Surface Areas and Volumes Class 10th Notes - Free NCERT Class 10 Maths Chapter 13 Notes - Download PDF

Surface Areas and Volumes Class 10th Notes - Free NCERT Class 10 Maths Chapter 13 Notes - Download PDF

Edited By Safeer PP | Updated on Mar 19, 2022 01:25 PM IST

NCERT Class 10 Maths chapter 13 notes based on Surface, areas, and volumes, in which all the solid shapes are included and how to calculate the areas, volumes, or surface areas, total surface areas of curved shape can be easily evaluated. The chapter Surface, areas, and volumes Class 10 Notes also include combined shapes formulas.

This Story also Contains
  1. NCERT Class 10 Chapter 13 Notes-
  2. Surface, Areas, and Volumes Class 10 Notes- Topic 1:
  3. Surface, Areas, and Volumes Class 10 Notes- Topic 2:
  4. Cuboid :
  5. Cube :
  6. Cylinder :
  7. Cone :
  8. Sphere :
  9. Hemisphere :
  10. Spherical Shell :
  11. Shapes of Frustum :
  12. Surface, Areas, and Volumes Class 10 Notes- Topic 3:
  13. Significance of NCERT Class 10 Maths Chapter 13 Notes
  14. Class 10 Chapter Wise Notes
  15. NCERT Solutions of Cass 10 Subject Wise
  16. NCERT Class 10 Exemplar Solutions for Other Subjects:

Introductory Part may include: The NCERT Class 10 chapter 13 notes Properties of Surface, areas, and volumes are the highlighting key feature of Surface, areas and volumes 10 notes. CBSE Class 10 Maths chapter 13 notes also cover the basic formulae in the chapter. CBSE Class 10 Maths chapter 13 notes contain systematic explanations of topics using examples and exercises. All these topics can be downloaded from Class 10 Maths chapter 13 notes pdf download.

Also, students can refer,

NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes

NCERT Exemplar for Class 10 Maths Chapter 13 Surface Area and Volumes

NCERT Class 10 Chapter 13 Notes-

Surface, Areas, and Volumes Class 10 Notes- Topic 1:

Surface area: Surface area can be defined as the measurement of the total area that has been occupied by the object.

Volume: Volume is defined as the measurement of space that is occupied by the object.

Surface, Areas, and Volumes Class 10 Notes- Topic 2:

Basic solid shapes and their respective formula will be discussed in this chapter.

Cuboid :

The surface area of a cuboid with the roof.

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The surface area of the cuboid is equal to 2(lb +bh +hl) sq. unit.

Here, l denotes the length, b denotes the breadth, h denotes the height.

Area of Cuboid:

Area of cuboid covering four walls= [Perimeter length is multiplied by height] sq. unit.

= 2(l+ b) h

Surface area of cuboid with no roof:

Surface area with no roof =lb+2[bh +hl]

Volume of cuboid:

Volume = l×b×h unit.

Cube :

In this let us consider that each edge of length is to be taken as “a” unit.

1644404242556

Surface area:

Area of a cube that covers all the side = 6side2

= 6a2

Area of covered by all the four walls = 4side2

= 4a2 sq. units.

The surface area of a cube with no lid:

Area of a cube with no roof or lid=5 side2

= 5a2 sq. unit

Cylinder :

1644404242333

Surface area:

Surface area is taken as curved = base perimeter× height

= 2rπ×h

Total surface area can be calculated as the sum of curved surface area, and the area of two circular ends of the cylinder

Total surface area = 2rπh×2πr2

= 2rπ(r+h)

Volume:

Cylinder’s Volume = πr2h

The other volume which is a material present in a hollow pipe

= External volume- Internal Volume

=πR2h-πr2h

=πh[R2-r2]

The surface area of Hollow cylinder:

The total surface area of the hollow cylinder can be calculated as the curved surface area of the outer cylinder and inner cylinder with the addition of two base rings.

2Rπh+ 2rπh+ 2[πR2-πr2]

Cone :

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Slant height or length:

Slant height (l2) = (r2+h2) units.

h= height of the cone

r =radius of the cone

l = slant height of the cone

Surface area:

Surface area covered by all sides or lateral surface area= rπl sq unit.

Total surface area is to be calculated as the sum of curved surface area and area of base which is circular.

[rπl+ πr2] sq. unit

Volume:

Cone’s Volume = πr2h/3 cubic unit.

Sphere :

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Surface area:

The surface area of the sphere can be calculated as 4πr2.

Volume:

Sphere’s Volume = 4πr3/3 cubic unit.

Hemisphere :

Surface area:

The surface area which is curved = 2πr2

Total surface area is calculated as = 2πr2+ πr2= 3πr2

Volume:

Hemisphere’s Volume = 2πr3/3 cubic unit.

Spherical Shell :

1644404240890

Surface area:

Taking r2=r and r1=R

Total surface area can be calculated as = 4πr2+4πR2.

= 4π(r2+R2) sq. unit.

Volume:

Spherical shell’s volume = 4π(R3-r3)/3 cubic unit.

Shapes of Frustum :

1644404242019

Surface area:

Curved surface area can be calculated using = πrl(R+r)

Here, l is the slant height which is calculated using the slant height of frustum formulae.

The total surface area of the frustum of the cone can be calculated using = πrl(R+r)+ πr2+πR2 sq. unit.

Slant height:

Slant height(l2) = (R-r)2+h2

Volume:

Frustum Cone’s volume = πh(r2+R2+Rr)/3 sq. unit.

Surface, Areas, and Volumes Class 10 Notes- Topic 3:

Calculation of volume of a combination of solids can be done in accordance with the joining of two basic solids and then evaluate the sum of volumes of two.

Conversion of solids:

Conversion of shapes from one state to another can be possible by melting one solid shape and joining this with another.

Significance of NCERT Class 10 Maths Chapter 13 Notes

Surfaces, areas, and volumes Class 10 notes, will help to understand the formulas, statements, rules in detail. Also, this NCERT Class 10 Maths chapter 13 contains previous year’s questions and NCERT TextBook pdf. Surfaces, areas, and volumes Class 10 notes contain FAQ or frequently asked questions along with a topic-wise explanation. The Surfaces, areas, and volumes of Class 10 notes can be helpful in offline mode as well after downloading the notes.

Class 10 Chapter Wise Notes

NCERT Solutions of Cass 10 Subject Wise

NCERT Class 10 Exemplar Solutions for Other Subjects:

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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