**NCERT Solutions Class 10 Maths Chapter 13 Surface Areas And Volumes**

**NCERT Solutions Class 10** **Maths Chapter 13** **Surface Areas And Volumes:** National Council of Educational Research and Training Class 10 Maths Chapter 13 Solutions – Surface Areas And Volumes. NCERT Solutions Class 10 Maths Chapter 13 PDF Download.

**NCERT Solutions Class 10** **Maths Chapter ****13****: Overview**

Board |
NCERT |

Class |
10 |

Subject |
Maths |

Chapter |
13 |

Chapter Name |
Surface Areas And Volumes |

Topic |
Exercise Solutions |

**NCERT Solutions Class 10** **Maths**

**Chapter ****13**** – ****Surface Areas And Volumes**

__Exercise ____13____.1__

__Exercise__

__13__

__.1__

(1)

(1) The volume of each cube = 64 cm^{3} = V

Length of each ides = 3√3 cm

= 3√64 cm

= 4 cm

If we draw two cube side wise them we get (6 – 1) = 5 surfaces are open

Now that resulting cuboid has (5 × 2) = 10

Squares surfaces having each sides 4 cm only

The surface area of the cuboid

= (10 × surface of each square) sq unit

= (10 × 4^{2}) cm^{2}

= 10 × 16 cm^{2}

= 160 cm^{2}

(2) The diameter of the hemisphere = 14 cm

The radius of the hemisphere = r = 7 cm

The height of the hollow cylinder

= r = 7 cm

The height of the cylinder (h)

= height of the vessel – radius of the hemisphere

= 13 cm – 7 cm = 6 cm = h

The inner surface area of the vessel

= Surface area of hemisphere + curved surface of hemisphere

= 2 π r^{2} + 2 π r h) sq unit

= [2 π r^{2} × 22/7 × (7)^{2} + 2 × 22/7 × 7 × 6] cm^{2}

= 2 × 22/7 (7 + 6) cm^{2}

= 2 × 22 × 13 cm^{2}

= 572 cm^{2}

The inner surface area of the vessel is 572 cm^{2}

(3) The radius of the cone = 3.5 cm

The radius of the hemisphere = 3.5 cm

The height of the hemisphere is the same of the radius

The Total height of the toy = 15.5 cm

The height of the cone = h

= Total height of the toy – the radius of the hemisphere

= 15.4 cm – 3.5 cm

h = 12 cm

The slant height of the cone = l = √h^{2} + r^{2} unit

= √12^{2} + 3.5^{2 }cm

= √144 + 12.25 cm

= √156.25 cm

= 12.5 cm

The total surface area of the toy

= CSA of the cone + CSA of the hemisphere

[CSA = Curved surface area]

= [π r l + 2 π r^{2}] sq unit

= π r [L + 2 r] cm2

= 22/7 × 3.5 [12.5 + 2 × 3.5] cm^{2}

= 11 × (12.5 + 7) cm^{2} = 11 × 19. 5 cm^{2} = 214.5 cm^{2}

The total surface area of the toy in 214.5 cm^{2}

(4) The sides of the cubical black a = 7 cm

The hemisphere is surmounted on the cubical black

The greatest diameter which the hemisphere can have is equals to the edge of the cube.

So it have diameter (d) = the side of the black (a) diameter (d) = 7 cm, radius (a) = 7/2 cm

The total surface of the solid

= Curved surface area of the hemisphere + surface area of the cubical black – Flat surface of the hemisphere

= (2 π r^{2} + 6a^{2} – π r^{2}) sq unit

= (π r^{2} + 6a^{2 }) sq unit

= [22/7 × (7/2)^{2} + 6 × 7^{2}] cm^{2}

= (22/7 × 7/2 × 7/2 + 6 × 49) cm^{2}

= (38.5 + 294) cm^{2}

= 332.5 cm^{2}

(5) The diameter of the hemisphere (d) = L unit

The radius of the hemisphere (r) = L/2 unit

The sides of the hemisphere

r = L/2 unit

The sides of the cube (a) = L unit

The surface area of the hemisphere + surface area of cube – Flat surface of the hemisphere

= (6a^{2} + 2 π r^{2 }– πr^{2}) sq unit

= (6a2 + π r^{2}) sq unit

= (6l2 + π r^{2}/4) sq unit

= L^{2}/4 (π + 24) sq unit

(6) The capsule is in the shape of a cylinder with hemisphere in each ends

The diameter of each hemisphere (d)

= The diameter of the cylinder = 5 mm

The radius of the hemisphere (r) = 5/2 mm

Length of the cylinder = Length of the capsule – 2 × radius of hemisphere

= 14 mm – 2 × 5/2 mm

= 14 mm – 5 mm

= 9 mm

Surface area of the capsule

= 2 × curved surface area of hemisphere + curved surface area cylindrical part

= (2 × 2 π r^{2} + 2 π r h) sq unit

= 2 π r (2 r + h) sq unit

= 2 × π × 5/2 (2 × 5/2 + 9) mm^{2}

= (22/7 × 5 × 14) mm^{2}

= 220 mm^{2}

Surface area of the capsule is 220 mm^{2}

(7) The height of the cylindrical h = 2.1 m

The diameter of the Conceal top = the diameter of the cylinder (d)

= 4 m

Therefore, The radius of the cylinder and the conical top (r) = 4/2

= 2 m

The slant height of the conical Lop (L) = 2.8 M

The area of canvas

= Curved surface area of the conical top + curved surface area of the cylinder

= (π r L + 2πrh) sq unit

= π r (l + 2h) m2

= π × 2 (2.8 + 2 × 2.1) m^{2}

= 22/7 × 2 × (2.8 + 4.2) m^{2}

= 22/7 × 2 × 7 m^{2}

= 44 m^{2}

The area of the canvas is 44 m^{2}

The cost of canvas of the tent for each sq meter = Rs 500

The cost of canvas for the tent

= 44m^{2} × Rs 500 per m^{2}

= Rs (44 × 500)

= Rs 22000

The total cost of canvas Rs – 22000

(8) The solid has a cylindrical size and a conical cavity

The height of the cylinder

= The height of the conical cavity (h) = 2.4 cm

The diameter of the cylinder = 1.4 cm

The radius of the cylinder = the radius of the conical cavity (r) = 1.4/2 cm

= 0.7 cm

The slant height of the conical cavity (l)

= √h^{2} + r^{2}

= √(2.4)^{2} + (0.7)^{2} cm

= √6.25 cm

= 2.5 cm

The surface area of the solid

= Total surface area of solid cylinder + Total surface area of the cylinder

= π r L + [2 π r h + r^{2}]

= π r (L + 2h + r)

= 22/7 × 0.7 (2.5 + 4.8 + 0.7)

= 22 × 0.1 × (8)

= 2.2 × 8

= 17.6 cm²

So, the total surface area of the remaining solid is 17.6 cm²

(9) The solid is mode from a solid cylinder and two hemisphere had been scooped out from each end.

∴ The radius of the hemisphere (r)

= the radius of the cylinder (r) = 3.5 cm

∴ The height of the cylinder (h)

= 10 cm

The surface area of the solid

= curved surface area of the cylinder + 2 × curved surface area of hemisphere

= (2 π r h + 2 × 2 π r²) cm²

= 2 π r (h + 2r) cm²

= 2 × 22/7 × 3.5 (10 + 2 × 3.5) cm²

= 22 × (10 + 7) cm²

= 22 × 17 cm²

= 371 cm²

∴ The total surface area of the solid is 374 cm²

**EXERCISE – 13.2**

(1)

The height of the cone (h)

= the radius of the cone (r) = 1 cm

The volume of the solid

= the volume of the cone + the volume of the hemisphere

= (1/3 π r² h + 2/3 π r) cm³

= 1/3 π r² (h + 2r) cm³

= 1/3 × 22/7 × (1)² (1 + 2) cm³

= 1/3 × π × 3 cm³

= π cm³

∴ Total volume of solid = π cm³

**In case you are missed :- NCERT Solution for Triangles**

(2) The made shaped like cylinder with two cone each side

The diameter of the solid

= 3 cm

The radius of the cylinder

(r) = The radius of each cone

(r) = 3/2 cm

Total height of the solid = 12 cm

Height of each cone (h) = 2 cm

Height of the cylinder (h₂) = height

= Height of the solid – 2 × height of each cone

= 12 cm – 2 × 2 cm

= 12 cm – 4 cm

= 8 cm

The volume of the solid model

= the volume of the cylinder + 2 × volume of the cone

= (π r² h₂ + 2 × 1/3 π r² h_{1}) cm³

= π r² (h₂ + 2/3 h_{1}) cm³

= π × (3/2)^{2} – (8 + 2/3 × 2) cm³

= 22/7 × 9/8 × (24 + 4/3) cm³

= 22/7 × 9/4 × 28/3 cm³

= 66 cm³

∴ The volume of the model

= the volume of air contained in the model

= 66 cm³

(3) For one gulab jamun it made by a cylinder having two hemisphere in each end

The radius of each hemisphere (r)

Stand

= the of the cylinder (r)

= 2.8/2 cm

= 1.4 cm

= (π r² h + 2 × 2/3 π r ³) cm³

= π r² (h + 4/3 r) cm³

= 22/7 × 1.4 × 1.4 (2.2 + 4/3 × 1.4) cm³

= 22/7 × 1.4 × 1.4 (6.6 + 5.6/3) cm³

= 22/7 × 1.4 × 1.4 × 12.2/3 cm³

= 25.05 cm³

Volume of 45 gulab jamun

= 45 × 25.05 cm³

It given that each gulab jamun contains 30% sugar syrup

So syrup quantity in 45 gulab jamun

= 30% × 45 × 25.05 cm³

= 30/100 × 45 × 25.05 cm³

= 3381.75/10= 338.175 cm³

∴ The quantity of 45 gulab jamun is 338.175 cm³

(4) The stand is in the shape of a cuboid with four cone shaped depression Stand

For the cuboid

Length = L

= 15 cm

Breadth = b

= 10 cm

Height = h

= 3.5 cm

For the cone shapes depression

radius = r

= 0.5 cm

depth = d

= 1.4 cm

∴ The volume of the wooden stand

= Volume of the cuboid – 4 × volume of each cone

= (L h b) cm³ – 4 × 1/3 π r² d cm³

= L h b cm³ – 4/3 π r² d cm³

= (15 × 10 × 3.5 – 4/3 π × (0.5)² × 1.4) cm³

= (525 – 4/3 × 22/7 × 0.25 × 1.4) cm³

= (525 – 22 × 0.2/3) cm³

= (525 – 4.4/3) cm³

= (525 – 1.47) cm³

= 523.53 cm³

∴ The volume of the wooden pen stand h

= 523.53 cm³

(5) The cone holds water its brim the radius of the cone (R) = 5 cm the height of the cone (h) = 8 cm

The water its hold is equivalent of its volume

The volume of the cone (v) = 1/3 π r² h cm³

= 1/3 × 22/7 × 5 × 5 × 8 cm³

= 22 × 25 × 8/3 × 7 cm³

How, the radius of each lead shots is

= 0.50 m

The volume of lead shot = 4/3 π r³

= 4/3 × 22/7 × (0.5)³ cm³

The number of lead shot

= The volume of water from out/volume of one lead shot

= 22/7 × 50/3/4/3 × 22/7 × 0.5 × 0.5

= 10 × 10 × 10 × 10/4 × 5 × 5

= 100

Number of lead shots is 100.

(6) In the pole one cylinder is surmounted on another cylinder

The diameter of base cylinder = 24 cm

The radius of base cylinder (r) = 24 / 2 = 12 cm

The Height of the base cylinder = 220 cm

The radius of the mounted cylinder (r) = 8 cm

The height of mounted cylinder (h) = 60 cm

The volume of the pole

= The volume of the base cylinder + the volume of the sur mounted cylinder

= (π r² h + π r² h) cm³

= π [r² h + r² h] cm³

= π [{12² × 220} + {(8)² × 60}] cm³

= π [(144 × 220) + (64 × 60)] cm³

= π [31680 + 3840] cm³

= 3.14 × 35520 cm³

= 111532.8 cm³

Now, we have give that 1 cm³ of iron has 8 g of mass

∴ Total mass of the iron pole is

= (8 × 11532.8) g

= 892262.4 g

= 892.262 kg

(7) The solid shape is made with a cone surmounted on a hemisphere

The water displaced by it is the same of its own volume

∴ The height of the cone (h) = 120 cm

The radius of the cone

= The radius of the hemisphere

= 60 cm

The radius of the cylinder = 60 cm

= (r)

The height of the cylinder (h) = 180 cm

The volume of the shape

= Volume of the cone + volume of the hemisphere

The volume of water left in the cylinder

= volume of the cylinder – the volume of the solid shape

= π r² h cm³ – (1/3 π r² h + 2/3 π r³) cm³

= {π r² h – 1/3 π r² (h + 2r)} cm³

= π r² (h – 1/3 h – 2/3 r) cm³

= 22/7 × (60)² (180 – 1/3 × 120 – 2/3 × 60) cm³

= 22/7 × 60 × 60 (180 – 40 – 40) cm³

= 22/7 × 60 × 60 × 100 cm³

= 792000/7 cm³

= 792000/7 × 1 L

= 1131.43 L (approx.)

(8) The shape is a spherical glass vessel and a cylinder surmounted on u

The diameter of the cylinder = 2 cm

The radius of the cylinder (r) = 1 cm

The height of the cylinder (h) = 8 cm

The diameter of the shape = 8.5 cm

The radius of the sphere (r) = 8.5/2 cm

The volume of the glass vessel

= volume of the cylinder + volume of the sphere

= (π r² h + 4/3 π r³) cm³

= {π × 1² × 8 + 4/3 π × (8.5/2)³} cm³

= π (8 + 4/3 × 8.5/2 × 8.5/2 × 8.5/2) cm³

= 3.14 (8 + 102.35) cm³

= 3.14 × 110.35 cm³

= 346.51 cm³

Volume obtained by child is incorrect, correct volume is 346.51 cm³

**In case you are missed :- NCERT Solution for Introduction To Trigonometry**