NCERT Exemplar Solutions for Class 10 Maths Chapter 12 Surface Areas and Volumes Exercise 12.2

Chapter 12 Surface Areas and Volumes NCERT Exemplar Solutions Exercise 12.2 Class 10 Maths

Chapter Name	NCERT Maths Exemplar Solutions for Chapter 12 Surface Areas and Volumes Exercise 12.2
Book Name	NCERT Exemplar for Class 10 Maths
Other Exercises	Exercise 12.1 Exercise 12.3 Exercise 12.4
Related Study	NCERT Solutions for Class 10 Maths

Exercise 12.2 Solutions

Short Answer Questions with Reasoning

Write ‘True’ or ‘False’ and justify your answer in the following:

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 hemispheres are joined together along their bases, a sphere of same base radius is formed. 
Curved surface Area of a sphere  = 4πr² .

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 one cylinder is placed over another, the base of first cylinder and top of other cylinder will not be covered in total surface area.
We know that,
Total surface area of cylinder = 2πrh + 2πr² h 
(Where r = base radius and h = height)
Total surface area of shape formed = 2(Total surface of single cylinder) – 2(Area of base of cylinder) = 2(2πrh + 2πr²) – 2(πr²)
= 4πrh + 2πr²

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 πr[√(r² + h² + 3r + 2h].

Solution

False 
Explanation :
When a solid cone is placed over a solid cylinder of same base radius, the base of cone and top of the cylinder will not be covered in total surface area.
Since the height of cone and cylinder is same,
We get,
Total surface area of cone = πrl + πr² , (where r = base radius and l = slant height)
Total surface area of shape formed = Total surface area of cone + Total Surface area of cylinder – 2(Area of base)

Total surface area of cylinder = 2πrh + 2πr²h
(Where r = base radius and h = height)
Total surface area of cylinder = πr(r + l) + (2πrh + 2πr² ) – 2(πr² )
= πr² + πrl + 2πrh + 2πr² – 2πr² 
= πr(r + l + 2h)
= πr [√(r² + h² + r + 2h]

4. A solid ball is exactly fitted inside the cubical box of side a. The volume of the ball is
(4/3)πa³.

Solution

False 
Explanation :
Let the radius of sphere = r
When a solid ball is exactly fitted inside the cubical box of side a,
We get,
Diameter of ball = Edge length of cube
2r = a
Radius,
r = a/2 
We also know that, 
Volume of sphere = (4/3)πr³ 
Volume of ball = (4/3)π(a/2)³ 
= (4/3)π(a³/8)
= (1/6)πa³

5. The volume of the frustum of a cone is (1/3)πh[r₁² + r₂² – r₁r₂], where h is vertical height of the frustum and r₁ , r₂ are the radii of the ends. 

Solution

False 
As volume of the frustum of a cone is  (1/3)πh[r₁² + r₂² – r₁r₂]

6. The capacity of a cylindrical vessel with a hemispherical portion raised upward at the bottom as shown in the fig. is (πr³/3)[3h - 2r]. 

Solution

True 
We know that, capacity of cylindrical vessel = πr²h - (2/3)πr³ 
= (πr²/3)[3h - 2r]

7. The curved surface area of a frustum of a cone is πl(r₁ + r₂ ), where l = √h² + (r₁ + r₂ )² , r₁ and r₂ are the radii of the two ends of the frustum and h is the vertical height. 

Solution

False 
We know that, 
Curved surface area of a frustum of a cone is πl(r₁ + r₂), 
Where, l = √h² + (r₁ + r₂)² , r₁ and r₂ are the radii of the two ends  of the frustum and h is the vertical height.

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 
Because the resulting figure is:

Here, ABCD is a frustum of a cone and CDEF is hollow cylinder.