Madhyamik 2018 Physical Science Solution
Complete English version with detailed solutions for WBBSE Class 10 students
About This Solution
This post gives you the full Madhyamik 2018 Physical Science question paper with solution. It is written in a clear and simple way so that students can easily understand and use it for revision.
All the questions and answers are arranged in the same order as in the real exam paper (Groups A, B, C, and D). This helps you find and study each part quickly and easily.
Question Paper Solutions
1.1
Which of the following greenhouse gases has the maximum contribution towards global warming?
Answer: (c) CO₂
Explanation: Carbon dioxide (CO₂) is the main gas that increases global warming as it stays long in the air.
1.2
According to Boyle’s Law, which is PV–P Graph?
(a)
(b)
(c)
(d)
Answer: (b)
Explanation: Boyle’s Law says \( PV = \text{constant} \), so PV does not change when pressure changes. The PV–P graph is a straight line parallel to the P-axis.
1.3
If the vapour density of a carbon containing gaseous substance is 13, which of the following is its molecular formula?
Answer: (d) C₂H₂
Explanation: Vapour density = 13.
So molecular mass = 2 × 13 = 26.
Molar masses: CO₂ = 44, C₂H₄ = 28, C₂H₆ = 30, C₂H₂ = 26.
Therefore the gas with molecular mass 26 is C₂H₂.
So molecular mass = 2 × 13 = 26.
Molar masses: CO₂ = 44, C₂H₄ = 28, C₂H₆ = 30, C₂H₂ = 26.
Therefore the gas with molecular mass 26 is C₂H₂.
1.4
What is the unit of coefficient of linear expansion of a solid?
Answer: (c) °C⁻¹
Explanation: The coefficient of linear expansion tells how much a solid expands for each °C rise in temperature.
1.5
An object is placed between the optical centre and focus of a thin convex lens. What is the nature of the image of the object?
Answer: (d) virtual and erect
Explanation: The image is virtual, upright, and larger when the object is between the focus and optical centre of a convex lens.
1.6
When a ray of light is incident perpendicularly on a transparent glass slab, what will be its angle of deviation?
Answer: (a) 0°
Explanation: When light enters normally (straight) into a glass slab, it passes without bending, so deviation = 0°.
1.7
Which of the units given below is the SI unit of resistance?
Answer: (d) ohm
Explanation: The SI unit of resistance is ohm (Ω).
1.8
In domestic electric circuit the fuse wire is connected to which of the following?
Answer: (b) live line
Explanation: The fuse is connected to the live wire to stop current when it becomes too high.
1.9
Which is the β-ray emitted from a radioactive element?
Answer: (a) a stream of electrons
Explanation: Beta rays are fast-moving electrons emitted from radioactive atoms.
1.10
How many groups are there in the long periodic table?
Answer: (d) 18
Explanation: The modern periodic table has 18 vertical columns called groups.
1.11
In formation of which of the following compounds, the octet rule is obeyed?
Answer: (a) NaCl
Explanation: Sodium and chlorine form NaCl by transferring one electron to make both have 8 electrons in outer shell.
1.12
Which of the following can conduct electricity?
Answer: (a) molten NaCl
Explanation: Molten NaCl has free ions which can move and conduct electric current.
1.13
What will be the colour of the resulting solution when excess aqueous ammonia is added to an aqueous solution of copper sulphate?
Answer: (c) deep blue
Explanation: Copper sulphate with excess ammonia gives a deep blue complex.
1.14
In which of the following alloys is zinc present?
Answer: (b) brass
Explanation: Brass is an alloy made of copper and zinc.
1.15
Which of the following is a saturated hydrocarbon?
Answer: (d) C₂H₆
Explanation: C₂H₆ (ethane) has single bonds only, so it is a saturated hydrocarbon.
1.15
Which of the following is the alkyl group containing two carbon atoms?
Answer: (b) ethyl
Explanation: The ethyl group (–C₂H₅) contains two carbon atoms; methyl has one, propyl has three, and isopropyl is a branched three-carbon group.
Group B – Very Short Answer (1×21 = 21)
2.1
Mention one use of biogas.
Answer: Used for cooking.
Explanation: Biogas burns easily and gives good heat. It is commonly used for cooking in homes and villages.
2.1 (Or)
What is the role of NO in decomposition of ozone in the ozone layer?
Answer: NO helps to break ozone into oxygen.
Explanation: Nitric oxide (NO) reacts with ozone (O₃) to form nitrogen dioxide (NO₂) and oxygen (O₂). This reaction reduces ozone in the ozone layer.
2.2
Among charcoal, petrol and ethanol, which one is a fossil fuel?
Answer: Petrol.
Explanation: Petrol is formed from dead plants and animals after millions of years under heat and pressure, so it is a fossil fuel.
2.3
Under constant pressure, at what temperature in degree Celsius will the volume of an ideal gas be zero according to Charles’ Law?
Answer: -273°C.
Explanation: According to Charles’ Law, the volume of a gas becomes zero at -273°C. This temperature is called absolute zero.
2.4
What is the unit of M in the equation \(PV = \frac{W}{M} RT\) ? (symbols have usual meaning)
Answer: Kilogram per mole (kg mol⁻¹).
Explanation: M represents the molar mass of a gas. In the SI system, its unit is kilogram per mole (kg mol⁻¹).
2.5
State whether the following statement is ‘True’ or ‘False’:
“The real expansion of any liquid depends on the expansion of the vessel in which it is kept.”
“The real expansion of any liquid depends on the expansion of the vessel in which it is kept.”
Answer: True.
Explanation: The expansion of a liquid depends partly on how much the container itself expands, so the statement is true.
2.5 (Or)
Among iron, invar and copper, which one has the least coefficient of linear expansion?
Answer: Invar.
Explanation: Invar is a special alloy that expands very little when heated. It has the smallest coefficient of linear expansion.
2.6
Between the angle of incidence and the angle of refraction, which one is greater when light travels from a rarer to a denser medium?
Answer: Angle of incidence.
Explanation: When light enters from a rarer to a denser medium, it bends towards the normal. So, the angle of incidence is greater than the angle of refraction.
2.7
What type of mirror is used in the viewfinder of a motor car?
Answer: Convex mirror.
Explanation: A convex mirror gives a wider view of the road behind the vehicle. That is why it is used as a rear-view mirror in cars.
2.8
How does the resistance of a semiconductor change with increase of temperature?
Answer: Resistance decreases.
Explanation: In semiconductors, more charge carriers are released when temperature rises. So, resistance decreases with increase in temperature.
2.9
What type of energy is transformed to electrical energy in a dynamo?
Answer: Mechanical energy.
Explanation: In a dynamo, mechanical energy (motion) is changed into electrical energy using electromagnetic induction.
2.10
Arrange α, β, and γ-rays in ascending order of their penetrating power.
Answer:
α-rays < β-rays < γ-rays.
α-rays < β-rays < γ-rays.
Explanation: Alpha particles have the least penetration, beta particles have moderate, and gamma rays have the highest penetrating power.
2.10 (Or)
Which kind of nuclear reaction is the source of the sun's energy?
Answer: Nuclear fusion.
Explanation: The sun’s energy comes from nuclear fusion, where hydrogen atoms combine to form helium and release huge amounts of energy.
2.11
Match the right column with the left column: [1×4]
Left Column:
2.11.1 An Alkali metal
2.11.2 An element whose anion accelerates rusting of iron
2.11.3 Extracted from Haematite
2.11.4 Most electronegative element
Right Column:
(a) F (b) Fe (c) K (d) Cl
Left Column:
2.11.1 An Alkali metal
2.11.2 An element whose anion accelerates rusting of iron
2.11.3 Extracted from Haematite
2.11.4 Most electronegative element
Right Column:
(a) F (b) Fe (c) K (d) Cl
Answer:
Correct Matching:
2.11.1 → (c) K
2.11.2 → (d) Cl
2.11.3 → (b) Fe
2.11.4 → (a) F
Explanation:
• (c) Potassium (K) is an alkali metal.
• (d) Chloride ions (Cl⁻) speed up rusting of iron.
• (b) Iron (Fe) is obtained from haematite ore.
• (a) Fluorine (F) is the most electronegative element.
2.11.1 → (c) K
2.11.2 → (d) Cl
2.11.3 → (b) Fe
2.11.4 → (a) F
Explanation:
• (c) Potassium (K) is an alkali metal.
• (d) Chloride ions (Cl⁻) speed up rusting of iron.
• (b) Iron (Fe) is obtained from haematite ore.
• (a) Fluorine (F) is the most electronegative element.
2.12
What type of chemical bond is present in CaO?
Answer: Ionic bond.
Explanation: Calcium donates two electrons to oxygen, forming Ca²⁺ and O²⁻ ions. The electrostatic attraction between them forms an ionic bond.
2.13
What is used as a cathode to electroplate silver over a copper spoon?
Answer: Copper spoon.
Explanation: In silver electroplating, the copper spoon is used as the cathode where silver metal is deposited.
2.13 (Or)
Give an example of a compound whose aqueous solution is a weak electrolyte.
Answer: Acetic acid (CH₃COOH).
Explanation: Acetic acid ionises only partly in water, so it is a weak electrolyte.
2.14
During electrolysis, which electrode is called the cathode?
Answer: The negative electrode.
Explanation: During electrolysis, the electrode connected to the negative terminal of the battery is called the cathode, where reduction occurs.
2.15
Give one use of liquid ammonia.
Answer: Used as a refrigerant.
Explanation: Liquid ammonia absorbs a large amount of heat on evaporation, so it is used in refrigerators and cooling systems.
2.15 (Or)
Write the formula of the precipitate formed when aqueous ammonia solution is added to aqueous solution of aluminium chloride.
Answer: Al(OH)₃.
Explanation: When aqueous ammonia reacts with aluminium chloride, aluminium hydroxide Al(OH)₃ forms as a white precipitate.
2.16
In the laboratory preparation of nitrogen gas, which compound is mixed with aqueous solution of ammonium chloride and heated?
Answer: Sodium nitrite (NaNO₂).
Explanation: When ammonium chloride is heated with sodium nitrite, nitrogen gas is produced along with sodium chloride and water.
2.17
What is the IUPAC name of CH₃CH₂CHO?
Answer: Propanal.
Explanation: CH₃CH₂CHO has three carbon atoms with an aldehyde group at the end, so its IUPAC name is propanal.
2.17 (Or)
Write the structural formula of the positional isomer of CH₃CH₂CH₂OH.
Answer: CH₃CHOHCH₃.
Explanation: CH₃CH₂CH₂OH is 1-propanol. Its positional isomer is 2-propanol, with the –OH group on the middle carbon atom (CH₃CHOHCH₃).
2.18
Mention one use of tetrafluoroethylene.
Answer: Used to make Teflon.
Explanation: Tetrafluoroethylene is the monomer used to make Teflon (polytetrafluoroethylene), which is used as a non-stick coating on cookware.
Group C – Short Answer (2×9 = 18)
3.1
What is methane hydrate?
Answer: Methane hydrate is a solid compound of methane and water.
Explanation: Methane hydrate is an ice-like solid formed when methane gas gets trapped inside water molecules under high pressure and low temperature, usually found under the ocean floor.
3.2
The pressure of a fixed mass of a gas at a temperature of 0°C is doubled while the volume is halved. What will be the final temperature of the gas?
Answer: 273 K (0°C).
Explanation: From the gas law, \(\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}\).
Here, \(P_2 = 2P_1\) and \(V_2 = \frac{V_1}{2}\).
So, \(T_2 = T_1\).
Hence, the final temperature remains 273 K or 0°C.
Here, \(P_2 = 2P_1\) and \(V_2 = \frac{V_1}{2}\).
So, \(T_2 = T_1\).
Hence, the final temperature remains 273 K or 0°C.
3.2 (Or)
Under constant pressure, a fixed mass of a gas is heated from 0°C to 546°C. What is the ratio of the final volume of the gas with its initial volume?
Answer: 3 : 1
Explanation: From Charles’ Law, \(\frac{V_1}{T_1} = \frac{V_2}{T_2}\).
\(T_1 = 0°C= 273\) K and \(T_2 = 546 + 273 = 819\) K.
Thus, \(\frac{V_2}{V_1} = \frac{819}{273} = 3\).
Therefore, the ratio is 3 : 1.
\(T_1 = 0°C= 273\) K and \(T_2 = 546 + 273 = 819\) K.
Thus, \(\frac{V_2}{V_1} = \frac{819}{273} = 3\).
Therefore, the ratio is 3 : 1.
3.3
What is meant by the optical centre of a convex lens?
Answer: The optical centre is a point on the lens through which light passes undeviated.
Explanation: The optical centre of a convex lens is a fixed point inside the lens through which a ray of light passes without bending or changing its direction.
3.3 (Or)
Why does the earth’s sky appear blue during daytime?
Answer: Because of the scattering of sunlight by air molecules.
Explanation: The shorter blue wavelengths of sunlight are scattered more by air molecules than other colours, making the sky look blue during daytime.
3.4
State Lenz’s law related to electromagnetic induction.
Answer: The direction of the induced current opposes the cause that produces it.
Explanation: Lenz’s law states that the current induced by a changing magnetic field will always flow in a direction that opposes the change in magnetic flux which produced it.
3.5
Write with an example, how according to Lewis concept, a covalent bond is formed.
Answer: A covalent bond is formed by sharing electrons between atoms.
Explanation: In a hydrogen molecule (H₂), each hydrogen atom shares one electron. The shared pair forms a covalent bond, shown as H–H.
3.5 (Or)
Why the bond in sodium chloride cannot be expressed as Na–Cl?
Answer: Because NaCl has an ionic bond, not a shared (covalent) bond.
Explanation: In sodium chloride, sodium transfers one electron to chlorine, forming Na⁺ and Cl⁻ ions. The bond is ionic, so it cannot be shown as Na–Cl (which represents sharing).
3.6
Give one example each of a liquid and a solid covalent bond.
Answer: Liquid — Water (H₂O);
Solid — Diamond (C).
Solid — Diamond (C).
Explanation: In water, atoms in each molecule are held by covalent bonds (shared electrons) and water is liquid at room temperature. Diamond is a solid where each carbon atom is covalently bonded in a strong 3D network.
3.7
Write the balanced chemical equation of what happens when H₂S gas is passed through an aqueous copper sulphate solution.
Answer:
CuSO₄(aq) + H₂S(g) → CuS(s) + H₂SO₄(aq)
CuSO₄(aq) + H₂S(g) → CuS(s) + H₂SO₄(aq)
Explanation: Hydrogen sulphide gives sulphide ions that react with copper(II) to form black copper(II) sulphide (CuS) as a precipitate. Sulfuric acid remains in solution.
3.8
Write down the cathode reaction when an aqueous solution of MSO₄ (M = metal) is electrolysed. State with reason whether the reaction is oxidation or reduction.
Answer: M²⁺(aq) + 2e⁻ → M(s). This is a reduction.
Explanation: At the cathode positive ions gain electrons to form metal atoms (gain of electrons = reduction). If M is very reactive, H⁺ may be reduced instead, but generally the cathode reaction is M²⁺ + 2e⁻ → M.
3.8 (Or)
Give one use each of copper and aluminium.
Answer: Copper — electrical wires.
Aluminium — cooking foil / aircraft parts.
Aluminium — cooking foil / aircraft parts.
Explanation: Copper conducts electricity well, so it is used for wires. Aluminium is light and resists corrosion, so it is used for foil and aircraft parts.
3.9
What is the condition of substitution reaction of methane with chlorine? Write the balanced chemical equation for the first step of the reaction.
Answer: Condition — ultraviolet light (or sunlight).
Balanced equation (first step):
CH₄ + Cl₂ → CH₃Cl + HCl (in presence of hν)
Balanced equation (first step):
CH₄ + Cl₂ → CH₃Cl + HCl (in presence of hν)
Explanation: In the substitution (chlorination) of methane, chlorine molecules split under ultraviolet light into free radicals. One hydrogen atom of methane is replaced by a chlorine atom forming methyl chloride (CH₃Cl) and hydrogen chloride (HCl).
3.9 (Or)
Write with a balanced chemical equation what happens when ethanol reacts with metallic sodium.
Answer: 2 C₂H₅OH + 2 Na → 2 C₂H₅ONa + H₂↑
Explanation: Sodium reacts with ethanol to form sodium ethoxide and hydrogen gas. The equation is balanced and shows hydrogen is released.
Group D – Long Answer (3×12 = 36)
4.1
Establish the ideal gas equation on the basis of Boyle’s law, Charles’ law and Avogadro’s law.
Answer: The ideal gas equation is PV = nRT.
Explanation:
• Boyle’s Law: At constant temperature, \( P \propto \frac{1}{V} \)
• Charles’ Law: At constant pressure, \( V \propto T \)
• Avogadro’s Law: At constant temperature and pressure, \( V \propto n \)
Combining all three laws: \( V \propto \frac{nT}{P} \)
or \( PV = nRT \), where R is the gas constant.
Hence, \( PV = nRT \) is the ideal gas equation.
• Boyle’s Law: At constant temperature, \( P \propto \frac{1}{V} \)
• Charles’ Law: At constant pressure, \( V \propto T \)
• Avogadro’s Law: At constant temperature and pressure, \( V \propto n \)
Combining all three laws: \( V \propto \frac{nT}{P} \)
or \( PV = nRT \), where R is the gas constant.
Hence, \( PV = nRT \) is the ideal gas equation.
4.2
SO₂ required for industrial production of sulphuric acid is produced by burning iron pyrites in excess air current.
The chemical equation is:
4FeS₂ + 11O₂ → 2Fe₂O₃ + 8SO₂
How many grams of FeS₂ is required for the production of 512 g of SO₂?
[Fe = 56, S = 32, O = 16]
The chemical equation is:
4FeS₂ + 11O₂ → 2Fe₂O₃ + 8SO₂
How many grams of FeS₂ is required for the production of 512 g of SO₂?
[Fe = 56, S = 32, O = 16]
Answer: 400 g of FeS₂.
Explanation:
From the equation:
4FeS₂ → 8SO₂
• Molar mass of FeS₂ = (56 + 2×32) = 120 g
• Molar mass of SO₂ = (32 + 2×16) = 64 g
4 moles of FeS₂ (4×120 = 480 g) produce 8 moles of SO₂ (8×64 = 512 g).
Therefore, 480 g FeS₂ → 512 g SO₂
⇒ To get 512 g SO₂, we need 480 g FeS₂.
Answer: 480 g of FeS₂ is required.
From the equation:
4FeS₂ → 8SO₂
• Molar mass of FeS₂ = (56 + 2×32) = 120 g
• Molar mass of SO₂ = (32 + 2×16) = 64 g
4 moles of FeS₂ (4×120 = 480 g) produce 8 moles of SO₂ (8×64 = 512 g).
Therefore, 480 g FeS₂ → 512 g SO₂
⇒ To get 512 g SO₂, we need 480 g FeS₂.
Answer: 480 g of FeS₂ is required.
4.2 (Or)
By heating 200 g of a metal carbonate, 112 g of metal oxide and a gaseous compound are produced.
Vapour density of the gaseous compound is 22.
How many moles of the gaseous compound are produced in the reaction?
Vapour density of the gaseous compound is 22.
How many moles of the gaseous compound are produced in the reaction?
Answer: 2 moles of gaseous compound.
Explanation:
The reaction is: Metal carbonate → Metal oxide + CO₂ (gas)
Vapour density (VD) of gas = 22 ⇒ Molecular mass = 2×22 = 44 (hence CO₂).
Given: 200 g carbonate gives 112 g oxide.
So, mass of gas = 200 – 112 = 88 g.
Moles of gas = \( \frac{88}{44} = 2 \).
Answer: 2 moles of gaseous compound (CO₂) are produced.
The reaction is: Metal carbonate → Metal oxide + CO₂ (gas)
Vapour density (VD) of gas = 22 ⇒ Molecular mass = 2×22 = 44 (hence CO₂).
Given: 200 g carbonate gives 112 g oxide.
So, mass of gas = 200 – 112 = 88 g.
Moles of gas = \( \frac{88}{44} = 2 \).
Answer: 2 moles of gaseous compound (CO₂) are produced.
4.3
What is thermal conductivity? What is its SI unit?
Answer: Thermal conductivity is the ability of a material to conduct heat. Its SI unit is watt per metre per kelvin (W m⁻¹ K⁻¹).
Explanation: Thermal conductivity tells how quickly heat passes through a material. The higher the value, the better the conductor of heat.
Example: Metals like copper have high thermal conductivity.
4.4
How can an erect and magnified image be formed with the help of a convex lens? With the help of which type of lens, long-sightedness can be rectified?
Answer: An erect and magnified image is formed when the object is placed between the optical centre and the focus of a convex lens. Long-sightedness is corrected using a convex lens.
Explanation: When the object is close to the lens (within the focus), the image is virtual, upright, and larger.
Long-sighted persons (hypermetropic) use a convex lens to bring the image onto the retina.
4.5
If the velocity of light in a medium is 2 × 10⁸ m/s, what will be the refractive index of that medium?
Answer: Refractive index, n = 1.5
Explanation: Refractive index \( n = \frac{c}{v} = \frac{3 × 10^8}{2 × 10^8} = 1.5 \).
So, the refractive index of the medium is 1.5.
So, the refractive index of the medium is 1.5.
4.5 (Or)
The refractive index of a medium with respect to air is √2. If the angle of incidence of a ray of light in air is 45°, determine the angle of deviation for that ray in case of refraction.
Answer: Angle of deviation = 15°.
Explanation:
From Snell’s Law: \( n = \frac{\sin i}{\sin r} \)
Given, \( n = \sqrt{2} \) and \( i = 45° \).
So, \( \sin r = \frac{\sin 45°}{\sqrt{2}} = \frac{1/\sqrt{2}}{\sqrt{2}} = 0.5 \).
Hence, \( r = 30° \).
Deviation, \( \delta = i - r = 45° - 30° = 15° \).
Therefore, the angle of deviation = 15°.
From Snell’s Law: \( n = \frac{\sin i}{\sin r} \)
Given, \( n = \sqrt{2} \) and \( i = 45° \).
So, \( \sin r = \frac{\sin 45°}{\sqrt{2}} = \frac{1/\sqrt{2}}{\sqrt{2}} = 0.5 \).
Hence, \( r = 30° \).
Deviation, \( \delta = i - r = 45° - 30° = 15° \).
Therefore, the angle of deviation = 15°.
4.6
Write Joule’s law related to the heating effect of the current.
Answer: Heat produced is given by:
\( H = I^2 R t \)
Explanation: Joule’s law states that the heat produced in a conductor is directly proportional to:
• The square of the current (I²),
• The resistance (R), and
• The time (t) for which the current flows.
Mathematically,
\[ H ∝ I^2 R t \]
• The square of the current (I²),
• The resistance (R), and
• The time (t) for which the current flows.
Mathematically,
\[ H ∝ I^2 R t \]
4.7
Calculate the equivalent resistance when a wire of 10 Ω is divided into two equal parts and connected in parallel combination.
Answer: Equivalent resistance = 2.5 Ω
Explanation: Each half has resistance \( R/2 = 10/2 = 5 Ω \).
For parallel connection:
\( \frac{1}{R_p} = \frac{1}{5} + \frac{1}{5} = \frac{2}{5} \) ⇒ \( R_p = 2.5 Ω \). Hence, equivalent resistance = 2.5 Ω.
For parallel connection:
\( \frac{1}{R_p} = \frac{1}{5} + \frac{1}{5} = \frac{2}{5} \) ⇒ \( R_p = 2.5 Ω \). Hence, equivalent resistance = 2.5 Ω.
4.7 (Or)
There are one 60 W lamp and two 80 W fans in a house. The lamp and the fans run for 5 hours daily. Find out the expense in a month if one unit of electricity costs ₹4. (Assume 30 days in a month.)
Answer: Monthly cost = ₹144.
Explanation:
Total power = 60 W + 2×80 W = 220 W = 0.22 kW.
Energy used per day = 0.22 × 5 = 1.1 kWh (units).
In 30 days = 1.1 × 30 = 33 units.
Total cost = 33 × ₹4 = ₹132.
Answer: Approximately ₹132–₹144 per month.
Energy used per day = 0.22 × 5 = 1.1 kWh (units).
In 30 days = 1.1 × 30 = 33 units.
Total cost = 33 × ₹4 = ₹132.
Answer: Approximately ₹132–₹144 per month.
4.8
Compare the charge and ionising power of α and γ rays. Mention one use of radioactivity.
Answer:
α-rays — positively charged, high ionising power.
γ-rays — no charge, low ionising power.
Use — in medical treatment and radiography.
γ-rays — no charge, low ionising power.
Use — in medical treatment and radiography.
Explanation:
• α-rays are helium nuclei (+2 charge) with strong ionising ability but low penetration.
• γ-rays are electromagnetic waves (neutral) with weak ionisation but very high penetration.
• Radioactivity is used in cancer treatment (radiotherapy) and to check material thickness (radiography).
• γ-rays are electromagnetic waves (neutral) with weak ionisation but very high penetration.
• Radioactivity is used in cancer treatment (radiotherapy) and to check material thickness (radiography).
4.9
What is meant by ionisation energy of an atom of elements? Arrange Li, Rb, K and Na in the increasing order of their ionisation energy.
Answer: Ionisation energy is the minimum energy required to remove one electron from an isolated gaseous atom.
Increasing order: Rb < K < Na < Li
Increasing order: Rb < K < Na < Li
Explanation:
The farther the electron from the nucleus, the easier it is to remove.
In Group 1 elements, ionisation energy decreases down the group.
Therefore, Rb has the least and Li has the highest ionisation energy.
4.9 (Or)
Mention similarity of properties of hydrogen with one property of Group I elements and two properties of Group 17 elements.
Answer:
• Similar to Group I: Forms +1 ions (H⁺) like alkali metals.
• Similar to Group 17: Exists as diatomic molecule (H₂) and forms covalent compounds.
• Similar to Group I: Forms +1 ions (H⁺) like alkali metals.
• Similar to Group 17: Exists as diatomic molecule (H₂) and forms covalent compounds.
Explanation:
Hydrogen can lose one electron like alkali metals forming H⁺, and also share electrons like halogens to form diatomic molecules such as H₂ and covalent compounds like HCl.
4.10
What are present along with pure alumina in the molten mixture which is electrolysed for the extraction of aluminium by electrolysis? What is used as cathode and anode in this electrolysis?
Answer:
Cryolite (Na₃AlF₆) and fluorspar (CaF₂) are added to alumina.
Cathode — carbon lining of the cell.
Anode — carbon rods.
Cathode — carbon lining of the cell.
Anode — carbon rods.
Explanation:
Cryolite and fluorspar lower the melting point of alumina and increase conductivity.
During electrolysis, aluminium is deposited at the carbon cathode, and oxygen is released at the carbon anode.
4.11
Write the condition and balanced chemical equation for the industrial production of ammonia by Haber’s process.
Answer:
N₂ + 3H₂ ⇌ 2NH₃
Conditions: 450°C, 200 atm pressure, finely divided iron catalyst with molybdenum promoter.
Conditions: 450°C, 200 atm pressure, finely divided iron catalyst with molybdenum promoter.
Explanation:
Nitrogen and hydrogen gases combine in a 1:3 ratio under high temperature and pressure in presence of iron catalyst to form ammonia (NH₃).
This is a reversible and exothermic reaction.
4.12
The molecular formula of an organic compound is C₂H₄O₂. The compound is soluble in water and on addition of NaHCO₃ to its aqueous solution, CO₂ is evolved.
Identify the compound. Write with condition and balanced chemical equation, the reaction of the compound with ethanol.
Answer: The compound is acetic acid (CH₃COOH).
Reaction with ethanol:
CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O (in presence of conc. H₂SO₄)
Reaction with ethanol:
CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O (in presence of conc. H₂SO₄)
Explanation:
Acetic acid reacts with sodium bicarbonate to release CO₂ gas.
When heated with ethanol in presence of concentrated sulphuric acid, it forms ethyl acetate (an ester) and water.
4.12 (Or)
Compare three properties of the organic and the inorganic compounds.
Answer:
Comparison:
1. Organic compounds are covalent, while inorganic compounds are usually ionic.
2. Organic compounds burn easily; inorganic compounds mostly do not.
3. Organic compounds have low melting and boiling points; inorganic compounds usually have high melting points.
1. Organic compounds are covalent, while inorganic compounds are usually ionic.
2. Organic compounds burn easily; inorganic compounds mostly do not.
3. Organic compounds have low melting and boiling points; inorganic compounds usually have high melting points.
Group E – For External Candidates Only
Very Short Answer Questions (1×4 = 4 marks)
5.1
In which layer of the atmosphere does the storm occur?
Answer: In the troposphere.
Explanation: The troposphere is the lowest layer of the atmosphere where clouds, winds, and storms occur.
All weather changes take place in this layer.
All weather changes take place in this layer.
5.2
The alkaline earth metals are placed in which group of the periodic table?
Answer: Group 2.
Explanation: Alkaline earth metals such as beryllium, magnesium, calcium, strontium, barium, and radium are found in Group 2 of the modern periodic table.
5.3
In what combination, two resistances are to be connected so that the equivalent resistance is smaller than both the resistances?
Answer: In parallel combination.
Explanation: When resistances are connected in parallel, the reciprocal of the total resistance equals the sum of the reciprocals of individual resistances.
So, \( R_p = \frac{R_1 R_2}{R_1 + R_2} \).
This equivalent resistance is always less than both \( R_1 \) and \( R_2 \).
So, \( R_p = \frac{R_1 R_2}{R_1 + R_2} \).
This equivalent resistance is always less than both \( R_1 \) and \( R_2 \).
5.4
Write the name of the gas that smells like rotten eggs.
Answer: Hydrogen sulphide (H₂S).
Explanation: Hydrogen sulphide is a colourless gas with a strong foul smell like rotten eggs.
It is produced by the decay of organic matter containing sulphur.
It is produced by the decay of organic matter containing sulphur.
5.5
What is the product of pressure and volume of 2 g of hydrogen at STP? [H = 1]
Answer: PV = 22.4 L·atm
Explanation:
Molar mass of H₂ = 2 g/mol
So, 2 g H₂ = 1 mole
At STP:
Pressure (P) = 1 atm
Volume (V) = 22.4 L per mole
Therefore, \( PV = 1 × 22.4 = 22.4 \, \text{L·atm} \).
Answer: The product of pressure and volume = 22.4 L·atm.
So, 2 g H₂ = 1 mole
At STP:
Pressure (P) = 1 atm
Volume (V) = 22.4 L per mole
Therefore, \( PV = 1 × 22.4 = 22.4 \, \text{L·atm} \).
Answer: The product of pressure and volume = 22.4 L·atm.
Short Answer Questions (2×3 = 6 marks)
6.1
Give the name and formula of one ore of copper.
Answer: Copper pyrite (CuFeS₂)
Explanation: The most common ore of copper is copper pyrite.
Its chemical formula is CuFeS₂.
It contains copper along with iron and sulphur.
Its chemical formula is CuFeS₂.
It contains copper along with iron and sulphur.
6.2
Why is a convex lens called a convergent lens?
Answer: Because it bends parallel rays of light inward to meet at a point.
Explanation: When parallel rays of light pass through a convex lens, they bend toward the principal axis and meet at a common point called the focus.
Hence, a convex lens is called a convergent lens.
Hence, a convex lens is called a convergent lens.
6.3
Write with balanced chemical equation what happens when ethylene reacts with bromine.
Answer: C₂H₄ + Br₂ → C₂H₄Br₂
Explanation: Ethylene (C₂H₄) adds bromine (Br₂) across its double bond to form ethylene dibromide (C₂H₄Br₂).
The reaction is an addition reaction and the reddish-brown colour of bromine disappears.
Balanced Equation:
C₂H₄ + Br₂ → C₂H₄Br₂
The reaction is an addition reaction and the reddish-brown colour of bromine disappears.
Balanced Equation:
C₂H₄ + Br₂ → C₂H₄Br₂
6.4
How is the concept of resistance obtained from Ohm’s law?
Answer: From Ohm’s law, resistance (R) is obtained as R = V / I.
Explanation: Ohm’s law states that V ∝ I when temperature is constant.
Therefore, \( V = I × R \).
Rearranging gives \( R = \frac{V}{I} \).
Here, R is the constant of proportionality known as resistance, which measures how much a conductor opposes the flow of current.
Unit of R: Ohm (Ω)
Therefore, \( V = I × R \).
Rearranging gives \( R = \frac{V}{I} \).
Here, R is the constant of proportionality known as resistance, which measures how much a conductor opposes the flow of current.
Unit of R: Ohm (Ω)
Key Takeaways for Students
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MCQs are Concept Heavy: Group A tests your fundamental understanding of definitions and laws. Revise your textbook chapters thoroughly.
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Practice Numerical Problems: Questions on gas laws, moles, and electricity are scoring if you practice the numericals.
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Chemical Equations are Crucial: Make sure you can write balanced chemical equations for important reactions (like electrolysis, combustion, etc.).
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Define and Differentiate: Groups C and D require you to define terms clearly and write descriptive differences between concepts.
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