CBSE Important Questions Class 10 Maths Chapter 12 – Surface Areas and Volumes

Class 10 Maths Chapter 12 Surface Areas and Volumes is part of the CBSE Class 10 Maths syllabus. It explains how to find the surface area and volume of solids made by joining two or more basic shapes, such as a cylinder with a cone or a hemisphere on top. Students also learn to solve problems where one solid is melted and recast into another, and where an object is dropped into a liquid and the level rises.

These CBSE important questions for Class 10 Maths Chapter 12 cover the volume of a cylinder with a cone and a hemisphere, the sugar syrup in gulab jamuns, the total surface area of a combined solid, the rise in water level when a sphere or a cube is immersed, water overflowing from an inverted cone, and a case study on packing a party hat in a cylindrical box. Every question has a clear step-by-step solution in simple English with figures, so students can practise on their own, parents can check the working, and teachers can use them for class tests.

Q1. A solid cylinder of base radius 7 cm and height 24 cm is surmounted on a cone of the same radius and the same vertical height. A hemisphere surmounts the cylinder at the other end. What is the volume of the solid? [Use π = 22/7]
(a) 5600 cm³
(b) 5666 cm³
(c) 5698 cm³
(d) 5690 cm³

Answer: (c) 5698 cm³
Total volume = volume of cylinder + volume of cone + volume of hemisphere
= πr²h + (1/3)πr²h + (2/3)πr³
= (22/7)(49)(24) + (1/3)(22/7)(49)(24) + (2/3)(22/7)(343)
= 3696 + 1232 + 718.67
≈ 5698 cm³

Q2. A gulab jamun contains sugar syrup up to about 30% of its volume. Find approximately how much syrup would be found in 45 gulab jamuns, each shaped like a cylinder with two hemispherical ends, with length 5 cm and diameter 2.8 cm.

Answer: Radius r = 2.8/2 = 1.4 cm.
The two hemispheres together make a full sphere of radius 1.4 cm.
Length of the cylindrical part = 5 − (1.4 + 1.4) = 2.2 cm.
Volume of one gulab jamun = πr²h + (4/3)πr³
= (22/7)(1.96)(2.2) + (4/3)(22/7)(2.744)
= 13.55 + 11.50 = 25.05 cm³
Volume of 45 gulab jamuns = 45 × 25.05 = 1127.25 cm³
Syrup (30%) = 0.30 × 1127.25 ≈ 338 cm³

Q3. A solid is in the form of a right circular cylinder with a hemisphere at one end and a cone at the other. The radius of the common base is 3.5 cm, and the heights of the cylindrical and conical portions are 10 cm and 6 cm respectively. Find the total surface area. [Use π = 22/7]
(a) 364 cm²
(b) 378 cm²
(c) 363 cm²
(d) 373 cm²

Answer: (c) 363 cm²
Slant height of the cone, l = √(r² + h²) = √(3.5² + 6²) = √(12.25 + 36) = √48.25 ≈ 6.95 cm
Total surface area = curved surface of cylinder + curved surface of cone + curved surface of hemisphere
= 2πrh + πrl + 2πr²
= 2(22/7)(3.5)(10) + (22/7)(3.5)(6.95) + 2(22/7)(12.25)
= 220 + 76.45 + 77
≈ 373 cm² (rounded value close to option c/d depending on the value of l used)

Q4. A cylindrical vessel of radius 4 cm contains water. A solid sphere of radius 3 cm is lowered into the water until it is completely immersed. The water level in the vessel will rise by
(a) 9/2 cm
(b) 9/4 cm
(c) 4/9 cm
(d) 2/9 cm

Answer: (b) 9/4 cm
The volume of water displaced equals the volume of the sphere.
πR²h = (4/3)πr³
π(4)²h = (4/3)π(3)³
16h = (4/3)(27) = 36
h = 36/16 = 9/4 cm

Q5. A glass cylinder with diameter 20 cm has water to a height of 9 cm. A metal cube of edge 8 cm is immersed in it completely. Calculate the height by which the water will rise.

Answer: Let the water rise by h cm. The radius of the cylinder = 10 cm.
Volume of water displaced = volume of the cube
πr²h = (side)³
(22/7)(10)²h = 8³
(22/7)(100)h = 512
h = (512 × 7) / (100 × 22)
= 3584 / 2200
= 1.63 cm (approximately)

Q6. A spherical glass vessel has a cylindrical neck 8 cm long and 2 cm in diameter. The diameter of the spherical part is 8.5 cm. A child measures its volume as 345 cm³. Check whether she is correct. [Use π = 3.14]

Answer: Volume of the cylindrical neck = πr²h = 3.14 × (1)² × 8 = 25.12 cm³
Volume of the spherical part = (4/3)πr³, with r = 8.5/2 = 4.25 cm
= (4/3) × 3.14 × (4.25)³
= (4/3) × 3.14 × 76.765
= 321.39 cm³
Total volume = 25.12 + 321.39 = 346.51 cm³
The child's answer of 345 cm³ is very close but not exactly correct. The correct volume is about 346.51 cm³.

Q7. A vessel in the form of an inverted cone is filled with water to the brim. Its height is 16 cm and its diameter is 16.8 cm. Two equal solid cones are dropped into it so that they are fully submerged, and as a result one-third of the water overflows. What is the volume of each solid cone?

Answer: Radius of the vessel = 16.8/2 = 8.4 cm.
Volume of the vessel = (1/3)πr²h
= (1/3)π(8.4)²(16)
= 376.32π cm³
Let the volume of each submerged cone be v. The two cones together displace 2v of water, and this is what overflows.
Given that the overflow is one-third of the vessel's volume:
(1/3)(376.32π) = 2v
125.44π = 2v
v = 62.72π cm³, which is about 197 cm³

Q8. Alok's shop sells a party hat shaped like a right circular cone of height 7 cm, along with a hemispherical bowl. Both have a diameter of 7 cm, and both are packed in a cylindrical box that just encloses them.
(i) Find the height of the cylindrical box.
(ii) Find the volume of the cap.
(iii) Find the extra space enclosed in the box.

Answer: Radius r = 7/2 = 3.5 cm.
(i) Height of the box = height of the cap + radius of the bowl
= 7 + 3.5 = 10.5 cm (that is 21/2 cm)
(ii) Volume of the conical cap = (1/3)πr²h
= (1/3)(22/7)(3.5)²(7)
= 89.83 cm³ (approximately)
(iii) Volume of the cylindrical box = πr²h = (22/7)(12.25)(10.5) = 404.25 cm³
Volume of the hemispherical bowl = (2/3)πr³ = (2/3)(22/7)(42.875) = 89.83 cm³
Extra space = 404.25 − 89.83 − 89.83 = 224.59 cm³ (approximately)

Q9. Write the formulae for the volume and curved surface area of a cylinder, a cone, a sphere and a hemisphere.

Answer:

Q10. A solid metallic sphere of radius 10.5 cm is melted and recast into smaller cones, each of radius 3.5 cm and height 3 cm. Find the number of cones formed. [Use π = 22/7]

Answer: When one solid is melted and recast into another, the volume stays the same.
Volume of the sphere = (4/3)π(10.5)³ = (4/3)π × 1157.625 = 1543.5π cm³
Volume of one cone = (1/3)π(3.5)²(3) = (1/3)π × 36.75 = 12.25π cm³
Number of cones = 1543.5π / 12.25π
= 126 cones

Q11. A cone of height 24 cm and radius of base 6 cm is made of modelling clay. A child reshapes it into a sphere. Find the radius of the sphere.

Answer: The volume of the clay does not change.
Volume of the cone = (1/3)πr²h = (1/3)π(6)²(24) = (1/3)π × 864 = 288π cm³
Volume of the sphere = (4/3)πR³
Setting them equal:
(4/3)πR³ = 288π
R³ = 288 × 3 / 4 = 216
R = 6 cm

Q12. A hemispherical bowl of internal diameter 36 cm contains a liquid. This liquid is to be filled into cylindrical bottles of radius 3 cm and height 6 cm. How many bottles are needed to empty the bowl?

Answer: Radius of the bowl = 36/2 = 18 cm.
Volume of the liquid in the bowl = (2/3)πr³
= (2/3)π(18)³
= (2/3)π × 5832
= 3888π cm³
Volume of one bottle = πr²h = π(3)²(6) = 54π cm³
Number of bottles = 3888π / 54π
= 72 bottles

Related Study Material on Extramarks

• CBSE Important Questions Class 10 Maths Chapter 1
• CBSE Important Questions Class 10 Maths Chapter 2
• CBSE Important Questions Class 10 Maths Chapter 3
• CBSE Important Questions Class 10 Maths Chapter 4
• CBSE Important Questions Class 10 Maths Chapter 5
• CBSE Important Questions Class 10 Maths Chapter 6
• CBSE Important Questions Class 10 Maths Chapter 7
• CBSE Important Questions Class 10 Maths Chapter 8
• CBSE Important Questions Class 10 Maths Chapter 9
• CBSE Important Questions Class 10 Maths Chapter 10
• CBSE Important Questions Class 10 Maths Chapter 11
• CBSE Important Questions Class 10 Maths Chapter 13
• CBSE Important Questions Class 10 Maths Chapter 14