NCERT Class 9 Solutions: Surface Areas and Volumes (Chapter 13) Exercise 13.6 – Part 1

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Formulas for volume of common solids

Volumes of Common Solids

Formulas for volume of common solids

Q-1 The circumference of the base of a cylindrical vessel is 132cm and its height is 25cm . How many liters of water can it hold? ( 1000cm3=1l )

Solution:

  • Let, the radius of the cylindrical vessel be r

  • Height (h) of vessel =25cm

  • Circumference of vessel =132cm

    We know that, circumference of the base of the cylindrical vessel

    • 2πrcm=132cm

    • 2×227×r=132

    • r=132×72×22

    • r=21cm

    Therefore, capacity of the cylindrical vessel

  • πr2h

  • 227×21×21×25cm3

  • 34650cm3

  • (346501000l) (Since, 1000cm3=1l )

  • 34.65l

Therefore, vessel can hold 34.65 liters of water.

Q-2 The inner diameter of a cylindrical wooden pipe is 24cm and its outer diameter is 28cm . The length of pipe is 35cm . Find the mass of the pipe, if 1cm3 of wood has a mass of 0.6g .

Solution:

  • Inner diameter =24cm

  • Therefore, inner radius (r) = 242cm (radius=diameter2)=12cm

  • Outer diameter =28cm

  • Therefore, outer radius(R) =282cm (radius=diameter2)=14cm

  • Outer volume =πR2h=(227×14×14×35)=21560cm3

  • Inner volume =πr2h=(227×12×12×35)=15840cm3

Since the inner volume is empty, volume of the wood used = outer volume - inner volume

  • =21560cm315840cm3

  • =5720cm3

Since, 1cm3 of wood has a mass of 0.6g , therefore mass of the pipe = 5720×0.6g=3.432kg

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