**Class 10 Maths NCERT Solutions of Chapter 13 exercise 13.1**

Class 10 maths NCERT solutions of chapter 13-Surace areas and volumes are presented here for helping the class 10 students in their preparation of maths paper in forthcoming assessments and CBSE board exam 2020-21. The NCERTquestions of the chapter 13-Surface areas and volume will help the students of 10 class in complete comprehension of the areas and volumes of three-dimensional figures like 3D shapes, cubes, cuboids, cylinders, cones, spheres and three-dimensional composite shapes made of at least two or more than two figures.All questions of the chapter 13 exercise 13.1 are solved by a specialist of maths subject according to the CBSE standards. The NCERT solutions published here by Future Study Point will benifit all categories of students in light of the fact that each question is explained by step by step strategy so every students could comprehend the concept unmistakably with no assistance of teacher or tutor.

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**Class 10 Maths NCERT Solutions of Chapter 13 exercise 13.1**

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**Q1. 2 cubes each of volume 64 cm ^{3} are joined end to end. Find the surface area of the resulting cuboid.**

Ans.

The volume of cubes = Side³ = 64

Side = 4 cm

Length of resulting cuboid = 4 + 4 = 8 cm

Breadth of cuboid = 4 cm

Height of resulting cuboid = 4 cm

Surface area of resulting cuboid = 2 ( lb + bh + hl)

l = 8 cm, b = 4 cm, h = 4 cm

Surface area of resulting cuboid = 2 ( 8 ×4 + 4×4 + 4×8) = 2(32 + 16 + 32) = 2× 80 = 160

Hence the required surface area of resulting cuboid = 160 cm²

**Q2. A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel.**

Solution:

Ans.Height of vessel is=13cm

Diameter of hemisphere = 14 cm, radius of hemishere = 14/2 = 7 cm

Height of cylinder = 13 – 7 = 6 cm

The inner surface area of the vessel = CSA of hemisphere + CSA of cylinder

CSA of hemisphere = 2πr²

The curved surface area of the hemisphere is = 308 cm²

CSA of cylinder = 2πrh

The curved surface area of cylinder = 264 cm²

Hence surface area of the vessel = 308 + 264 = 572 cm²

**Q3.A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of the same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy.**

Ans.

Height of cone = 13.5 – 3.5= 12 cm, radius of hemishere = 3.5 cm

The total surface area of the toy = CSA of hemisphere + CSA of cone

CSA of hemisphere = 2πr²

CSA of cone = πrl, where l is slant height

CSA of cone

The total surface area of the toy = 77 + 137.5 = 214.5

Hence the total surface area of the toy is 214.5 sq.cm

**Q4.A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have? Find the surface area of the solid.**

Ans. The greatest diameter of the hemisphere surmounted on a cube will be side of cube i.e 7 cm

The surface area of so formed solid = Total surface area of cube + curved surface area of the hemisphere – area of the base of the hemisphere

Total surface area of cube = 6×side² = 6 × 7 × 7 = 294

The total surface area of cube = 294 sq.cm

The curved surface area of hemisphere = 2πr²

Area of base (i.e circle) = πr²

Surface area of given solid = 294 + 77 – 38.5 = 332.5 sq.cm

Hence the surface area of the solid is 332.5 sq.cm

**Q5.A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter d of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid.**

Solution:

Ans. Let the side of the cubical wooden block is = a unit

Diameter d of the hemisphere = a unit

The radius of hemisphere depression = a/2

The remaining surface area of the solid = Total surface area of cube + curved surface area of the hemisphere – area of the base of the hemisphere

The total surface area of cube = 6a²

The curved surface area of the hemisphere = 2πr²

Area of the circular base of hemisphere =πr²

The remaining surface area of the solid

**Q6.A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends. The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area.**

Ans. The length of the capsule = 14 mm, length of the cylindrical portion of capsule = Length of the capsule – 2×2.5 = 14 – 5 = 9 mm

The surface area of the capsule = CSA of the cylindrical portion of the capsule + CSA of two hemispheres

CSA of the cylindrical portion of the capsule = 2πrh

CSA of two hemispheres = 2× 2πr² = 4πr²

r = 2.5 mm

CSA of two hemispheres

The surface area of the capsule

The surface area of the capsule is =220 mm²

**Q7.A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 2.1 m and 4 m respectively, and the slant height of the top is 2.8 m, find the area of the canvas used for making the tent. Also, find the cost of the canvas of the tent at the rate of Rs 500 per m ^{2}. (Note that the base of the tent will not be covered with canvas.)**

Ans.

Area of the canvas used for making the tent = Curved surface area of the cylindrical part of tent + Curved surface area of the conical part of the tent

The curved surface area of the cylindrical part of tent = 2πrh

Radius r of the cylindrical shape = 4/2 = 2 m

Height h of the cylindrical part = 2.1 m

The curved surface area of the cylindrical part of the tent

The curved surface area of cylindrical shape = 26.4 m²

The curved surface area of the conical part of the tent = πrl

Radius r of the conical shape = 4/2 = 2 m

Slant height l of the conical part = 2.8 m

The curved surface area of the conical part of the tent

The curved surface area of conical shape = 17.6 m²

Area of the canvas used for making the tent = 26.4 + 17.6 = 44 m²

The rate of canvas = Rs 500 per m²

∴The cost of the canvas = 500 × 44 = Rs 22000

**Q8.From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest cm ^{2}.**

Ans. Height of the given cylinder = 2.4 cm

The diameter of the given cylinder = 1.4 cm

Radius of the given cylinder = 1.4/2 = 0.7 cm

The height of the hollowed-out conical cavity = 2.4 cm

The remaining surface area of the cylinder = Curved surface area of cylinder + area of the circular base of cylinder + Curved surface area of the conical cavity

= 2πrh + πr² + πrl

= πr( 2h + r + l)

Therefore the remaining surface area of the cylinder = 17.6 ≈ 18 cm²

**Q9.A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in the figure. If the height of the cylinder is 10 cm, and its base is of radius 3.5 cm, find the total surface area of the** article.

Ans.

The total surface area of the article = Curved surface area of the cylinder + Curved surface area of scooped out hemispheres

The radius of cylinder and hemispheres = 3.5 cm

Height of the cylinder = 10 cm

The curved surface area of the cylinder = 2πrh

The curved surface area of scooped out hemispheres = 2× 2πr² = 4πr²

The total surface area of the article = 220 + 154 = 374 cm²

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**NCERT Solutions of Science and Maths for Class 9,10,11 and 12**

**NCERT Solutions of class 9 maths**

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Chapter 3- Coordinate Geometry | Chapter 11-Construction |

Chapter 4- Linear equations in two variables | Chapter 12-Heron’s Formula |

Chapter 5- Introduction to Euclid’s Geometry | Chapter 13-Surface Areas and Volumes |

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Chapter 2- Relations and functions | Chapter 10- Straight Lines |

Chapter 3- Trigonometry | Chapter 11-Conic Sections |

Chapter 4-Principle of mathematical induction | Chapter 12-Introduction to three Dimensional Geometry |

Chapter 5-Complex numbers | Chapter 13- Limits and Derivatives |

Chapter 6- Linear Inequalities | Chapter 14-Mathematical Reasoning |

Chapter 7- Permutations and Combinations | Chapter 15- Statistics |

Chapter 8- Binomial Theorem | Chapter 16- Probability |

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Chapter 1-Relations and Functions | Chapter 9-Differential Equations |

Chapter 2-Inverse Trigonometric Functions | Chapter 10-Vector Algebra |

Chapter 3-Matrices | Chapter 11 – Three Dimensional Geometry |

Chapter 4-Determinants | Chapter 12-Linear Programming |

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Chapter 8-Application of Integrals |