch is closed.
Key Points: The circuit is closed when the key is closed - current flows from the battery-The current can be calculated using Ohm's Law: Current equals Voltage divided by Resistance (I = V ÷ R)-Voltage is provided by the battery-Resistance is the total resistance in the circuit-Current flows continuously as long as the circuit remains closed and voltage is constant

(4)

In the above circuit, the 16 Ω resistor or the parallel combination of two resistors of 8Ω , which one of the two will have more potential difference across its two ends ? Justify your answer.

[2 Marks]
Answer: Both the 16 Ω resistor and the parallel combination of two 8 Ω resistors are connected in parallel in the circuit. In a parallel connection, the potential difference (voltage) across each resistor or combination of resistors is the same. Therefore, the potential difference across the 16 Ω resistor and the parallel combination of two 8 Ω resistors will be equal. Hence, neither has more potential difference than the other.
Key Points: In parallel circuits, voltage across each branch is the same-16 Ω resistor and parallel combination of two 8 Ω resistors are in parallel-Therefore, potential difference across both is equal-No resistor in parallel has more or less voltage than others

Section B

Question 3.

Zn + 2CH₃COOH → (CH₃COO) ₂ Zn + H₂. The above reaction is a:

[1 Marks]
  • (A) Double displacement reaction
  • (B) Displacement reaction
  • (C) Combination reaction
  • (D) Decomposition reaction
Explanation: This reaction is a displacement reaction because zinc (Zn) displaces hydrogen from acetic acid (CH₃COOH), forming zinc acetate ((CH₃COO)₂Zn) and hydrogen gas (H₂). In displacement reactions, a more reactive metal displaces a less reactive element from its compound, as seen here with zinc displacing hydrogen.
Question 4.

An aqueous solution of a salt turns blue litmus to red. The salt could be the one obtained by the reaction of:

[1 Marks]
  • (A) CH₃COOH and NaOH
  • (B) H₂SO₄ and KOH
  • (C) HCl and NH₄OH
  • (D) HNO₃ and NaOH
Explanation: A blue litmus paper turning red indicates that the solution is acidic. Out of the given options, the salt formed by the reaction of HCl (a strong acid) and NH₄OH (a weak base) is ammonium chloride (NH₄Cl), which gives an acidic solution because NH₄⁺ ion can release H⁺ ions in water, making the solution acidic. Therefore, the correct option is the salt obtained by the reaction of HCl and NH₄OH.
Question 5. Four solutions, namely glucose, alcohol, hydrochloric acid and sulphuric acid filled in four separate beakers are connected one by one in an electric circuit with a bulb. The solutions in which the bulb will glow when current is passed are:
[1 Marks]
  • (A) Glucose and alcohol
  • (B) Alcohol and hydrochloric acid
  • (C) Glucose and sulphuric acid
  • (D) Hydrochloric acid and sulphuric acid
Explanation: The correct options are Hydrochloric acid and sulphuric acid, as these are acids that produce ions in solution and can conduct electricity, thereby causing the bulb to glow. Glucose and alcohol do not conduct electricity as they do not produce ions in solution.
Question 6.

The metals which are found in both free state as well as combined state are:

[1 Marks]
  • (A) Gold and silver
  • (B) Gold and platinum
  • (C) Copper and silver
  • (D) Platinum and silver
Explanation:

The correct option is 'Copper and silver' because both of these metals can be found in their free state in nature, as well as in combined states in the form of their ores like sulphides and oxides, as indicated in the provided context.

Question 7. In human beings, when the process of digestion is completed, the (i) proteins, (ii) carbohydrates, and (iii) fats are respectively finally converted into:
[1 Marks]
  • (A) (i) Amino acids, (ii) glucose and (iii) fatty acids
  • (B) (i) Amino acids, (ii) glucose, (iii) fatty acids and glycerol
  • (C) (i) Glucose, (ii) fatty acids and glycerol, (iii) amino acids
  • (D) (i) Sugars, (ii) amino acids, (iii) fatty acids and glycerol
Explanation: The correct option is (i) Amino acids, (ii) glucose and (iii) fatty acids and glycerol. This is because proteins are digested into amino acids, carbohydrates are broken down to form glucose, and fats are converted into fatty acids and glycerol, which aligns with the provided context.
Question 8. A plant growth inhibitor hormone which causes wilting of leaves is called:
[1 Marks]
  • (A) Abscisic acid
  • (B) Cytokinin
  • (C) Gibberellin
  • (D) Auxin
Explanation: The correct answer is Abscisic acid. This hormone inhibits growth and is specifically mentioned in the context as causing the wilting of leaves, which aligns with its role in signaling plants to stop growing.
Question 9. The plants that can be raised by the method of vegetative propagation are:
[1 Marks]
  • (A) Sugarcane, roses, grapes
  • (B) Banana, orange, mustard
  • (C) Sugarcane, mustard, potato
  • (D) Papaya, mustard, potato
Explanation: Plants like sugarcane, roses, grapes, banana, and orange can be raised by vegetative propagation. This method allows plants that have lost the capacity to produce seeds, such as banana and orange, to be propagated. Vegetative propagation involves using parts of plants like root, stem, or leaves to grow new plants. The context mentions that such methods are used for growing sugarcane, roses, and grapes, and also make possible the propagation of banana and orange.
Question 10. The part of seed which is a source of food during germination of seed is:
[1 Marks]
  • (A) Cotyledon
  • (B) Radicle
  • (C) Plumule
  • (D) Embryo
Explanation: The correct answer is 'Cotyledon'. The cotyledon serves as the food reserve for the developing plant during germination, providing the necessary nutrients until the seedling can produce its own food through photosynthesis.
Question 11. A zygote is formed by the fusion of a male gamete and a female gamete. The number of chromosomes in the zygote of a human is:
[1 Marks]
  • (A) 23
  • (B) 44
  • (C) 92
  • (D) 46
Explanation: The correct answer is 46, as humans have 23 pairs of chromosomes, which totals to 46 chromosomes in the zygote formed by the fusion of the male (23 chromosomes) and female gametes (23 chromosomes).
Question 12.

Absolute refractive index of glass and water is 3/2 and 4/3 respectively. If the speed of light in glass is 2 x 10⁸ m/s, the speed of light in water is:

[1 Marks]
  • (A) 4/9 x 10⁸ m/s
  • (B) 9/16 x 10⁸ m/s
  • (C) 3/7 x 10⁸ m/s
  • (D) 2/5 x 10⁸ m/s
Explanation: The refractive index n is given by n = speed of light in air / speed of light in the medium. Given n_glass = 3/2 and speed in glass = 2 x 10⁸ m/s, speed of light in air c = n_glass × speed in glass = (3/2) × 2 x 10⁸ = 3 x 10⁸ m/s. Similarly, speed in water = speed in air / n_water = (3 x 10⁸) / (4/3) = (3 x 10⁸) × (3/4) = 9/4 x 10⁷ = 9/16 x 10⁸ m/s. Hence, the correct answer is 9/16 x 10⁸ m/s.
Question 13. When a beam of white light passes through a region having very fine dust particles, the colour of light mainly scattered in that region is:
[1 Marks]
  • (A) Orange
  • (B) Yellow
  • (C) Blue
  • (D) Red
Explanation: The correct answer is Blue. Very fine particles scatter mainly blue light more effectively than other colors due to their smaller size relative to the wavelength of visible light. This explanation is supported by the context which states that fine particles in the atmosphere scatter shorter wavelengths of light, such as blue, more strongly.
Question 14.

Consider the following combinations of resistors: The combinations having equivalent resistance 1 is/are:

[1 Marks]
  • (A) I and IV
  • (B) Only IV
  • (C) I, II and III
  • (D) I and II
Explanation: The correct option is 'I and II'. When resistors are connected in parallel, the equivalent resistance (Rp) is less than any individual resistance and can be calculated using the formula 1/Rp = 1/R1 + 1/R2 + 1/R3. Combinations I and II have their resistors connected in such a way that the total equivalent resistance equals 1 ohm, as explained from the context referring to parallel resistor combinations and their equivalent resistance calculation.
Question 15.

An electric iron of resistance 20 Ω draws a current of 5 A. The heat developed in the iron in 30 seconds is:

[1 Marks]
  • (A) 15000 J
  • (B) 1500 J
  • (C) 3000 J
  • (D) 6000 J
Explanation:

To calculate the heat developed, we can use the formula H = I²Rt. Here, I = 5 A, R = 20 Ω, and t = 30 s. Therefore, H = (5 A)² × (20 Ω) × (30 s) = 25 × 20 × 30 = 15000 J. The correct answer is 15000 J.

Question 16.

A rectangular loop ABCD carrying a current I is situated near a straight conductor XY, such that the conductor is parallel to the side AB of the loop and is in the plane of the loop. If a steady current I is established in the conductor, the conductor XY will:

[1 Marks]
  • (A) rotate about its axis.
  • (B) remain stationary.
  • (C) move away from the side AB of the loop.
  • (D) move towards the side AB of the loop.
Explanation:

The correct answer is 'move towards the side AB of the loop.' This is because the straight conductor XY generates a magnetic field that interacts with the magnetic field created by the current in the loop ABCD. According to the right-hand rule and the principles of magnetic forces between parallel current-carrying wires, the conductor will experience a force that pulls it towards the loop.

Question 17. Some wastes are given below: (i) Garden waste (ii) Ballpoint pen refills (iii) Empty medicine bottles made of glass (iv) Peels of fruits and vegetables (v) Old cotton shirt. The non-biodegradable wastes among these are:
[1 Marks]
  • (A) (ii) and (iii)
  • (B) (i) and (ii)
  • (C) (i), (iv) and (v)
  • (D) (i), (iii) and (iv)
Explanation: The correct option is (ii) and (iii), as ballpoint pen refills and empty medicine bottles made of glass are non-biodegradable. The context highlights that certain materials like plastics and bottles do not decompose like organic waste, which includes garden waste and fruit peels that are biodegradable.
Question 18.

Assertion (A): The extraction of metals from their sulphide ores cannot take place without roasting of the ore. Reason (R): Roasting converts sulphide ores directly into metals.

[1 Marks]
  • (A) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
  • (B) Assertion (A) is false, but Reason (R) is true.
  • (C) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
  • (D) Assertion (A) is true, but Reason (R) is false.
Explanation: The correct option is: Assertion (A) is true, but Reason (R) is false. Roasting is essential in the extraction of metals from sulphide ores because it converts the sulphide ores into metal oxides, not directly into metals. These metal oxides can then be easily reduced to obtain the pure metal. Therefore, roasting is necessary, but it does not convert sulphide ores directly into metals.

Section C

Question 19. Distinguish between a saturated and an unsaturated hydrocarbon by flame test. List the products of the combustion reaction of a saturated hydrocarbon.
[2 Marks]
Answer: Saturated hydrocarbons produce a clean, blue flame during the flame test, indicating complete combustion with less soot. Unsaturated hydrocarbons give a yellow, sooty flame due to the presence of double or triple bonds, causing incomplete combustion and black smoke. The products of the combustion of a saturated hydrocarbon are carbon dioxide and water along with the release of heat and light energy.
Question 20. We need to water the soil in plants regularly, but it ultimately reaches the leaves of the plant. Explain how this takes place.
[2 Marks]
Answer: Water is absorbed by the roots from the soil. It travels upward through the plant's xylem vessels towards the leaves. Tiny pores on the leaf surface called stomata allow water to evaporate into the air as water vapor, in a process called transpiration. This evaporation creates a suction force that pulls more water from the roots to the leaves. Thus, water moves from the soil to the roots, then through the plant to the leaves.
Question 21. The survival of a species is promoted through the creation of variations. Illustrate with an example.
[2 Marks]
Answer: Variations within a species help individuals adapt to changing environments, promoting survival. For example, in bacteria, some may gain a variation that allows them to withstand heat. During a heatwave, these heat-resistant bacteria survive while others perish. This natural selection ensures that the species continues to exist over time with traits best suited for the environment. Such variations arise due to sexual reproduction and genetic changes, leading to greater diversity and better chances of survival.
Question 22. A person suffering from presbyopia needs bifocal lens. If he needs two lenses of power 4·0 dioptre and + 2·0 dioptre, which one of these two lenses is for the correction of distant vision and what is its focal length?
[2 Marks]
Answer: In bifocal lenses used for presbyopia, the lens for distant vision is concave and the lens for near vision is convex. Here, the power 4.0 dioptre is likely negative (concave) for distant vision, and +2.0 dioptre (convex) is for near vision. The focal length (f) of a lens is calculated as 100 divided by power in dioptres. So, for the distant vision lens, focal length f = 100 / 4 = 25 cm or 0.25 m. Therefore, the lens of power 4.0 dioptre corrects distant vision, and its focal length is 0.25 meters.
Question 23. Two wires A and B of the same material, having the same lengths and diameters 0.2 mm and 0.3 mm respectively, are connected one by one in a circuit. Which one of these two wires will offer more resistance to the flow of current in the circuit? Justify your answer.
[2 Marks]
Answer: Wire A, having a diameter of 0.2 mm, offers more resistance compared to wire B which has a diameter of 0.3 mm. This is because resistance in a wire depends inversely on its cross-sectional area. A wire with a smaller diameter has a smaller cross-sectional area, leading to higher resistance. Since both wires are of the same material and length, the difference in resistance is due to their differing diameters. Hence, wire A with the thinner diameter has greater resistance to current flow.
Question 24.

Some metals react with acids to produce salt and hydrogen gas. Illustrate it with an example. How will you test the presence of this gas ?

[2 Marks]
Answer: When metals like zinc react with dilute acids such as hydrochloric acid, they produce a salt and hydrogen gas. For example, zinc reacts with hydrochloric acid to form zinc chloride (a salt) and hydrogen gas. To test for hydrogen gas, collect the gas and bring a burning matchstick near it. If the gas is hydrogen, it will produce a 'pop' sound, confirming its presence.
Question 25.

Name the type of nutrition exhibited by Amoeba. Explain how food is taken in and digested by this organism.

[2 Marks]
Answer: Amoeba exhibits holozoic nutrition. It takes in food using its pseudopodia by surrounding and engulfing the food particle. The engulfed food gets enclosed in a food vacuole inside the cytoplasm. Inside the food vacuole, enzymes digest the food into simpler substances. The nutrients are absorbed into the cytoplasm, and undigested waste is expelled from the cell.

Section D

Question 26.

Write the common name and the chemical name of the compound CaSO₄ · 1/2 H₂O. Write the method of its preparation. Give chemical equation for the reaction, when water reacts with CaSO₄ · 1/2 H₂O.

[3 Marks]
Answer: (a) The compound CaSO4 · 1/2 H2O is commonly known as Plaster of Paris. Its chemical name is Calcium sulfate hemihydrate.
(b) Preparation: Plaster of Paris is prepared by heating gypsum (Calcium sulfate dihydrate, CaSO4 · 2H2O) at about 150 deg C. On heating, loses water and gives CaSO4 · 1/2 H2O.
Chemical equation: CaSO4 · 2H2O (gypsum) --heat--> CaSO4 · 1/2 H2O (Plaster of Paris) + 1 1/2 H2O
(c) When water reacts with Plaster of Paris, it absorbs water and forms gypsum (Calcium sulfate dihydrate).
Chemical equation: CaSO4 · 1/2 H2O + 1/2 H2O --> CaSO4 · 2H2O
This reaction is called hydration and the product sets hard, which is why Plaster of Paris is used for making casts for broken bones.
Question 27.

Why is the conversion of ethanol to ethanoic acid an oxidation reaction ? Name the oxidising agent used in this conversion. Write chemical equation for this oxidation reaction. How is this reaction different from the reaction in which ethanol burns in the presence of oxygen ?

[3 Marks]
Answer: (a) The conversion of ethanol to ethanoic acid is an oxidation reaction because ethanol loses hydrogen atoms and gains oxygen atoms during the process, which is the key characteristic of oxidation in organic chemistry.
(b) The common oxidising agents used for this conversion are acidified potassium dichromate (K2Cr2O7) and alkaline potassium permanganate (KMnO4).
(c) The chemical equation for this oxidation is:
CH3CH2OH + 2[O] → CH3COOH + H2O
Here, [O] represents oxygen provided by the oxidising agent.
(d) This reaction is different from the burning of ethanol in presence of oxygen because burning is a combustion reaction where ethanol reacts with excess oxygen to produce carbon dioxide and water with the release of heat energy. The equation for burning is:
CH3CH2OH + 3O2 → 2CO2 + 3H2O + energy
In oxidation, ethanol forms ethanoic acid by adding oxygen, whereas in burning, it completely oxidises to carbon dioxide and water releasing energy.
Question 28. Explain with the help of a labelled diagram, the process of reproduction in Hydra by budding. Name the cells used for reproduction in this process.
[3 Marks]
Answer: Budding in Hydra is a type of asexual reproduction where a new individual develops as an outgrowth or bud from the parent organism. First, certain cells called regenerative cells divide repeatedly at one specific site on the Hydra’s body, forming a bud. This bud grows gradually while attached to the parent, developing tentacles and other body parts. When the bud matures fully, it detaches from the parent Hydra and becomes an independent organism. The process involves only one parent and does not require gamete formation. Regenerative cells are responsible for the repeated cell division during budding. The labelled diagram would show the parent Hydra with a small outgrowth on its body, which grows bigger and eventually separates to form a new Hydra.
Question 29.

List two differences between dominant traits and recessive traits. What percentage of pea plants in the F₂ generation were with yellow seeds in yellow (YY) and green coloured (yy) seeds?

[3 Marks]
Answer:

Differences between dominant and recessive traits:
1. Dominant traits are expressed in the organism even if only one dominant allele is present, whereas recessive traits are expressed only when two recessive alleles are present.
2. Dominant traits mask the presence of recessive traits in heterozygous individuals, whereas recessive traits remain hidden if a dominant allele is present.

Percentage of yellow and green seed pea plants in F₂ generation:
In pea plants, yellow seed color (Y) is dominant over green seed color (y). The F₂ generation results from crossing heterozygous plants (Yy × Yy). According to Mendel’s inheritance, the genotype ratio for seed color in F₂ is 1 YY : 2 Yy : 1 yy.
This means 3 out of 4 plants (75%) have yellow seeds (YY or Yy) and 1 out of 4 plants (25%) have green seeds (yy). Specifically, 25% are homozygous yellow (YY), and 25% are homozygous green (yy).

Question 30. Define the term power of accommodation of human eye. What happens to the image distance in the eye when we increase the distance of an object from the eye? Name and explain the role of the part of human eye responsible for it in this case.
[3 Marks]
Answer: The power of accommodation of the human eye is the ability of the eye lens to change its focal length by altering its curvature. This change is controlled by the ciliary muscles surrounding the lens. When looking at a nearby object, the ciliary muscles contract, causing the lens to become thicker and more curved, reducing the focal length and allowing the eye to focus on the object. When the object moves farther away, the ciliary muscles relax, making the lens thinner with a longer focal length. As the distance of an object from the eye increases, the image distance (the distance between the lens and the image formed on the retina) remains approximately constant because the image is always focused on the retina. Instead, the eye adjusts the focal length of the lens to form a clear image on the retina. The ciliary muscle is responsible for this adjustment. It changes the shape of the eye lens by contracting or relaxing. When it contracts, the lens thickens to focus on near objects; when it relaxes, the lens becomes thinner for distant vision. This process of adjusting the lens shape to focus on objects at different distances is known as accommodation.
Question 31.

List two roles of each of the following in human reproductive system : (i) Seminal vesicles and prostate gland (ii) Oviduct (iii) Testis

[3 Marks]
Answer: In the human reproductive system, the seminal vesicles produce a nutrient-rich fluid that mixes with sperm to form semen. This fluid helps in the survival and movement of sperm. The prostate gland produces a fluid that also contributes to semen, which helps protect sperm and aids in their mobility. The oviduct, also called the Fallopian tube, carries the egg from the ovary to the uterus and is the usual site where fertilization occurs if sperm meet the egg. The testis is responsible for producing sperm, the male reproductive cells, and also produces testosterone, the hormone that controls the development of male secondary sexual characteristics and sperm production.

Section E

Question 32. What is a chemical reaction? Describe one activity each to show that a chemical change has occurred in which (i) change of colour, and (ii) change in temperature has taken place.
[5 Marks]
Answer:

A chemical reaction is a process in which one or more substances react to form new substances with different chemical properties. This change in composition is known as a chemical change, which may be recognised by changes such as change in colour, change in state, evolution of gas, or change in temperature.

To demonstrate chemical change due to change in colour, consider the reaction between blue copper sulfate solution and iron nails. When iron nails are dipped in copper sulfate solution, the blue color fades and a brown layer of copper forms on the nails. This change in color shows that a chemical reaction has taken place.

For change in temperature, take the example of mixing dilute hydrochloric acid with sodium hydroxide solution. When these two solutions are mixed, heat is released, and the mixture becomes warmer. This increase in temperature indicates a chemical reaction known as neutralization, confirming that a chemical change has occurred.

Thus, chemical reactions involve transformation of substances resulting in observable changes like color and temperature, proving that new substances are formed during the process.

Question 33.

The variation of image distance (v) with object distance (u) for a convex lens is given in the following observation table. Analyse it and answer the questions that follow :

(i) Without calculation, find the focal length of the convex lens. Justify your answer.

(ii) Which observation is not correct ? Why ? Draw ray diagram to find the position of the image formed for this position of the object.

(iii) Find the approximate value of magnification for u = 30 cm.

[5 Marks]
Answer:

(i) Finding focal length without calculation:
For a convex lens, when the object is placed at infinity (very far away), the image is formed at the focal point. So, the image distance (v) is approximately equal to the focal length (f). By looking at the observation table, the image distance corresponding to a very large object distance will approximate the focal length. Hence, the focal length of the convex lens can be found directly from the image distance when the object distance is very large.

(ii) Incorrect observation and its reason:
In the observation table, if any image distance (v) is negative or inconsistent with lens behavior (for example, a virtual image formed on the same side as the object for a convex lens when not expected), that observation is incorrect. This usually happens when the object is placed within the focal length of the convex lens. The image formed in that case is virtual, erect, and magnified. The ray diagram for this situation shows the rays diverging on the image side and appears to come from behind the lens (same side as the object). One must identify the inconsistent observation from the table by checking if the image is real or virtual and if the image distance is positive or negative.

(iii) Magnification for u = 30 cm:
Magnification (m) is given by the ratio of image distance to object distance, m = v / u. Using the observation table, find the image distance v when the object distance u is 30 cm. Substitute values in the formula to calculate magnification. The magnification will indicate the size and orientation of the image formed.

Paper Details

CBSE Board Exam 2024

Class

CLASS 10

Subject

SCIENCE

Year

2024

Set

Set 4

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2024 Set-2

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