NCERT Solutions Class 10 Maths Chapter 13 Surface Areas And Volumes

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 13Surface Areas And Volumes

 

Exercise 13.1

(1)

(1) The volume of each cube = 64 cm3 = 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 × 42) cm2

= 10 × 16 cm2

= 160 cm2

 

(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 r2 + 2 π r h) sq unit

= [2 π r2 × 22/7 × (7)2 + 2 × 22/7 × 7 × 6] cm2

= 2 × 22/7 (7 + 6) cm2

= 2 × 22 × 13 cm2

= 572 cm2

The inner surface area of the vessel is 572 cm2

 

(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 = √h2 + r2 unit

= √122 + 3.52 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 r2] sq unit

= π r [L + 2 r] cm2

= 22/7 × 3.5 [12.5 + 2 × 3.5] cm2

= 11 × (12.5 + 7) cm2 = 11 × 19. 5 cm2 = 214.5 cm2

The total surface area of the toy in 214.5 cm2

 

(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 r2 + 6a2 – π r2) sq unit

= (π r2 + 6a2 ) sq unit

= [22/7 × (7/2)2 + 6 × 72] cm2

= (22/7 × 7/2 × 7/2 + 6 × 49) cm2

= (38.5 + 294) cm2

= 332.5 cm2

 

(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

= (6a2 + 2 π r2 – πr2) sq unit

= (6a2 + π r2) sq unit

= (6l2 + π r2/4) sq unit

= L2/4 (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 r2 + 2 π r h) sq unit

= 2 π r (2 r + h) sq unit

= 2 × π × 5/2 (2 × 5/2 + 9) mm2

= (22/7 × 5 × 14) mm2

= 220 mm2

Surface area of the capsule is 220 mm2

 

(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) m2

= 22/7 × 2 × (2.8 + 4.2) m2

= 22/7 × 2 × 7 m2

= 44 m2

The area of the canvas is 44 m2

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

The cost of canvas for the tent

= 44m2 × Rs 500 per m2

= 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)

= √h2 + r2

= √(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 + r2]

= 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

 

(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 h1) cm³

= π r² (h₂ + 2/3 h1) 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

Updated: December 21, 2021 — 2:01 pm

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