CBSE Board Class 10 Science Question Paper 2025 Set 3 (31/4/3) Available- Download Solution PDF with Answer Key

The second shell (L-shell) can hold a maximum of 8 electrons. Assuming the second shell is completely filled for an element progressing to the third shell, it has 8 electrons.
Number of electrons in the third shell = 25% of electrons in the second shell
Number of electrons in the third shell = 0.25 \(\times\) 8 = 2 electrons.
The electronic configuration of element 'M' is:
First shell (K-shell): 2 electrons
Second shell (L-shell): 8 electrons
Third shell (M-shell): 2 electrons
So, the electronic configuration is 2, 8, 2.
The atomic number of element 'M' = Total number of electrons = 2 + 8 + 2 = 12.
The element with atomic number 12 is Magnesium (Mg).
Electronic configurations of the given options:
(a) Sodium (Na, Z=11): 2, 8, 1
(b) Magnesium (Mg, Z=12): 2, 8, 2
(c) Aluminium (Al, Z=13): 2, 8, 3
(d) Calcium (Ca, Z=20): 2, 8, 8, 2
Element 'M' is Magnesium. \[ \boxed{Magnesium} \] Quick Tip: \textbf{Quick Tip:} The maximum number of electrons in the first shell (K) is 2, second shell (L) is 8, and third shell (M) is 18 (but it starts filling the fourth shell after 8 electrons in the third shell for elements like K, Ca). Calculate the number of electrons in the third shell based on the second shell's count (usually assumed to be full if the third shell is being filled). Determine the atomic number and identify the element.

In a bisexual flower, the male reproductive part is the stamen, which consists of the anther and the filament.

(a) Anther: The anther is the part of the stamen that produces pollen grains. Pollen grains contain the male gametes.
(b) Ovary: The ovary is part of the pistil (female reproductive part) and contains ovules, which house the female gamete (egg cell).
(c) Stigma: The stigma is the receptive tip of the pistil where pollen lands.
(d) Filament: The filament is the stalk that supports the anther.

The male gametes are developed within the pollen grains, which are produced in the anther. \[ \boxed{anther} \] Quick Tip: \textbf{Quick Tip:} Stamen = Anther + Filament (Male part) Pistil/Carpel = Stigma + Style + Ovary (Female part) Anthers produce pollen grains, which contain male gametes. Ovules (inside the ovary) contain the female gamete (egg cell).

Assertion (A): In large animals, oxygen cannot reach different parts of the animal's body "easily" just by simple diffusion from the surface. The surface area to volume ratio is too small, and diffusion distances are too large for efficient oxygen supply to all cells. Specialized respiratory and circulatory systems are required. So, the statement "easily" makes the assertion false.
Reason (R): Respiratory pigments (like hemoglobin in vertebrates or hemocyanin in some invertebrates) bind to oxygen in areas of high oxygen concentration (e.g., lungs, gills) and release it in areas of low oxygen concentration (body tissues). They play a crucial role in transporting oxygen. This statement is true.
Explanation: Assertion (A) is false because oxygen transport in large animals is a complex process, not "easy" by simple means. Reason (R) is a true statement about how oxygen is transported, which is precisely why such pigments are needed in large animals where simple diffusion is insufficient. The reason highlights a mechanism that makes oxygen transport possible but doesn't support the idea that it's "easy" without these mechanisms. \[ \boxed{Assertion (A) is false, but Reason (R) is true.} \] Quick Tip: \textbf{Quick Tip:} Simple diffusion is only efficient for oxygen transport over very short distances or in small organisms with a high surface area to volume ratio. Large, complex animals require specialized respiratory surfaces (lungs, gills) and circulatory systems with respiratory pigments to transport oxygen effectively to all body tissues.

Assertion (A): When white light passes through a glass prism, it splits into its constituent spectrum of colors (typically observed as violet, indigo, blue, green, yellow, orange, red - VIBGYOR). This phenomenon is called dispersion. The assertion states it splits into seven colors, which is a common way to describe the visible spectrum. This is true.
Reason (R): Different colors of light have different wavelengths. The refractive index of glass varies slightly with the wavelength of light. Violet light (shorter wavelength) has a higher refractive index in glass than red light (longer wavelength). Due to this difference in refractive index, violet light bends (deviates) more than red light when passing through a prism. Red light bends the least. This statement is true.
Explanation: The reason (R) correctly explains why dispersion (Assertion A) occurs. The differential bending of different colors of light due to their different speeds (and hence refractive indices) in the prism material is the cause of white light splitting into its spectrum. \[ \textbf{Both (A) and (R) are true, and (R) correctly explains (A).} \] Quick Tip: \textbf{Quick Tip:} Dispersion of light is the splitting of white light into its constituent colors. This occurs because the refractive index of the prism material is different for different wavelengths (colors) of light. Shorter wavelengths (like violet) bend more, and longer wavelengths (like red) bend less.

The compound commonly used in black and white photography is silver chloride (AgCl) or silver bromide (AgBr).
The reaction that occurs is a photochemical decomposition reaction, which is endothermic.
Justification:
The decomposition of silver chloride (or silver bromide) into silver and chlorine (or bromine) occurs when light energy is absorbed. For example: \[ 2AgCl(s) \xrightarrow{Sunlight} 2Ag(s) + Cl_2(g) \]
Since the reaction requires energy (in the form of light) to proceed, it is an endothermic reaction. Energy is absorbed from the surroundings (light) to break the bonds in AgCl. \[ \textbf{Compound: Silver Chloride (AgCl) or Silver Bromide (AgBr).}
\textbf{Reaction: Endothermic.}
\textbf{Justification: Requires light energy for decomposition.} \] Quick Tip: \textbf{Quick Tip:} Photochemical reactions are those initiated by the absorption of light. Endothermic reactions absorb energy from their surroundings. Exothermic reactions release energy. Decomposition reactions often require an input of energy (heat, light, or electricity).

N/A Quick Tip: \textbf{Quick Tip:} For ionic bond formation, show electron loss by metal and electron gain by non-metal leading to ion formation and then the ionic compound. Amphoteric oxides react with both acids and bases. Common examples include Al\(_2\)O\(_3\), ZnO, PbO, SnO. When writing reactions of amphoteric oxides, show one reaction with an acid and one with a base.

The tissues that form the vascular bundle in plants are:
1. Xylem
2. Phloem

Functions:
1. Xylem:

Its primary function is the transport of water and dissolved minerals from the roots to all other parts of the plant (stems, leaves).
It also provides mechanical support to the plant due to its lignified cell walls.


2. Phloem:

Its primary function is the transport of food (mainly sugars like sucrose), synthesized during photosynthesis in the leaves, to other parts of the plant where it is needed for growth or storage (e.g., roots, fruits, seeds, growing tips). This process is called translocation.
\[ \textbf{Plant Vascular Tissues:}
\textbf{Xylem (water/minerals/support), Phloem (sugars).} \] Quick Tip: \textbf{Quick Tip:} Vascular bundles are the transport system in vascular plants. Xylem transports "sap" upwards (water and minerals). Phloem transports "food" (sugars) both upwards and downwards to where it's needed.

Flow Chart Explaining Father's Role in Sex Determination:


\fbox{Human Parents
\(\downarrow\)

\begin{tabular{c c c
\fbox{\parbox{3cm{\centering Mother (Female)
Genotype: XX
Produces eggs with X chromosome only & & \fbox{\parbox{3cm{\centering Father (Male)
Genotype: XY
Produces sperms:
50% with X chromosome
50% with Y chromosome

\end{tabular
\(\swarrow \quad \searrow\)

Fertilization
\(\↙ \quad \quad \quad \↘\)

\begin{tabular{c c
\fbox{\parbox{4.5cm{\centering If egg (X) is fertilized by sperm with X chromosome:
Child's Genotype: XX
\(\downarrow\)
Sex of Child: Female & \fbox{\parbox{4.5cm{\centering If egg (X) is fertilized by sperm with Y chromosome:
Child's Genotype: XY
\(\downarrow\)
Sex of Child: Male

\end{tabular

Explanation:
Human females have two X chromosomes (XX) and produce eggs that all carry an X chromosome. Human males have one X and one Y chromosome (XY) and produce two types of sperm: half carry an X chromosome and half carry a Y chromosome.
During fertilization, if an egg (X) is fertilized by a sperm carrying an X chromosome, the resulting zygote will have an XX genotype, developing into a female child.
If an egg (X) is fertilized by a sperm carrying a Y chromosome, the resulting zygote will have an XY genotype, developing into a male child.
Since the mother can only contribute an X chromosome, the sex of the child is determined by the type of chromosome (X or Y) carried by the sperm from the father that fertilizes the egg. Therefore, the father is responsible for the sex of the child. \[ \begin{tabular}{p{\linewidth}} \textbf{Parental Contribution:}
Mother (XX) \(\rightarrow\) Only X eggs
Father (XY) \(\rightarrow\) X or Y sperm (50% each)
\textbf{Offspring Sex Determination:}
\begin{tabular}{cc} X (egg) + X (sperm) & \(\rightarrow\) XX \(\female\)
X (egg) + Y (sperm) & \(\rightarrow\) XY \(\male\)
\end{tabular}
\textbf{Critical Factor:} Paternal sperm chromosome
\end{tabular} \] Quick Tip: \textbf{Quick Tip:} Human sex determination is based on XX (female) and XY (male) chromosomes. Females are homogametic (produce only X-bearing gametes). Males are heterogametic (produce X-bearing and Y-bearing gametes). The Y chromosome carries genes responsible for male development.

The most widely used method for refining impure metals is electrolytic refining.

Description of Electrolytic Refining:
Electrolytic refining is a process used to obtain high-purity metals from impure metals. It is based on the principle of electrolysis.

Setup:

Anode: A thick block of the impure metal is made the anode.
Cathode: A thin strip of pure metal is made the cathode.
Electrolyte: A solution of a soluble salt of the metal being refined is used as the electrolyte.

Process:
When an electric current is passed through the electrolytic solution:

At the anode (impure metal), metal atoms from the impure block lose electrons (oxidize) and go into the electrolyte solution as positive ions.
Example: \(M(impure) \rightarrow M^{n+}(aq) + ne^-\)
These metal ions (\(M^{n+}\)) from the electrolyte move towards the cathode (pure metal strip).
At the cathode, the metal ions gain electrons (reduce) and get deposited as pure metal on the cathode.
Example: \(M^{n+}(aq) + ne^- \rightarrow M(pure)\)

Impurities:

Soluble impurities from the anode dissolve in the electrolyte.
Insoluble impurities from the anode settle down below the anode as anode mud or anode slime. These often contain valuable metals like gold, silver, platinum.


This method is very effective for refining metals like copper, zinc, tin, nickel, silver, gold, etc., to a high degree of purity. \[ \begin{tabular}{p{\linewidth}} \textbf{Electrolytic Refining Process:}
Anode (Impure): M → M\(^{n+}\) + ne\(^-\) (Oxidation)
Cathode (Pure): M\(^{n+}\) + ne\(^-\) → M (Reduction)
\textbf{Impurity Fate:}
• Soluble: Remain in electrolyte
• Insoluble: Anode mud (Au, Ag, Pt in Cu refining)
\end{tabular} \] Quick Tip: \textbf{Quick Tip:} Remember \textbf{OIL RIG}: Oxidation Is Loss (of electrons), Reduction Is Gain (of electrons). Or \textbf{LEO says GER}: Loss of Electrons is Oxidation, Gain of Electrons is Reduction. In electrolytic refining: Anode = Impure metal (gets thinner), Cathode = Pure metal (gets thicker). Anode mud can be a valuable byproduct.

The first step of cellular respiration is Glycolysis.
It occurs in the cytoplasm of the cell.

Equation for the process of breakdown of glucose in a human cell:

(i) In the presence of oxygen (Aerobic Respiration):
Glucose (C\(_6\)H\(_{12}\)O\(_6\)) is first broken down into pyruvate through glycolysis in the cytoplasm. Then, in the presence of oxygen, pyruvate enters the mitochondria and is completely oxidized.
The overall equation for aerobic respiration is: \[ C_6H_{12}O_6 (Glucose) + 6O_2 (Oxygen) \xrightarrow{In mitochondria} 6CO_2 (Carbon dioxide) + 6H_2O (Water) + Energy (ATP) \]
More detailed steps:

Glycolysis (in cytoplasm): Glucose \(\rightarrow\) 2 Pyruvate + Energy (ATP, NADH)
Link reaction & Krebs Cycle (in mitochondria): Pyruvate \(\rightarrow\) Acetyl-CoA \(\rightarrow\) CO\(_2\) + Energy (ATP, NADH, FADH\(_2\))
Oxidative Phosphorylation (in mitochondria): Uses O\(_2\) to produce a large amount of ATP.


(ii) Due to lack of oxygen (Anaerobic Respiration in human muscle cells - Lactic Acid Fermentation):
When there is a lack of oxygen, such as during strenuous exercise, human muscle cells undergo anaerobic respiration. Pyruvate formed during glycolysis is converted into lactic acid.
Equation: \[ C_6H_{12}O_6 (Glucose) \xrightarrow{Glycolysis in cytoplasm} 2 Pyruvate \xrightarrow{Lack of O_2 in cytoplasm} 2 Lactic Acid + Small amount of Energy (ATP) \]
Specifically for the conversion of pyruvate: \[ Pyruvic acid \xrightarrow{Lack of O_2} Lactic acid + Energy \]
The overall from glucose would be: \[ C_6H_{12}O_6 (Glucose) \xrightarrow{In cytoplasm, lack of O_2} 2C_3H_6O_3 (Lactic Acid) + Energy (2 ATP) \] \[ \begin{tabular}{p{\linewidth}} \textbf{1. Glycolysis:} Cytoplasmic glucose breakdown
\textbf{2. Aerobic:} C\(_6\)H\(_{12}\)O\(_6\) + 6O\(_2\) \(\rightarrow\) 6CO\(_2\) + 6H\(_2\)O + ATP
\textbf{3. Anaerobic:} C\(_6\)H\(_{12}\)O\(_6\) \(\rightarrow\) 2C\(_3\)H\(_6\)O\(_3\) (lactate) + 2ATP
\end{tabular} \] Quick Tip: \textbf{Quick Tip:} Glycolysis is common to both aerobic and anaerobic respiration. Aerobic respiration yields significantly more ATP than anaerobic respiration. Anaerobic respiration in human muscles produces lactic acid, which can cause muscle fatigue.

Definition of Potential Difference:
Potential difference (V) between two points in an electric circuit carrying current is defined as the work done (W) to move a unit positive charge (q) from one point to the other.
Mathematically, \(V = \frac{W}{q}\).

S.I. Unit of Potential Difference:
The S.I. unit of potential difference is the Volt (symbol: V).

Definition of Volt:
One volt is defined as the potential difference between two points in a current-carrying conductor when 1 joule of work is done to move a charge of 1 coulomb from one point to the other.
So, \(1 Volt = \frac{1 Joule}{1 Coulomb}\) or \(1 V = 1 J/C\).

Expression in terms of S.I. unit of work and charge:
As per the definition, potential difference (V) is work done (W) per unit charge (q).
The S.I. unit of work is the Joule (J).
The S.I. unit of charge is the Coulomb (C).
Therefore, the S.I. unit of potential difference (Volt) can be expressed as: \[ Volt (V) = \frac{Joule (J)}{Coulomb (C)} \] \[ [\(\bullet\)] \textbf{Definition:} Work done per unit charge [\(\bullet\)] \textbf{Mathematical Form:} \(\Delta V = \frac{W}{q}\) [\(\bullet\)] \textbf{S.I. Unit:} Volt (V) [\(\bullet\)] \textbf{Dimensional Formula:} [ML\(^2\)T\(^{-3}\)I\(^{-1}\)] [\(\bullet\)] \textbf{Base Unit Equivalence:} 1 V = 1 J/C \] Quick Tip: \textbf{Quick Tip:} Potential difference is also known as voltage. It is the "electrical pressure" or "force" that causes current to flow. Remember the relationship: \(V = W/q\).

N/A

N/A Quick Tip: \textbf{Quick Tip:} For 35(A): Pollination \(\rightarrow\) Fertilization \(\rightarrow\) Ovule to Seed \(\rightarrow\) Ovary to Fruit. For 35(B): Placenta is the lifeline between mother and fetus. If no fertilization, the uterine lining sheds (menstruation).

N/A

N/A Quick Tip: \textbf{Quick Tip:} For 36(A): Work backwards from combustion products. Dehydration of alcohols gives alkenes. Hydrogenation of alkenes gives alkanes. For 36(B): Ethanoic acid is C\(_2\)H\(_4\)O\(_2\). Esterification: Acid + Alcohol \(\rightleftharpoons\) Ester + Water. Hydrolysis is the reverse. Acids react with carbonates to give CO\(_2\).