NCERT Exemplar Solution for Class 10 Maths: Surface Areas and Volumes, Exercise 12.2

Here you get the CBSE Class 10 Mathematics chapter 12, Surface Areas and Volumes: NCERT Exemplar Problems and Solutions (Part-II). This part of the chapter includes solutions for Exercise 12.2 of NCERT Exemplar Problems for Class 10 Mathematics Chapter: Surface Areas and Volumes. This exercise comprises of only the Very Short Answer Type Questions framed from various important topics in the chapter. Each question is provided with a detailed solution.

CBSE Class 10 Mathematics Syllabus 2017-2018

NCERT Exemplar problems are a very good resource for preparing the critical questions like Higher Order Thinking Skill (HOTS) questions. All these questions are very important to prepare for CBSE Class 10 Mathematics Board Examination 2017-2018 as well as other competitive exams.

Find below the NCERT Exemplar problems and their solutions for Class 10 Mathematics Chapter, Constructions:

Exercise 12.2

Very Short Answer Type Questions:

Write whether True or False and justify your answer.

Question. 1 Two identical solid hemispheres of equal base radius r cm are stuck together along their bases. The total surface area of the combination is 6πr².

Solution. False

Explanation:

When two solid hemispheres of same base radius r are joined together along their bases then the new solid formed is a sphere with the same radius r.

∴Total surface area of sphere = 4πr²

Question. 2 A solid cylinder of radius r and height h is placed over other cylinder of same height and radius. The total surface area of the shape so formed is 4πrh + 4πr².

Solution. False

Explanation:

When a solid cylinder of radius r and height h is placed over other cylinder of same height and radius, we will egt a new cylinder with same radius r and height 2h.

Question. 3 A solid cone of radius r and height h is placed over a solid cylinder having same base radius and height as that of a cone. The total surface area of the combined solid is

Solution. False

Explanation:

Here a solid cone of radius r and height h is placed over a solid cylinder having same base radius and height as that of a cone.

                                                                                

Total surface area of the combined solid = Curved surface area of cone + Curved surface area of cylinder + Area of the base of cylinder      

                                                                   

Question. 4 A solid ball is exactly fitted inside the cubical box of side a. The volume of the ball is4/3πr³.

Solution. False

Explanation:

Solution. False

Explanation:

Given, a frustum of cone with h its vertical height and r₁ and r₂ are the radii of its two ends.

 

Solution. True

Explanation:

Solution. False

Explanation:

Given, a frustum with its vertical height h and r₁ and r₂ the radii of its two ends.

Question. 8 An open metallic bucket is in the shape of a frustum of a cone, mounted on a hollow cylindrical base made of the same metallic sheet. The surface area of the metallic sheet used is equal to curved surface area of frustum of a cone + area of circular base + curved surface area of cylinder.'

Solution. True

Explanation:

Given, a bucket which is in shape of a frustum is mounted on a hollow cylindrical base.

                                                                                      

The surface area of the metallic sheet used to form given vessel will be equal to the total surface area of cylinder excluding the top and the curved surface area of frustrum of a cone.

∴ Total surface area of vessel = Curved surface area of frustrum + Curved surface area of cylinder + Area of base cylinder (Circular base of cylinder)

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