CBSE EXAMINATION PAPER-2025
SCIENCE
(Solved)
General Instructions :
Read the following instructions carefully and follow them :
- This question paper contains 35 questions. All questions are compulsory.
- This question paper is divided into 5 sections.
- Section A – questions number 1 to 1 are case based questions
- Section B – questions number 2 to 16 are multiple choice questions
- Section C – questions number 17 to 23 are very short answer
- Section D – questions number 24 to 30 are short answer
- Section E – questions number 31 to 35 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
A person while climbing up a rocky hill comes into a panic state and fear. His body starts reacting in a ‘flight-or-flight’ condition to adjust to the dangerous and stressful situation.
Based on the above facts, answer the questions that follow.
(1) Name the hormone secreted in the blood of the person in this situation.
[2 Marks](2) State any two responses in the body of the person as a result of the secretion of this hormone.
[1 Marks](3) How does the action of the chemical signal in terms of hormones differ from the electrical impulses via nerve cells?
[1 Marks](4) Name the source gland of the hormone secreted in this condition.
Section B
The most common method of extraction of metals from their oxide ores is :
Choose the incorrect statement about the common reaction used in hydrogenation of vegetable oils.
Match Column-I with Column-II and select the correct option from the choices provided.
The correct option is a-(iii), b-(ii), c-(i), d-(iv) because it accurately matches the items listed in Column-I with their corresponding descriptions in Column-II based on the context of bonding in carbon and the distinctions between mixtures, solutions, suspensions, and colloidal solutions.
The part of the brain which maintains the posture and balance of the body is :
The plant hormone present in greater concentration in the areas of rapidly dividing cells is :
Select a pair of bisexual flowers from the following :
The gastric glands present in the wall of the stomach release :
Absolute refractive index of water and glass is ⁴⁄₃ and ³⁄₂ respectively. If the speed of light in glass is 2 × 10⁸ m/s, the speed of light in water is:
n=vc⇒v=nc
So,
vw=c/nw=3×10power8/4/3=3×10power8×3/4=9/4×10power8
When a beam of white light passes through a region of very fine dust particles, the colour of light that scatters the most in that region is :
A wire of length ‘l’ is gradually stretched so that its length increases to 3l. If its original resistance is R, then its new resistance will be :
Which one of the following statements is not true about a bar magnet ?
Other than the abiotic components, which of the given biotic components are not required to make an aquarium with small herbivorous fishes a self-sustaining system ?
(i) Aquatic plants and aquatic animals
(ii) Terrestrial plants and terrestrial animals
(iii) Decomposers as bacteria and fungi
(iv) Consumers as clown fishes and sea urchins
Assertion (A) : Hydrogen gas is not evolved when a metal reacts with nitric acid.
Reason (R) : Nitric acid is a strong reducing agent and reduces the hydrogen produced in the reaction to water.
The correct option is 'Assertion (A) is true, but Reason (R) is false.' Assertion (A) is true because hydrogen gas is indeed not evolved when metals react with nitric acid due to its strong oxidizing properties. However, Reason (R) is false as nitric acid acts as an oxidizing agent, not a reducing agent, and it oxidizes the hydrogen gas produced to water.
Assertion (A) : In our actions of writing or talking, our nervous system communicates with the muscles.
Reason (R) : Cranial nerves and spinal nerves form the peripheral nervous system.
Assertion (A) : Use of jute bags for shopping reduces pollution.
Reason (R) : Jute is biodegradable and its bag may be reused as and when needed.
Section C
(a) Write the formula of the ions which (i) acids, and (ii) bases generate in water solutions.
(b) Dry HCl gas does not change the colour of dry litmus paper. Why ?
(a) Explain how the proteins control the ‘characteristics’ in an organism with the help of an example of ‘tallness’ trait in pea plant.
(b) Name the section of DNA that controls the ‘characteristics’ in an organism.
Section D
(a) What is a reactivity series of elements ? How is it developed ? Arrange the following elements as they are arranged in the reactivity series : Aluminum, Calcium, Copper, Lead
(b) Write balanced chemical equation to show the reaction of iron (III) oxide (Fe₂O₃)with aluminium.
On the basis of the characteristics of the processes given in the brackets in each case, differentiate between the following:
(a) Products of breakdown of pyruvate in aerobic and anaerobic respiration in human beings (product(s) of the processes)
(b) Respiration and photosynthesis in plants (gas released)
(c) Respiration in terrestrial animals and fishes (organs involved)
(b) In respiration, plants release carbon dioxide gas as a byproduct. In photosynthesis, plants release oxygen gas while making food.
(c) Terrestrial animals respire using lungs which allow exchange of gases with the air. Fishes respire using gills which extract oxygen dissolved in water.
A pure pea plant having round (R), yellow (Y) seeds is crossed with another pure pea plant having wrinkled (r), green (y) seeds. Subsequently F₁ progeny is self-pollinated to obtain F₂ progeny.
(a) What do the seeds of F₁ generation look like?
(b) Give the possible combinations of traits in seeds of F₂ generation. Also give their ratio.
(c) State the reason of obtaining seeds of new combination of traits in F₂ generation.
(b) The F₂ generation shows four types of seed traits combinations: round yellow (RY), round green (Ry), wrinkled yellow (rY), and wrinkled green (ry). Their phenotypic ratio is 9 : 3 : 3 : 1, respectively.
(c) The new combinations of traits in F₂ generation occur due to independent assortment of alleles during gamete formation. Alleles for seed shape and seed color segregate independently and recombine randomly during fertilization, producing new trait combinations.
Consider a rectangular cardboard having two holes P and Q through which a current carrying circular loop has been inserted as shown in the diagram.
(a) Make this diagram on your answer sheet and draw three magnetic field lines, one each passing through the points 1 (near P), 2 (at the centre of the loop) and 3 (near Q).
(b) List two factors on which the intensity of the magnetic field produced at the centre of the loop depends.
(c) Name the rule you will apply to determine the direction of magnetic field produced due to a current carrying straight conductor.
(b) The two factors on which the intensity of the magnetic field at the centre of the loop depends are:
1. The magnitude of the current flowing through the loop.
2. The radius of the circular loop.
(c) The rule used to determine the direction of the magnetic field produced by a current carrying straight conductor is the Right-Hand Thumb Rule. According to this rule, if the thumb of the right hand points in the direction of the current, then the curled fingers show the direction of the magnetic field around the conductor.
(a) ‘‘In a food chain energy flow is unidirectional.’’ Give two reasons for the given statement.
(b) If 10,000 J energy is available at the producer level, how much energy will be available to the secondary consumers ? Give reason to justify your answer.
In a food chain, energy flow is unidirectional due to two main reasons: Firstly, autotrophs or producers convert solar energy into chemical energy through photosynthesis. This energy is not returned to the original source but continues to flow to the next trophic levels. Secondly, as energy transitions from one trophic level to another, approximately 90% is lost at each step due to metabolic processes, heat production, and waste, which limits energy availability for subsequent consumers. For part (b), if 10,000 J of energy is available at the producer level, only about 1,000 J will be available to secondary consumers. This is derived from the 10% rule, where only 10% of the energy from one trophic level is passed to the next level, illustrating the inefficiency in energy transfer through the food chain.
Design an experimental set-up to demonstrate that ‘‘Alcohol and glucose contain hydrogen but are not categorised as acids’’. Also give the reason to justify this fact.
Section E
(i) Consider the following metals : K, Ca, Al, Cu, Ag, Fe Select from the above metals, a metal which
I. does not react with oxygen even at high temperature.
II. reacts with oxygen at ordinary temperature and forms a protective oxide layer which prevents the metal from further oxidation.
III. catches fire when kept in the open.
IV. does not burn in oxygen but the hot metal is coated with a black coloured oxide layer.
(ii) What are amphoteric oxides ? With the help of balanced chemical equations show that aluminium oxide is an amphoteric oxide.
(iii) What are alkalis ? Give one example.
I. Metal which does not react with oxygen even at high temperature is Ag (Silver). Silver is less reactive and resistant to oxidation.
II. Metal which reacts with oxygen at ordinary temperature and forms a protective oxide layer is Al (Aluminium). Aluminium oxide layer protects the metal from further oxidation.
III. Metal which catches fire when kept in the open is K (Potassium). It is highly reactive and reacts vigorously with oxygen.
IV. Metal which does not burn in oxygen but the hot metal is coated with a black coloured oxide layer is Cu (Copper). Copper forms black copper(II) oxide on heating.
(ii) Amphoteric oxides:
Amphoteric oxides are metal oxides that react with both acids and bases to form salt and water, showing both acidic and basic properties.
Aluminium oxide (Al2O3) is amphoteric because it reacts with acids and bases as shown below:
Reaction with acid (HCl): Al2O3 + 6HCl → 2AlCl3 + 3H2O
Reaction with base (NaOH): Al2O3 + 2NaOH → 2NaAlO2 + H2O
These reactions prove aluminium oxide behaves both as acid and base.
(iii) Alkalis:
Alkalis are bases that dissolve in water to give hydroxide ions (OH-) making the solution basic. An example of an alkali is sodium hydroxide (NaOH).
(i) Identify the parts ‘X’ and ‘Y’ in the figure given below :
(ii) Name the yellowish coloured structures produced by the part labelled as ‘Y’.
(iii) Write the name of the process by which these are transferred to the part labelled as ‘X’.
(iv) Explain the process of seed formation in a flowering plant.
(b) The yellowish coloured structures produced by the anther (part 'Y') are called pollen grains.
(c) The process by which pollen grains are transferred from the anther (Y) to the stigma of the pistil (X) is called pollination.
(d) Process of seed formation in a flowering plant: After pollination, a pollen grain lands on the stigma of the pistil (X). It germinates and forms a pollen tube that grows down through the style to reach the ovary. The male gametes travel through the pollen tube to the ovule inside the ovary. One male gamete fuses with the egg cell in the ovule thereby fertilization takes place and a zygote is formed. The zygote develops into an embryo. The ovule develops into a seed containing the embryo, and the ovary develops into a fruit surrounding the seed. This process of fertilization leading to seed formation ensures the formation of new plants.
(i) ‘‘In refraction of light through a rectangular glass slab, the emergent ray is always parallel to the direction of the incident ray.’’ Why ? Explain with the help of a ray diagram. What happens when a ray of light falls normally on one of the faces of a rectangular glass prism ? Draw diagram.
(ii) An object is placed at a distance of 30 cm from the optical centre of a concave lens of focal length 20 cm. Use Lens formula to determine the position of the image formed in this case.
If a ray of light falls normally (at 90 deg) on one face of a rectangular glass prism, it passes straight without any refraction because the angle of incidence is zero and light does not bend when passing through a boundary at normal incidence.
Ray diagram for refraction through glass slab:
Incident ray -> Refracted ray bending towards normal inside glass slab -> Emergent ray parallel to incident ray but shifted laterally.
Ray diagram for ray falling normally on prism:
Incident ray -> passes straight through without bending.
(ii) Given: Object distance u = -30 cm (since object distance is taken opposite to the direction of incident light), focal length f = -20 cm (concave lens has negative focal length).
Using lens formula: 1/f = 1/v - 1/u
=> 1/v = 1/f + 1/u = 1/(-20) + 1/(-30) = -1/20 - 1/30 = -(3/60 + 2/60) = -5/60 = -1/12
Therefore, v = -12 cm.
This means the image is formed at 12 cm on the same side as the object (virtual image). The image is erect, diminished, and virtual as is typical of a concave lens.
Hence, the position of the image is 12 cm in front of the lens on the same side as the object.
(i) Name the type of asexual mode of reproduction shown in the given figure.
(ii) Identify the unicellular organism in the diagram.
(iii) List any two advantages of asexual reproduction over sexual reproduction.
(iv) Name and explain any one mode of asexual reproduction observed in Hydra.
(b) The unicellular organism in the diagram is amoeba.
(c) Two advantages of asexual reproduction over sexual reproduction are:
1. It does not require a mate, so reproduction can occur rapidly and easily.
2. Offspring are genetically identical to the parent, ensuring stability of traits.
(d) One mode of asexual reproduction in Hydra is budding. In budding, a new small outgrowth called a bud develops on the body of the parent Hydra. The bud grows by cell division and develops tentacles and a mouth. Once fully formed, the bud detaches from the parent and lives independently as a new Hydra. This process allows Hydra to reproduce quickly without the need for fertilization.
(i) A student wishes to study the image formation by a concave mirror using candle flame as object. State the type of the image formed by the mirror and mention the change in the image formed, if any, that he observes when the candle flame is gradually moved away from the pole of the mirror. Draw a ray diagram to show the image formation when the object distance is nearly equal to the radius of curvature of the mirror.
(ii) A convex mirror used for rear-view on an automobile has a focal length of 3∙0 m. If a bus is located at 6∙0 m from this mirror, use mirror formula to find the position of the image of the bus as seen in the mirror.
As the candle flame is gradually moved away from the pole of the mirror from a position closer than the focal point, the image changes from being virtual, erect and enlarged (when object is closer than F), to real, inverted and magnified (between F and C), then to real, inverted and same size (at C), and then to real, inverted and diminished (beyond C).
Ray diagram: For the object at C, draw two rays from the top of the candle flame: one parallel to the principal axis which after reflection passes through the focal point, and another passing through C which reflects back on itself. The point where rays meet after reflection is the image location.
(ii) Given focal length, f = +3.0 m (for convex mirror focal length is positive), object distance, u = -6.0 m (object distances are taken negative in mirror formula conventions). Using mirror formula:
1/f = 1/v + 1/u
=> 1/v = 1/f - 1/u = 1/3.0 - 1/(-6.0) = 1/3 + 1/6 = (2 + 1)/6 = 3/6 = 1/2
=> v = 2 m
The positive value of v indicates that the image is virtual and formed behind the mirror, at 2 m from the mirror.
So, the image of the bus is formed at 2 m behind the convex mirror.
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