CBSE EXAMINATION PAPER-2024
SCIENCE
(Solved)
General Instructions :
Read the following instructions carefully and follow them :
- This question paper contains 33 questions. All questions are compulsory.
- This question paper is divided into 5 sections.
- Section A – questions number 1 to 2 are case based questions
- Section B – questions number 3 to 18 are multiple choice questions
- Section C – questions number 19 to 25 are very short answer
- Section D – questions number 26 to 31 are short answer
- Section E – questions number 32 to 33 are long answer
- There is no overall choice given in the question paper. However, an internal choice has been provided in few questions.
- Use of calculator is NOT allowed.
Section A
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](2) Which metal did not react with dilute hydrochloric acid? Give reason.
[1 Marks](3) Why is hydrogen gas not evolved when a metal reacts with dilute nitric acid? Name the ultimate products formed in the reaction
(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?
(1) Find the value of total resistance between the points A and B.
[1 Marks](2) Find the resistance between the points B and C.
(3) Calculate the current drawn from the battery, when the key is closed.
(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.
Section B
Zn + 2CH₃COOH → (CH₃COO) ₂ Zn + H₂. The above reaction is a:
An aqueous solution of a salt turns blue litmus to red. The salt could be the one obtained by the reaction of:
The metals which are found in both free state as well as combined state are:
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.
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:
Consider the following combinations of resistors: The combinations having equivalent resistance 1 is/are:
An electric iron of resistance 20 Ω draws a current of 5 A. The heat developed in the iron in 30 seconds is:
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.
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:
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.
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.
Section C
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 ?
Name the type of nutrition exhibited by Amoeba. Explain how food is taken in and digested by this organism.
Section D
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.
(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.
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 ?
(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.
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?
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).
List two roles of each of the following in human reproductive system : (i) Seminal vesicles and prostate gland (ii) Oviduct (iii) Testis
Section E
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.
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.
(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.
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