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Exam study plan

SSC CGL Preparation: Concepts and Practice

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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
71 / 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 71 of 100 | Quantitative Aptitude / Solid Mensuration / ठोस क्षेत्रमिति

Surface area and volume of cylinders

Learning outcome

Find curved area, total area and capacity of right circular cylinders, distinguishing radius from diameter and open from closed surfaces.

Concepts and assumptions

A right circular cylinder has congruent circular bases separated by perpendicular height h. Radius r runs from a base centre to its rim; diameter = 2 × r.

Unrolling the curved wall gives a rectangle with width equal to circumference, 2 × π × r, and height h. Thus curved surface area, CSA = 2 × π × r × h, also called lateral area.

Each base has area π × r². A closed cylinder has total surface area, TSA = CSA + 2 × π × r². A top-open container includes only one base: area = CSA + π × r². An opening is not a material disk.

Every horizontal section has the same circular area. Stacking sections gives volume V = π × r² × h. Capacity requires internal radius and height.

Use π = 22/7 as a prescribed approximation, retaining it throughout without additional rounding. Lengths below are exact inputs in centimetres. Containers are thin-walled; inner coating counts one surface only. Ignore seams and overlap. Use 1000 cm³ = 1 L.

Worked examples

Example 1 — Closed cylinder. A solid cylinder has radius 7 cm and height 12 cm. Base area = (22/7) × 49 = 154 cm². CSA = 2 × (22/7) × 7 × 12 = 528 cm². TSA = 528 + 2 × 154 = 836 cm². Volume = 154 × 12 = 1848 cm³.

Example 2 — Open container. A top-open cylindrical container has internal diameter 28 cm and height 30 cm. Radius = 28 ÷ 2 = 14 cm. Inner curved area = 2 × (22/7) × 14 × 30 = 2640 cm²; base area = 616 cm². Inner coating area = 2640 + 616 = 3256 cm². Brimful capacity = 616 × 30 = 18480 cm³ = 18.48 L.

Example 3 — Recover height. A cylinder has volume 5390 cm³ and radius 7 cm. Base area = 154 cm², so height = 5390 ÷ 154 = 35 cm. A rectangular label covering its curved surface exactly once needs area = 2 × (22/7) × 7 × 35 = 1540 cm². Neither base is labelled.

Common mistakes

Halve a diameter before squaring. A closed solid needs both bases. A thick hollow pipe requires inner and outer dimensions; its material volume excludes the hollow space.

Practice questions

  1. A cylinder has diameter 14 cm and height 9 cm. Find curved area and volume.
  2. A closed cylinder has radius 3.5 cm and height 16 cm. Find total area and volume.
  3. A top-open container has internal radius 7 cm and height 25 cm. Find inner coating area and capacity.
  4. A cylinder has diameter 28 cm and volume 12320 cm³. Find its height and curved area.

Worked solutions

  1. Radius = 7 cm. CSA = 2 × (22/7) × 7 × 9 = 396 cm². Volume = 154 × 9 = 1386 cm³.
  2. Base area = (22/7) × 3.5² = 38.5 cm². CSA = 2 × (22/7) × 3.5 × 16 = 352 cm². TSA = 352 + 77 = 429 cm². Volume = 38.5 × 16 = 616 cm³.
  3. CSA = 2 × (22/7) × 7 × 25 = 1100 cm². Coating area = 1100 + 154 = 1254 cm². Capacity = 154 × 25 = 3850 cm³ = 3.85 L.
  4. Radius = 14 cm; base area = 616 cm². Height = 12320 ÷ 616 = 20 cm. CSA = 2 × (22/7) × 14 × 20 = 1760 cm².
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