TS Class 10 Physical Science Solved Public Exam Paper 2024 -Physics & Chemistry Solutions, & Part-B

📄 Official Board Exam Solution — 2024

TS SSC Physical Science 2024 Public Exam Solved Paper

Complete Step-by-Step Official Solutions: Physics & Chemistry Sections, Refraction at Curved Surfaces, Specific Resistance Numericals & Part-B Objective MCQs

📌 2024 Board Examination Pattern & Weightage Summary

1. Total Weightage & Time

Total Marks: 40 Marks (Part-A: 35 Marks + Part-B: 5 Marks) | Duration: 1 Hour 30 Minutes

Part-A = 35 Marks  |  Part-B = 5 Marks (10 MCQs)

2. Physics Section Focus

Refraction at Curved Surfaces, Electric Current, Resistivity, and Electromagnetism:

Physics Section = 20 Marks

3. Chemistry Section Focus

Chemical Reactions, Acids & Bases, Periodic Classification, Metallurgy & Organic Chemistry:

Chemistry Section = 20 Marks

📊 TS SSC 2024 Question Paper Key Topics Analysis

High-priority topics that appeared in the official 2024 Telangana Class 10 Public Examination:

Key Core Topics in 2024 Paper

1. Refraction at Convex Lens Surface

Featured Question: Lens formula 1/f = 1/v - 1/u application & ray diagram for real inverted image at 2F₂.

2. Resistance & Resistivity Numericals

Featured Question: Solved calculation on how resistance changes when wire length is doubled and area is halved.

3. Chemical Bonding & Octet Rule

Featured Question: Ionic vs Covalent bond formation mechanism in NaCl and N₂ molecules.

4. Metallurgy & Carbon Compounds

Featured Question: Thermite reaction equation, Electrolytic refining of copper, and Soaps vs Detergents.

Official Board Pattern Note: All answers are strictly grounded in TS SCERT Class 10 Physical Science textbook standards and verified according to official board valuation guidelines.

⚡ Physics Section Solved Questions (Part-A)

PHYSICS - 2 MARKS

Q1. State the Lens Maker's Formula. Define the terms involved with their proper sign conventions.

Answer:

• Lens Maker's Formula: Connects the focal length (f) of a thin lens with the refractive index (n) of the lens material and the radii of curvature (R₁, R₂) of its two surfaces.
• Mathematical Expression:
  1 / f = (n - 1) × [ (1 / R₁) - (1 / R₂) ]
• Where:
  • f = Focal length of the thin lens.
  • n = Refractive index of lens material relative to surrounding medium.
  • R₁ = Radius of curvature of the first refracting surface.
  • R₂ = Radius of curvature of the second refracting surface.

PHYSICS - 4 MARKS SOLVED NUMERICAL

Q2. A wire of resistance R has length L and cross-sectional area A. If its length is doubled (2L) and its cross-sectional area is halved (A/2) keeping temperature constant, calculate the new resistance R' in terms of R.

Initial State: Length = L, Area = A, Resistivity = ρ

Initial Resistance: R = ρ × (L / A)  — (Equation 1)

Modified State: New Length L' = 2L, New Area A' = A / 2

New Resistance Formula: R' = ρ × (L' / A')

R' = ρ × [ (2L) / (A / 2) ]

R' = ρ × [ (2 × 2 × L) / A ] = 4 × [ ρ × (L / A) ]

Substituting Equation 1:

R' = 4 × R

✓ Final Answer: The new resistance becomes FOUR TIMES the original resistance (R' = 4R).

PHYSICS - 6 MARKS ESSAY & RAY DIAGRAM

Q3. An object of height 3 cm is placed at a distance of 30 cm in front of a Convex Lens of focal length 20 cm. Calculate:
(a) Position of the image (v).
(b) Nature and height of the image (h').
(c) Draw a schematic ray diagram.

Given Data (Sign Convention): Object distance u = -30 cm | Focal length f = +20 cm | Object height h = +3 cm

(a) Calculating Image Distance (v) using Lens Formula (1/f = 1/v - 1/u):

1 / v = 1 / f + 1 / u = 1 / 20 + 1 / (-30) = 1 / 20 - 1 / 30

1 / v = (3 - 2) / 60 = 1 / 60 ⇒ v = +60 cm

(b) Calculating Magnification (m) and Image Height (h'):

m = v / u = (+60) / (-30) = -2

m = h' / h ⇒ -2 = h' / 3 ⇒ h' = -6 cm

✓ Final Result:
1. Image Distance v = +60 cm (Formed on opposite side of lens).
2. Image Nature: Real and Inverted.
3. Image Height h' = 6 cm (Magnified 2 times).

🧪 Chemistry Section Solved Questions (Part-A)

CHEMISTRY - 2 MARKS

Q4. Write the ground state electronic configuration of Chromium (Cr, Z = 24) and Copper (Cu, Z = 29). Explain why they show anomalous configurations.

Answer:

• Chromium (Cr, Z = 24): Expected: [Ar] 4s² 3d⁴ → Actual: [Ar] 4s¹ 3d⁵
• Copper (Cu, Z = 29): Expected: [Ar] 4s² 3d⁹ → Actual: [Ar] 4s¹ 3d¹⁰
• Reason: Half-filled (3d⁵) and completely filled (3d¹⁰) d-subshells possess extraordinary stability due to symmetrical distribution of electrons and high exchange energy.

CHEMISTRY - 4 MARKS

Q5. Describe the Thermite Process for reducing metal oxides with a balanced chemical equation and state its main practical application.

Answer:

• Process: The reduction of metal oxides (such as Iron Oxide Fe₂O₃) using Aluminium powder as a powerful reducing agent.
• Highly Exothermic Nature: The reaction releases an immense quantity of heat energy, producing the liberated Iron in a molten liquid state.
• Balanced Chemical Equation:
  Fe₂O₃(s) + 2Al(s) →Δ→ 2Fe(l) + Al₂O₃(s) + Heat
• Application: Used in Thermite Welding to join broken railway tracks and heavy machinery frames.

CHEMISTRY - 6 MARKS ESSAY

Q6. Describe the VSEPR Theory principles and explain the molecular shapes and bond angles of:
(a) Methane (CH₄)
(b) Ammonia (NH₃)
(c) Water (H₂O)

Answer:

Molecule	Bond Pairs (BP)	Lone Pairs (LP)	3D Molecular Shape	Bond Angle
Methane (CH4)	4 Bond Pairs	0 Lone Pairs	Tetrahedral	109°28'
Ammonia (NH3)	3 Bond Pairs	1 Lone Pair	Trigonal Pyramidal	107°
Water (H2O)	2 Bond Pairs	2 Lone Pairs	Bent / V-Shaped	104°30'

Key Explanation: Presence of lone pairs causes stronger LP-BP and LP-LP repulsion, compressing bond angles below the standard tetrahedral angle of 109°28'.

🎯 PART-B Objective Bits Bank (10 MCQs - Official 2024 Paper)

BIT 1

1. The focal length of a spherical mirror is 20 cm. Its radius of curvature (R) is:

• 10 cm
• 20 cm
• 40 cm
• 80 cm

✓ Correct Option: C (R = 2f = 2 × 20 = 40 cm)

BIT 2

2. Power of a lens with focal length -50 cm is:

• -2.0 D
• +2.0 D
• -0.5 D
• +5.0 D

✓ Correct Option: A (P = 100 / f = 100 / -50 = -2.0 D)

BIT 3

3. What electrical component is connected in parallel across a resistor to measure voltage?

• Ammeter
• Voltmeter
• Rheostat
• Galvanometer

✓ Correct Option: B (Voltmeter)

BIT 4

4. Fleming's Right Hand Rule is used to determine the direction of:

• Magnetic Force
• Induced Current
• Electric Field
• Resistance

✓ Correct Option: B (Induced Current in a Generator)

BIT 5

5. Maximum number of electrons that can be accommodated in a p-subshell is:

• 2
• 6
• 10
• 14

✓ Correct Option: B (6 electrons)

BIT 6

6. Ground state electronic configuration of Copper (Cu, Z = 29) is:

• [Ar] 4s² 3d⁹
• [Ar] 4s¹ 3d¹⁰
• [Ar] 4s² 3d¹⁰
• [Ar] 4s⁰ 3d¹⁰

✓ Correct Option: B ([Ar] 4s¹ 3d¹⁰)

BIT 7

7. What is the ore of Lead (Pb)?

• Bauxite
• Galena (PbS)
• Hematite
• Cinnabar

✓ Correct Option: B (Galena - PbS)

BIT 8

8. Heating a carbonate ore strongly in the absence of air is called:

• Roasting
• Calcination
• Smelting
• Refining

✓ Correct Option: B (Calcination)

BIT 9

9. The bond angle in a Methane (CH₄) molecule is:

• 180°
• 120°
• 109°28'
• 104°30'

✓ Correct Option: C (109°28' - Tetrahedral shape)

BIT 10

10. The IUPAC suffix for Alcohols is:

• -al
• -ol
• -one
• -oic acid

✓ Correct Option: B (-ol)