ParikshaPDF logo
ParikshaPDFAnalogy-rich Hindi & English exam notes
ExamsMock TestsNotes
हिन्दीSwitch language
LoginRegister

We use cookies for analytics. Optional analytics cookies help us understand usage - privacy notice.

Back to SSC CGL

Exam study plan

SSC CGL Preparation: Concepts and Practice

Free

Build your SSC CGL foundations with English and Hindi lessons, worked examples and explained practice across Quantitative Aptitude, General Intelligence and Reasoning, English Comprehension and General Awareness. Study arithmetic, algebra, geometry and trigonometry; practise analogy, classification, series, directions, ranking and clocks; strengthen grammar and constitutional basics. Use the linked topic tests to check understanding and review mistakes. Coverage is expanding subject by subject and does not yet represent the complete SSC CGL syllabus. See the module list and mock-test section for currently available material.

Lessons

Subject → Module → Lesson

English ComprehensionGeneral AwarenessGeneral Intelligence and ReasoningQuantitative Aptitude
Course outline 100
73 / 100
Lessons 51–100 · Page 2 of 2
Previous page

1 lesson
  1. 51
    Replacement and repeated dilution1 practice test

5 lessons
  1. 52
    Work, rate and efficiency
  2. 53
    Combined work and remaining work
  3. 54
    Efficiency ratios and worker equivalence
  4. 55
    Alternate-day and changing-team work
  5. 56
    Work and wages

6 lessons
  1. 57
    Speed, distance, time and unit conversions
  2. 58
    Average speed for unequal times and distances
  3. 59
    Relative speed and meeting or overtaking
  4. 60
    Trains crossing people, platforms and other trains
  5. 61
    Boats and streams
  6. 62
    Races and circular tracks

2 lessons
  1. 63
    Length, area and volume unit conversions
  2. 64
    Measurement accuracy and dimensional checks

5 lessons
  1. 65
    Perimeter and area of squares and rectangles
  2. 66
    Area and perimeter of triangles
  3. 67
    Parallelogram, rhombus and trapezium areas
  4. 68
    Circumference, circle and semicircle areas
  5. 69
    Composite figures, paths and shaded regions

6 lessons
  1. 70
    Surface area and volume of cubes and cuboids
  2. 71
    Surface area and volume of cylinders
  3. 72
    Surface area and volume of right circular cones
  4. 73
    Surface area and volume of spheres and hemispheres
  5. 74
    Right prisms and right pyramids with triangular or square bases
  6. 75
    Composite solids and volume-preserving conversions

7 lessons
  1. 76
    Variables, expressions and algebraic operations
  2. 77
    Standard algebraic identities
  3. 78
    Elementary factorisation
  4. 79
    Linear equations in one variable
  5. 80
    Pairs of linear equations
  6. 81
    Surds and simplification
  7. 82
    Graphs of linear equations

9 lessons
  1. 83
    Lines, angles and parallel-line relationships
  2. 84
    Triangle angle and side properties
  3. 85
    Medians, altitudes, angle bisectors and triangle centres
  4. 86
    Congruence and similarity of triangles
  5. 87
    Pythagoras theorem and elementary applications
  6. 88
    Properties of quadrilaterals
  7. 89
    Interior and exterior angles of polygons
  8. 90
    Circle chords and angle properties
  9. 91
    Tangents and common tangents to circles

6 lessons
  1. 92
    Trigonometric ratios in a right triangle
  2. 93
    Standard-angle values
  3. 94
    Basic trigonometric identities
  4. 95
    Complementary-angle relationships
  5. 96
    Degrees and radians
  6. 97
    Elementary heights and distances

3 lessons
  1. 98
    Perfect squares and elementary square patterns
  2. 99
    Square roots by factorisation and division
  3. 100
    Estimating square roots

50 lessons across 10 modules

Lesson 73 of 100 | Quantitative Aptitude / Solid Mensuration / ठोस क्षेत्रमिति

Surface area and volume of spheres and hemispheres

Learning outcome

Calculate sphere and hemisphere areas and volumes, separating a curved surface from a flat circular face or an open mouth.

Concepts and assumptions

A sphere’s surface lies at distance r from its centre. Its diameter is 2 × r. It has no flat base: its entire surface is curved.

A geometric surface comparison gives the same area as the curved wall of a cylinder of radius r and height 2 × r. Thus sphere area = 4 × π × r². Comparing circular slices shows its volume is two-thirds of that enclosing cylinder: sphere volume = (4/3) × π × r³.

Cutting through the centre produces two hemispheres. Each has half the sphere’s volume and half its curved area:

Hemisphere volume = (2/3) × π × r³; curved area = 2 × π × r².

A solid hemisphere also has the circular cut face, area π × r². Its total area is therefore 3 × π × r². A bowl’s open mouth is not a surface to coat: its inner curved area excludes that disk. A solid hemisphere resting flat on a surface has only its curved area exposed if the base is hidden.

Use π = 22/7 as the prescribed approximation. Dimensions are exact; bowl capacity uses internal radius and negligible wall thickness. Keep intermediate fractions unrounded. Round only the explicitly requested final capacity to the nearest 0.001 L; 1000 cm³ = 1 L.

Worked examples

Example 1 — Sphere. A sphere has radius 21 cm. Area = 4 × (22/7) × 21² = 5544 cm². Volume = (4/3) × (22/7) × 21³ = 38808 cm³. Area is measured in square units, while volume is measured in cubic units.

Example 2 — Solid hemisphere. Radius is 10.5 cm. The circular face has area (22/7) × 10.5² = 346.5 cm². Curved area = 2 × 346.5 = 693 cm²; total area = 693 + 346.5 = 1039.5 cm². Volume = (2/3) × 346.5 × 10.5 = 2425.5 cm³.

Example 3 — Recover bowl capacity. A hemispherical bowl has inner curved area 1232 cm². Find its radius and brimful capacity to the nearest 0.001 L. From 2 × (22/7) × r² = 1232, r² = 196 and r = 14 cm. Capacity = (2/3) × (22/7) × 14³ = 17248/3 cm³ = 17248/3000 L ≈ 5.749 L. Do not round before converting units.

Common mistakes

A hemisphere’s total area is not half a sphere’s area: the cut face is extra. Do not coat an imaginary disk across a bowl’s opening or substitute diameter for radius.

Practice questions

  1. A sphere has radius 10.5 cm. Find surface area and volume.
  2. A solid hemisphere has radius 7 cm. Find curved and total areas.
  3. An open hemispherical bowl has internal diameter 42 cm. Find its capacity in litres.
  4. A sphere with surface area 616 cm² is cut into two equal solid hemispheres. Find the sum of their total surface areas.

Worked solutions

  1. Area = 4 × (22/7) × 10.5² = 1386 cm². Volume = (4/3) × (22/7) × 10.5³ = 4851 cm³.
  2. Circular face area = (22/7) × 49 = 154 cm². Curved area = 308 cm²; total area = 308 + 154 = 462 cm².
  3. Radius = 42 ÷ 2 = 21 cm. Capacity = (2/3) × (22/7) × 21³ = 19404 cm³ = 19.404 L.
  4. Each new circular face has area 616 ÷ 4 = 154 cm². The original curved surface is retained, so combined area = 616 + 2 × 154 = 924 cm².
73 / 100
Loading footer…