1. Chapter 13 Class 10 Surface Areas and Volumes (Term 2)
  2. Serial order wise
  3. Ex 13.4

Ex 13.4, 1 - Chapter 13 Class 10 Surface Areas and Volumes (Term 2)

Last updated at July 18, 2019 by

Ex 13.4, 1 - A drinking glass is in shape of a frustum - Frustum of a cone - Volume

Ex 13.4, 1 - Chapter 13 Class 10 Surface Areas and Volumes - Part 2

  1. Chapter 13 Class 10 Surface Areas and Volumes (Term 2)
  2. Serial order wise

    3. Ex 13.4

    ExamplesΒ β†’

Transcript

Ex 13.4, 1 A drinking glass is in the shape of a frustum of a cone of height 14 cm. The diameters of its two circular ends are 4 cm and 2 cm. Find the capacity of the glass. Since glass is in from of frustum Capacity of glass = Volume of frustum = 1/3 πœ‹β„Ž(π‘Ÿ12+π‘Ÿ22+π‘Ÿ1π‘Ÿ2) Hence , h = height of frustum = 14 cm r1 = (π·π‘–π‘Žπ‘šπ‘’π‘‘π‘’π‘Ÿ π‘œπ‘“ 1^𝑠𝑑 𝑒𝑛𝑑)/2 = 4/2 = 2 cm r2 = (π·π‘–π‘Žπ‘šπ‘’π‘‘π‘’π‘Ÿ π‘œπ‘“ 2^𝑛𝑑 𝑒𝑛𝑑)/2 = 2/2 = 1 cm Volume of glass = 1/3 πœ‹β„Ž(π‘Ÿ12+π‘Ÿ22+π‘Ÿ1π‘Ÿ2) = 1/3Γ—22/7Γ—14(22+12+2Γ—1) = 1/3 Γ— 22Γ—2(4+1+2) = (22 Γ— 2 Γ— 7)/3 = 308/3 = 102.66 cm3 Hence, volume of glass = 102.66 cm3

Chapter 13 Class 10 Surface Areas and Volumes (Term 2)
Serial order wise

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