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Madhyamik Class 10 Physical Science Solved Paper 2025

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Madhyamik Class 10 Physical Science Question Paper Solved 2025

Physical Science

Time – Three Hours Fifteen Minutes

(First fifteen minutes for reading the Question paper only)

Full Marks – 90

(For Regular and Sightless Regular Candidates)

Full Marks – 100

(For External and sightless External Candidates)


Special credits will be given for answers which are brief and to the point.

Marks will be deducted for spelling mistakes, untidiness and bad handwriting.


Figures in the margin indicate full marks for each Question.

Only the External Candidates will answer Group -E

Figures in the margin indicate full marks for each Question.

Question 1 | Group A
Multiple choice Questions. Four alternative answers are given for each of the following Questions. Write the correct one. [1 × 15 = 15]
1.1

Which of the following rays has the highest capacity to ionise gases?

(a) α-ray

(b) γ-ray

(c) β-ray

(d) Light ray

EasyAtomic Nucleus01 Mark marks
Answer

(a) α-ray

Explanation

Alpha (α) rays have the highest ionising power because they possess a +2 charge and comparatively large mass.

1.2

The characteristics of a fuse wire are:

(a) High resistance, low melting point

(b) Low resistance, high melting point

(c) Low resistance, low melting point

(d) High resistance, high melting point

EasyCurrent Electricity01 Mark marks
Answer

(a) High resistance, low melting point

Explanation

A fuse wire is a safety device used to protect electrical circuits from damage caused by overloading or short circuits.

It should have:

  • High resistance, so that it heats up quickly when excessive current flows.
  • Low melting point, so that it melts easily and breaks the circuit before the appliances or wiring are damaged.
1.3

How many groups are there in the modern long form of the periodic table?

(a) 7

(b) 17

(c) 15

(d) 18

EasyPeriodic Table01 Mark marks
Answer

(d) 18

Explanation

The modern long form of the periodic table consists of 18 vertical columns, called groups, and 7 horizontal rows, called periods.

1.4

Which of the following compounds has no existence as a separate and distinct molecule?

(a) H₂S

(b) CHCl₃

(c) NO₂

(d) NaCl

EasyIonic and Covalent Bond01 Mark marks
Answer

(d) NaCl

Explanation

Sodium chloride (NaCl) is an ionic compound. It does not exist as individual molecules; instead, it consists of a three-dimensional ionic crystal lattice made up of Na⁺ and Cl⁻ ions held together by strong electrostatic forces.

1.5

Which acid is a weak electrolyte?

(a) CH₃COOH

(b) H₂SO₄

(c) HNO₃

(d) HCl

EasyElectrolysis01 Mark marks
Answer

(a) CH₃COOH (Acetic acid)

Explanation

A weak electrolyte is a substance that partially ionises in aqueous solution, producing a relatively small number of ions. Therefore, it conducts electricity poorly.

1.6

What colour is produced when H₂S gas is passed through a solution of potassium dichromate acidified with dilute sulphuric acid?

(a) Orange

(b) Violet

(c) Green

(d) Deep blue

ModerateInorganic Chemistry01 Mark marks
Answer

(c) Green

Explanation

Hydrogen sulphide (H₂S) acts as a reducing agent. When it is passed through acidified potassium dichromate (K₂Cr₂O₇) solution, the orange-coloured dichromate ions (Cr₂O₇²⁻) are reduced to green-coloured chromium(III) ions (Cr³⁺).

1.7

Which of the following is an ore of Aluminium?

(a) Hematite

(b) Bauxite

(c) Malachite

(d) Chalcocite

EasyMetallurgy01 Mark marks
Answer

(b) Bauxite

Explanation

Bauxite is the chief ore of aluminium. It mainly contains hydrated aluminium oxide (Al₂O₃·2H₂O) along with impurities such as iron oxide and silica.

1.8

Identify the triple bond unsaturated hydrocarbon:

(a) CH₄

(b) C₂H₆

(c) C₂H₄

(d) C₂H₂

EasyOrganic Chemistry01 Mark marks
Answer

(d) C₂H₂

Explanation

A triple bond unsaturated hydrocarbon is known as an alkyne. Alkynes contain at least one carbon-carbon triple bond (C≡C).

Among the given compounds:

  • CH₄ (Methane) → Saturated hydrocarbon (single bonds only)
  • C₂H₆ (Ethane) → Saturated hydrocarbon (single bonds only)
  • C₂H₄ (Ethene) → Unsaturated hydrocarbon (double bond)
  • C₂H₂ (Ethyne/Acetylene) → Unsaturated hydrocarbon (triple bond)
1.9

Identify the greenhouse gas:

(a) Oxygen

(b) Hydrogen

(c) Water vapour

(d) Nitrogen

EasyConcerns about our Environment01 Mark marks
Answer

(c) Water vapour

Explanation

A greenhouse gas is a gas that absorbs and traps heat (infrared radiation) in the Earth’s atmosphere, causing the greenhouse effect and contributing to global warming.

1.10

Volume of 22 g of CO₂ at S.T.P. is [C = 12, O = 16]:

(a) 22.4 litre

(b) 11.2 litre

(c) 2.24 litre

(d) 1.12 litre

EasyChemical Calculations01 Mark marks
Answer

(b) 11.2 litre

Explanation

Molar mass of CO₂ = 12 + (16 × 2) = 44 g/mol

Number of moles = \text{Mass} \over \text{Molar mass} = 22 \over 44  = 0.5 mol

Volume = Number of moles × 22.4 L

= 0.5 × 22.4

= 11.2 L

Therefore, the volume of 22 g of CO₂ at S.T.P. is 11.2 litres.

1.11

Find how many g of calcium oxide will be obtained by intensely heating 10 g of calcium carbonate. Consider that calcium carbonate has completely decomposed. [Ca = 40, C = 12, O = 16]

(a) 4.4 g

(b) 5.6 g

(c) 10 g

(d) 100 g

ModerateChemical Calculations01 Mark marks
Answer

(b) 5.6 g

Explanation

Chemical Equation: CaCO₃ → CaO + CO₂

Masses →                    100            56

From the balanced equation,

100 g of CaCO₃ produces 56 g of CaO.

Therefore, Mass of CaO obtained from 10 g of CaCO₃

= \frac{56}{100}\times10

= 5.6 g

1.12

The coefficient of thermal conduction depends on:

(a) The difference in temperature between the two ends of the conductor

(b) Property of the material of the conductor

(c) The length of the conductor

(d) The area of cross-section of the conductor

ModerateThermal phenomena01 Mark marks
Answer

(b) Property of the material of the conductor

Explanation

The coefficient of thermal conductivity (k) is a property of the material that indicates how easily heat flows through it.

1.13

Which parameter remains unaltered during refraction of light?

(a) Velocity

(b) Amplitude

(c) Frequency

(d) Wavelength

ModerateLight – Reflection and Refraction01 Mark marks
Answer

(c) Frequency

Explanation

When light travels from one transparent medium to another, its frequency remains unchanged because it is determined by the source of light.

1.14

Which type of mirror produces a virtual, diminished and erect image?

(a) Plane mirror

(b) Concave mirror

(c) Convex mirror

(d) Parabolic mirror

EasyLight – Reflection and Refraction01 Mark marks
Answer

(c) Convex mirror

Explanation

A convex mirror always forms an image that is:

  • Virtual
  • Erect
  • Diminished (smaller than the object)
1.15

Which one of the following is the unit of electrical energy?

(a) Watt

(b) Ohm

(c) Kilowatt-hour

(d) Volt

EasyCurrent Electricity01 Mark marks
Answer

(c) Kilowatt-hour (kWh)

Question 2 | Group B
Answer the following Questions (alternatives are to be noted): [1 × 21 = 21]
2.1

State the S.I. unit of radioactivity.

EasyAtomic Nucleus01 Mark marks
Answer

SI unit of radioactivity: Becquerel (Bq)

What change will occur in atomic number if one β-particle is emitted from ²³⁵₉₂U?

EasyAtomic Nucleus01 Mark marks
Answer

The atomic number increases by 1.

Explanation

During β⁻-decay, a neutron changes into a proton, emitting a β-particle (electron) and an antineutrino.

_0^1n \rightarrow _1^1p + _{-1}^0\beta + \bar{\nu}

As a result:

  • Atomic number increases by 1.
  • Mass number remains unchanged.
2.2

Match the Right Column with the Left Column:

Left Column Right Column
2.2.1 Metal present in German silver (a) Li
2.2.2 A transuranium element (b) Zn
2.2.3 The metal extracted from calamine (c) Pu
2.2.4 Alkali metal present in second period of periodic table (d) Ni
EasyMetallurgy04 Mark marks
Answer

Left Column Correct Match
2.2.1 Metal present in German silver (d) Ni
2.2.2 A transuranium element (c) Pu
2.2.3 The metal extracted from calamine (b) Zn
2.2.4 Alkali metal present in second period of periodic table (a) Li
2.3

Write whether the following statement is True or False:

Electrolysis always involves the process of oxidation–reduction.

EasyElectrolysis01 Mark marks
Answer

True

Explanation

Electrolysis is a chemical process in which an electric current brings about a chemical change in an electrolyte. During electrolysis, both oxidation and reduction occur simultaneously.

2.4

Which of the following is/are electrolyte(s) — Sugar solution in water, Ethanol and Acetic Acid?

EasyElectrolysis01 Mark marks
Answer

Acetic acid (CH₃COOH)

2.5

Which gas is produced at the cathode during electrolysis of acidified water?

EasyElectrolysis01 Mark marks
Answer

Hydrogen gas (H₂)

What happens when an iron spoon is dipped into acidified copper sulphate solution?

EasyElectrolysis01 Mark marks
Answer

A reddish-brown layer of copper is deposited on the iron spoon, and the blue colour of the copper sulphate solution gradually fades to light green.

2.6

Which metal is used as catalyst in Haber’s process for preparation of ammonia?

EasyInorganic Chemistry01 Mark marks
Answer

Iron (Fe)

Fill up the blank: CaC₂ + N₂ → __________ + C

EasyInorganic Chemistry01 Mark marks
Answer

CaCN₂ (Calcium cyanamide)

2.7

Name the metal which is extracted in thermite process.

EasyMetallurgy01 Mark marks
Answer

Iron (Fe)

Explanation

The thermite process is used to extract or weld iron by reducing iron(III) oxide (Fe₂O₃) with aluminium powder. Aluminium, being more reactive than iron, removes oxygen from iron oxide to produce molten iron.

Chemical Equation: Fe2O3 + 2Al → Al2O3 + 2Fe

2.8

Write down the IUPAC name of the following organic compound:

2 Bromo Propane

ModerateOrganic Chemistry01 Mark marks
Answer

2-Bromopropane

Give an example of positional isomerism

ModerateOrganic Chemistry01 Mark marks
Answer

1-Propanol and 2-Propanol

2.9

Write the name of monomer of a polymer which is used to prepare non-stick utensils.

EasyOrganic Chemistry01 Mark marks
Answer

Tetrafluoroethylene (TFE)

Chemical Formula: CF₂ = CF₂

2.10

Which layer of atmosphere is susceptible to storm and rain?

EasyConcerns about our Environment01 Mark marks
Answer

Troposphere

What is the unit of concentration of ozone layer?

EasyConcerns about our Environment01 Mark marks
Answer

Dobson Unit (DU)

2.11

Name the greenhouse gas discharged from a refrigerator

EasyConcerns about our Environment01 Mark marks
Answer

Chlorofluorocarbon (CFC)

2.12

Write whether the following statement is True or False:

According to Charles’ law, at (−) 273°C, the volume of any gas is infinite.

EasyBehaviour of Gases01 Mark marks
Answer

False

Explanation

At −273°C (0 K), the volume of an ideal gas is theoretically zero, not infinite.

2.13

What is the number of molecules present in 16 g of O₂ at S.T.P.?

ModerateBehaviour of Gases01 Mark marks
Answer

3.01 × 10²³ molecules

Explanation

Molar mass of O₂ = 16 × 2 = 32 g/mol

Number of moles = \frac{\text{Mass}}{\text{Molar Mass}}

= \frac{16}{32}

= 0.5 mol

Number of molecules = 0.5 × 6.022 × 10²³ molecules

= 3.011 × 10²³ molecules ≈ 3.01 × 10²³ molecules

2.14

Write whether the following statement is True or False:

If coefficient of linear expansion of iron is 1.2 × 10⁻⁵ /°C, then coefficient of cubical expansion of iron is 3.6 × 10⁻⁵ /°C.

EasyThermal phenomena01 Mark marks
Answer

True

Explanation

The relationship between the coefficients of expansion for an isotropic solid is: γ = 3α

∴ γ = 3α = 3 × 1.2 × 10⁻⁵ = 3.6 × 10⁻⁵ /°C

Write whether the following statement is True or False:

Which physical quantity has the unit W m⁻¹ K⁻¹?

EasyThermal phenomena01 Mark marks
Answer

Coefficient of Thermal Conductivity (Thermal Conductivity)

2.15

What type of lens is used as a magnifying glass?

EasyLight – Reflection and Refraction01 Mark marks
Answer

Convex lens

Explanation

A convex lens is used as a magnifying glass because it forms a virtual, erect, and magnified image when the object is placed between the optical centre and the principal focus of the lens.

2.16

For which value of the angle of incidence in refraction of light is Snell’s law not applicable?

EasyLight – Reflection and Refraction01 Mark marks
Answer

Angle of incidence = 0° (Normal incidence)

2.17

What is now used in electrical circuits instead of fuse wire?

EasyCurrent Electricity01 Mark marks
Answer

Miniature Circuit Breaker (MCB)

2.18

State the resistance of a 220 V – 100 W bulb.

EasyCurrent Electricity01 Mark marks
Answer

Voltage (V) = 220 V

Power (P) = 100 W

R = \frac{220^2}{100}

R = \frac{48400}{100} = 484 Ω

Question 3 | Group C
Answer the following Questions (alternatives are to be noted):  [2 × 9 = 18]
3.1

Write down the formula of Fire ice. How is methane gas obtained from it? [1+1]

EasyConcerns about our Environment02 Mark marks
Answer

Formula of Fire Ice: CH₄·nH₂O (Methane Hydrate)

Methane gas is obtained by heating Fire Ice or by reducing the pressure.

3.2

A certain amount of gas occupies 750 cc at –3°C. The gas is heated at constant pressure till its volume becomes 1 litre. What is its final temperature?

ModerateBehaviour of Gases02 Mark marks
Answer

Initial Volume (V1) = 750 cc

Initial Temperature (T1) = –3°C = 270 K

Final Volume (V2) = 1000 cc

Final Temperature (T2) = ?

From Charles Law: \text{V₁}\over \text{T₁} = \text{V₂}\over \text{T₂}

T₂ = \text{V₂ × T₁}\over \text{V₁} = \text{1000 × 270}\over \text{750} = 360 K

T₂ = 360 K – 273 K = 87 K

Hence, the final temperature of the gas is 87°C.

What will be the volume of 8 g of H₂ gas (H = 1) at a pressure of 4 atmospheres and a temperature of 27°C?

[R = 0.082 litre atmosphere mole⁻¹ K⁻¹]

Moderate02 Mark marks
Answer

Mass of H₂ = 8 g

Pressure (P) = 4 atm

Temperature (T) = 27°C = 300 K

Gas Constant (R) = 0.082 L atm mol⁻¹ K⁻¹

Volume (V) = ?

Molar mass of H₂ = 2 g/mol

Number of moles (n) = \frac{\text{Mass}}{\text{Molar Mass}}

= \frac{8}{2}

= 4 mol

Ideal Gas Equation : PV=nRT

Volume (V) = \frac{\text{nRT}}{\text{P}}

= \frac{4 × 0.082 × 300}{4}

= 24.6 L

Hence, the volume of 8 g of H₂ gas is 24.6 L

3.3

Explain why the ozone layer is called ‘natural sunscreen’.

EasyConcerns about our Environment02 Mark marks
Answer

The ozone layer is called the natural sunscreen of the Earth because it absorbs most of the harmful ultraviolet (UV) rays coming from the Sun and protects living organisms.

How are global warming and greenhouse effect related?

EasyConcerns about our Environment02 Mark marks
Answer

The greenhouse effect is the process by which greenhouse gases trap heat in the Earth’s atmosphere. When this effect increases due to excessive greenhouse gases, it causes a rise in the Earth’s average temperature, known as global warming.

3.4

The volume of a certain amount of gas at a temperature of 27°C and at pressure 76 cm of Hg is 200 c.c. Find the volume of the gas when pressure of the gas changes to 38 cm of Hg and temperature changes to 127°C.

ModerateBehaviour of Gases02 Mark marks
Answer

Initial Pressure (P₁) = 76 cm Hg

Initial Volume (V₁) = 200 c.c.

Initial Temperature (T₁) = 27°C = 300 K

Final Pressure (P₂) = 38 cm Hg

Final Temperature (T₂) = 127°C = 400 K

Final Volume (V₂) = ?

From the Combined Gas Law, \frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

V₂ = \frac{P_1\times V_1\times T_2}{P_2\times T_1} = \frac{76\times200\times400}{38\times300} = 533.3 c.c.

Hence, the final volume of the gas is 533.3 c.c.

Find the volume of 14 g Nitrogen gas at a temperature of 227°C and at pressure of 83.14 cm of Hg.

[R = 8.314 J mol⁻¹ K⁻¹]

HardBehaviour of Gases02 Mark marks
Answer

Mass of N₂ = 14 g

Molar mass of N₂ = 28 g/mol

Pressure (P) = 83.14 cm Hg = 1.094 atm

Temperature (T) = 227°C = 500 K

Gas Constant (R) = 0.082 L atm mol⁻¹ K⁻¹

Volume (V) = ?

Number of moles = \frac{\text{Mass}}{\text{Molar Mass}}

= \frac{14}{28}

= 0.5 mol

Using the Ideal Gas Equation, PV = nRT

Volume (V) = \frac{nRT}{P}

= \frac{0.5\times0.082\times500}{1.094}

≈ 18.74 L

≈ 19 L

Hence, the volume of 14 g of nitrogen gas is approximately 19 L.

3.5

Explain with a suitable ray diagram the process of formation of image by a concave lens.

EasyLight – Reflection and Refraction02 Mark marks
Answer

A concave lens always forms a virtual, erect and diminished image of an object. The image is formed between the optical centre (O) and the principal focus (F) on the same side of the lens as the object.

concave lens ray diagram notes

Explain with a diagram why the sky appears blue.

EasyLight – Reflection and Refraction02 Mark marks
Answer

Sunlight consists of seven colours. As it passes through the Earth’s atmosphere, the molecules of air scatter light.

Since blue light has a shorter wavelength, it is scattered much more than the other colours. As a result, blue light reaches our eyes from all directions, making the sky appear blue.

3.6

A thin wire of resistance 4 Ω is bent to form a circle. Find the resistance across any diameter.

ModerateCurrent Electricity02 Mark marks
Answer

Resistance of the complete wire = 4 Ω

The wire is bent into a circle.

Therefore, the resistance of each semicircle = 4 ÷ 2 = 2 Ω

The two semicircles are connected in parallel between the ends of the diameter.

Equivalent resistance = \frac{R_1 \times R_2}{R_1 + R_2}

= \frac{2 \times 2}{2 + 2}

= \frac{4}{4}

= 1 Ω

Hence, the resistance across any diameter is 1 Ω.

3.7

Draw electron-dot structure of calcium oxide.

[Atomic numbers of Ca and O are 20 and 8 respectively]

ModerateIonic and Covalent Bond02 Mark marks
Answer

Electron-dot structure of Calcium Oxide (CaO):

exec 091617f7 8908 428b a13f ca635499a38b

 

Explain why the melting point of sodium chloride is much greater than that of glucose.

ModerateIonic and Covalent Bond02 Mark marks
Answer

Sodium chloride has a much higher melting point than glucose because NaCl is an ionic compound, whereas glucose is a covalent compound.

3.8

Identify the electrovalent and covalent compounds from among the following compounds:

LiH, NH₃, KCl, C₂H₆

EasyIonic and Covalent Bond02 Mark marks
Answer

Electrovalent (Ionic) Compounds:

  • LiH
  • KCl

Covalent Compounds:

  • NH₃
  • C₂H₆
3.9

Write with balanced chemical equation, what happens when dry ammonia gas is passed over heated sodium.

ModerateInorganic Chemistry02 Mark marks
Answer

When dry ammonia gas is passed over heated sodium, sodamide (sodium amide) and hydrogen gas are formed.

Balanced Chemical Equation:

2Na + 2NH₃ → 2NaNH₂ + H₂↑

Question 4 | Group D
Answer the following Questions (alternatives are to be noted): [3 × 12 = 36]
4.1

State Modern Periodic Law. Discuss the periodic trends of atomic radii for elements of Group 1 to 2 and Group 13 to 17. [1+2]

ModeratePeriodic Table03 Mark marks
Answer

Modern Periodic Law

The physical and chemical properties of elements are periodic functions of their atomic numbers.

Periodic Trend of Atomic Radius

  • Across a period (Group 1 → Group 2 and Group 13 → Group 17), the atomic radius decreases from left to right.
  • This is because the nuclear charge increases, pulling the electrons closer to the nucleus while the number of electron shells remains the same.

Explain the position of hydrogen in the modern periodic table. Which group of the periodic table contains solid, liquid and gaseous elements? [2+1]

ModeratePeriodic Table03 Mark marks
Answer

Position of Hydrogen

Hydrogen has the electronic configuration 1s¹.

  • Like Group 1 (Alkali metals), it has one valence electron and can lose one electron to form H⁺.
  • Like Group 17 (Halogens), it requires one electron to become stable and can gain one electron to form H⁻.

Therefore, hydrogen shows properties of both Group 1 and Group 17, and hence occupies a unique position in the modern periodic table.

Group 17 (Halogens) contains solid, liquid and gaseous elements:

  • Fluorine (F₂) and Chlorine (Cl₂) – Gases
  • Bromine (Br₂) – Liquid
  • Iodine (I₂) and Astatine (At) – Solids
4.2

What materials are used as cathode, anode and electrolyte during electroplating of a brass spoon with nickel?

EasyElectrolysis03 Mark marks
Answer

During the electroplating of a brass spoon with nickel:

  1. Cathode: Brass spoon
  2. Anode: Pure nickel plate
  3. Electrolyte: Aqueous solution of nickel sulphate (NiSO₄)
4.3

Write with balanced chemical equation, what happens when hydrogen sulphide is passed through aqueous solution of lead nitrate.

ModerateInorganic Chemistry03 Mark marks
Answer

When hydrogen sulphide (H₂S) is passed through an aqueous solution of lead nitrate [Pb(NO₃)₂], a black precipitate of lead sulphide (PbS) is formed and nitric acid (HNO₃) is produced.

Balanced Chemical Equation:

Pb(NO₃)₂ + H₂S → PbS↓ + 2HNO₃

Observation:

A black precipitate of lead sulphide (PbS) is formed.

4.4

Mention one use of CNG. Discuss one each of the harmful effects of methanol and ethanol. [1+2]

EasyOrganic Chemistry03 Mark marks
Answer

Use of CNG

  • Compressed Natural Gas (CNG) is used as a clean fuel in automobiles because it produces less air pollution.

Harmful Effects

(i) Methanol: (a) Highly poisonous. (b) Consumption may cause blindness or even death.

(ii) Ethanol: (a) Excessive consumption affects the brain and liver. (b) It can lead to addiction and impaired judgment.

Write the structural formulae of 1,2-dibromoethane and 1,1,2,2-tetrabromoethane. Write the name of the organic compound which is produced when ethyl alcohol is heated with concentrated sulphuric acid. [2+1]

ModerateOrganic Chemistry03 Mark marks
Answer

Structural Formulae of 1,2-dibromoethane and 1,1,2,2-tetrabromoethane

dibromoethane tetrabromoethane combined 4x3 1

When ethyl alcohol is heated with concentrated sulphuric acid, it undergoes dehydration to form Ethene (C₂H₄).

4.5

State Avogadro’s law. Use simple calculation to show that moist air is lighter than dry air. [1+2]

ModerateBehaviour of Gases03 Mark marks
Answer

Avogadro’s Law : Equal volumes of all gases at the same temperature and pressure contain an equal number of molecules.

To show that moist air is lighter than dry air

According to Avogadro’s Law, equal volumes of gases contain an equal number of molecules.

Molecular mass of dry air ≈ 29

Molecular mass of H₂O = (2 × 1) + 16 = 18

When water vapour mixes with dry air, some of the heavier air molecules (average molecular mass = 29) are replaced by lighter water vapour molecules (molecular mass = 18).

Since 18 < 29, the average molecular mass of moist air decreases.

Therefore, moist air is lighter than dry air.

4.6

In a closed vessel, 1 g of magnesium is burnt with 0.5 g of oxygen gas. Which reactant remains in excess? Find the quantity of excess reactant. [Mg = 24, O = 16]

ModerateChemical Calculations03 Mark marks
Answer

Mass of Mg = 1 g

Mass of O₂ = 0.5 g

Balanced Chemical Equation:

2Mg + O₂ → 2MgO

Molar mass of Mg = 24 g/mol

Molar mass of O₂ = 32 g/mol

From the balanced equation,

48 g of Mg reacts with 32 g of O₂.

Therefore, Mg required to react with 0.5 g of O₂

= \frac{48 \times 0.5}{32}

= 0.75 g

Available Mg = 1 g

Excess Mg = 1 − 0.75 = 0.25 g

Hence, magnesium remains in excess and the excess quantity is 0.25 g.

How many grams of CaCO₃ will react with excess of dilute HCl to produce 66 g of CO₂?

[Ca = 40, C = 12, O = 16]

ModerateChemical Calculations03 Mark marks
Answer

Molar Mass of CO₂ = 12 + (16 × 2) = 44 g

Molar Mass of CaCO₃ = 40 + 12 + (16 × 3) = 100 g

Balanced Chemical Equation:

CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂

100 g                                                    44 g

From the balanced equation,

100 g of CaCO₃ produces 44 g of CO₂.

Therefore, Mass of CaCO₃ required to produce 66 g of CO₂

= \frac{100 \times 66}{44}

= 150 g

Hence, 150 g of CaCO₃ is required to produce 66 g of CO₂.

4.7

Why are there gaps at regular intervals between consecutive rails in a railway track? State the relation between linear expansion coefficient (α), superficial expansion coefficient (β) and volume expansion coefficient (γ). [2+1]

EasyThermal phenomena03 Mark marks
Answer

Gaps are left between railway rails to provide space for thermal expansion during hot weather. This prevents the rails from bending or buckling, thereby avoiding accidents.

The relation between the coefficients of expansion is:

α : β : γ= 1 : 2 : 3

Find the length of an iron rod at 110°C if at 10°C, the length of the rod is 20 cm. [Volume expansion coefficient of iron = 36 × 10⁻⁶ /°C]

ModerateThermal phenomena03 Mark marks
Answer

Initial length (L₁) = 20 cm

Initial temperature = 10°C

Final temperature = 110°C

Rise in temperature (ΔT) = 110 − 10 = 100°C

Volume expansion coefficient (γ) = 36 × 10⁻⁶ /°C

Linear expansion coefficient (α) = \frac{\gamma}{3}

= \frac{36 \times 10^{-6}}{3}

= 12 × 10⁻⁶ /°C

Using the formula, L₂ = L₁ (1 + αΔT)

L₂ = 20 (1 + 12 × 10⁻⁶ × 100)

L₂ = 20 (1 + 0.0012)

L₂ = 20 × 1.0012

L₂ = 20.024 cm

Hence, the length of the iron rod at 110°C is 20.024 cm.

4.8

Prove that with a rectangular glass slab, incident ray and emergent ray are parallel to each other.

HardLight – Reflection and Refraction03 Mark marks
Answer

refraction-through-glass-slab-enhanced

At the first surface O, the ray enters the glass slab from air and is refracted.

According to Snell’s law, ₁μ₂ = sin i / sin r

where,

  • i = angle of incidence,
  • r = angle of refraction,
  • ₁μ₂ = refractive index of glass with respect to air.

At the second surface B, the ray emerges from the glass into air.

Again, by Snell’s law, ₂μ₁ = sin r / sin θ (where θ is the angle of emergence)

Since, ₂μ₁ = 1 / ₁μ₂

⇒ sin i / sin r = sin θ / sin r

⇒ sin i = sin θ

⇒ θ = i

Thus, the angle of emergence is equal to the angle of incidence. Therefore, the emergent ray is parallel to the incident ray. However, it is displaced sideways by a small distance called the lateral shift.

A ray of light is incident normally on one of the faces of a prism of refracting angle A and refractive index μ. If δ = angle of deviation, then establish the relation between μ, A and δ.

HardLight – Reflection and Refraction03 Mark marks
Answer

The principal section ABC of a prism is shown in Fig. A ray of light OP is incident at P on the face AB at an angle of incidence i₁. The angle of refraction at P is r₁. The refracted ray moves along PQ and is incident on the face AC of the prism at an angle of incidence r₂ at Q, and emerges along QR at an angle of emergence i₂. So, ∠LMR = δ is the angle of deviation, i.e., the angle through which the emergent ray is deviated from the incident ray.

 

Now, from quadrilateral APNQ,

∠APN + ∠AQN = 90°

∴ ∠PNQ + ∠PAQ = 180°

But in ΔPNQ,

∠PNQ = 180° − (r₁ + r₂)

or, 180° − (r₁ + r₂) + A = 180°

∴ A = r₁ + r₂

Again, the angle of deviation δ is given by,

δ = ∠MPQ + ∠MQR

δ = (∠MPN − r₁) + (∠MQN − r₂)

δ = (i₁ − r₁) + (i₂ − r₂)

δ = (i₁ + i₂) − (r₁ + r₂)

δ = i₁ + i₂ − A

∴ δ = i₁ + i₂ − A

4.9

What is Hypermetropia? Which type of lens is used to rectify it? [2+1]

EasyLight – Reflection and Refraction03 Mark marks
Answer

Hypermetropia (Long-sightedness) is a defect of vision in which a person can see distant objects clearly but cannot see nearby objects distinctly.

It is corrected by using a convex lens.

4.10

The series combination of three 20 Ω resistances is connected in parallel combination to a 30 Ω resistance. Determine the equivalent resistance of the final combination.

ModerateCurrent Electricity03 Mark marks
Answer

series-parallel-resistors-circuit-only(1)

Three resistors of 20 Ω each are connected in series.

Resistance in series = 20 + 20 + 20 = 60 Ω

This 60 Ω combination is connected in parallel with a 30 Ω resistor.

Equivalent resistance = \frac{R_1 \times R_2}{R_1 + R_2}

= \frac{60 \times 30}{60 + 30}

= \frac{1800}{90}

= 20 Ω

Hence, the equivalent resistance of the final combination is 20 Ω.

4.10

Three equal resistors connected in series across a source of e.m.f. together dissipate 10 W of power. What would be the power dissipated if the same resistors were connected in parallel across the same source of e.m.f.?

HardCurrent Electricity03 Mark marks
Answer

Let each resistor be R.

When connected in series,

Equivalent resistance = 3R

Given, Power dissipated = 10 W

Using the formula, P = \frac{V^2}{R}

⇒ 10 = \frac{V^2}{3R}

⇒ V² = 10 × 3R = 30 R

When the same resistors are connected in parallel,

Equivalent resistance = \frac{R}{3}

Power (P) = \frac{V^2}{R/3} = \frac{3V^2}{R}

Substituting V² = 30R,

P = \frac{3 \times 30R}{R} = 90 W

Hence, the power dissipated in the parallel combination is 90 W.

4.11

State Lenz’s law. “Lenz’s law actually follows the principle of conservation of energy.” — Justify. [1+2]

ModerateCurrent Electricity03 Mark marks
Answer

According to Lenz’s law, whenever there is a change in magnetic flux, the induced current always flows in a direction that opposes the change in magnetic flux.

For example, if a magnet is moved towards a coil, the induced current produces a magnetic field that opposes the approaching magnet. Therefore, an external force is required to keep the magnet moving.

If the induced current did not oppose the change, the magnet would continue to move on its own, producing electrical energy without any external work. This would violate the principle of conservation of energy.

Hence, Lenz’s law is consistent with the principle of conservation of energy.

4.12

What do you mean by mass defect? Actual mass of proton, neutron and helium nucleus are 1.00728 amu, 1.00867 amu and 4.0015 amu, respectively. What is the mass defect of ⁴₂He (helium) nucleus? [1+2]

ModerateAtomic Nucleus03 Mark marks
Answer

Mass Defect : Mass defect is the difference between the sum of the masses of the individual nucleons (protons and neutrons) and the actual mass of the nucleus.

Given,

Mass of one proton = 1.00728 amu

Mass of one neutron = 1.00867 amu

Mass of helium nucleus = 4.0015 amu

A helium nucleus contains: 2 protons and 2 neutrons

Mass of 2 protons = 2 × 1.00728 = 2.01456 amu

Mass of 2 neutrons = 2 × 1.00867 = 2.01734 amu

Total mass of nucleons

= 2.01456 + 2.01734

= 4.03190 amu

Mass defect = Total mass of nucleons − Actual mass of nucleus

= 4.03190 − 4.00150

= 0.03040 amu

Hence, the mass defect of the helium nucleus is 0.0304 amu.

Question 5 | Group E
5.1

Which gaseous hydrocarbon is used for ripening of fruits?

EasyOrganic Chemistry01 Mark marks
Answer

Ethene (Ethylene) – C₂H₄

Explanation

Ethene (ethylene) is a gaseous hydrocarbon that is used for the artificial ripening of fruits such as bananas, mangoes, and tomatoes. It acts as a plant hormone, accelerating the natural ripening process.

5.2

State one harmful effect of ultraviolet rays.

EasyConcerns about our Environment01 Mark marks
Answer

Ultraviolet (UV) rays can cause skin cancer.

5.3

What is the S.I. unit of specific resistance or resistivity?

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Answer

The SI unit of resistivity (specific resistance) is ohm metre (Ω m).

5.4

Write the volume of one mole oxygen gas at S.T.P.

EasyBehaviour of Gases01 Mark marks
Answer

At Standard Temperature and Pressure (S.T.P.), one mole of any gas occupies 22.4 litres.

5.5

Which one among the radioactive rays is an electromagnetic wave?

EasyAtomic Nucleus01 Mark marks
Answer

Gamma (γ) rays

Question 6 | Group E
6.1

CH₃COOH is an organic compound but NaHCO₃ is not an organic compound — give reason.

EasyOrganic Chemistry02 Mark marks
Answer

CH₃COOH (Acetic acid) is an organic compound because it contains carbon bonded directly to hydrogen (C–H bond), which is a characteristic feature of organic compounds.

6.2

State Fleming’s Left Hand Rule

EasyCurrent Electricity02 Mark marks
Answer

Fleming’s left-hand rule states that when you stretch the thumb, forefinger, and middle finger of your left hand to be at right angles to each other, the forefinger points to the magnetic field, the middle finger points to the current, and the thumb shows the force.

6.3

Why is the red colour selected for danger signal lights?

EasyLight – Reflection and Refraction02 Mark marks
Answer

Red colour is selected for danger signal lights because it has the longest wavelength and is scattered the least by the particles present in the atmosphere.

As a result, red light can travel a greater distance and remains clearly visible even in fog, mist, smoke, or dust. Therefore, it is used for danger signals, traffic lights, and stop signals.

6.4

Give one example of reducing property of H₂S.

EasyOrganic Chemistry02 Mark marks
Answer

Hydrogen sulphide (H₂S) reduces acidified potassium dichromate solution from orange to green.

Chemical Equation:

K₂Cr₂O₇ + 4H₂SO₄ + 3H₂S → Cr₂(SO₄)₃ + K₂SO₄ + 7H₂O + 3S

End of this paper
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