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CBSE Class 10  >  Class 10 Notes  >  Mathematics (Maths)   >  NCERT Solutions: Surface Areas & Volumes (Exercises 12.1, 12.2)

NCERT Solutions: Surface Areas & Volumes (Exercises 12.1, 12.2)

Exercise 12.1

Q1. 2 cubes each of volume 64 cm3 are joined end to end. Find the surface area of the resulting cuboid.

Sol.  Volume of each cube = 64 cm3
Exercise 12.1Let the edge of each cube = x
∴ x3 = 64 = 43
⇒ x =4 cm
⇒ Now, the Length of the resulting cuboid 'l' = 2x cm = 8cm
⇒ Breadth of the resulting cuboid 'b' = x cm = 4 cm
⇒ Height of the resulting cuboid 'h' = x cm = 4 cm
∴ Surface area of the cuboid = 2 (lb + bh + hl)
= 2 [(8 × 4) + (4 × 4) + (4 × 8)]
= 2 [32 + 16 + 32] cm2 = 2 [80] cm2 = 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.

Sol.  For the cylindrical part
 Radius (r) = 7 cm
 Height (h) = 6 cm
∴ Curved surface area
= 2πrh
Exercise 12.1
Exercise 12.1For hemispherical part
  Radius (r) = 7 cm
∴ Surface area = 2πr2
Exercise 12.1∴ Total surface area = CSA of cylinder + CSA of hemisphere
= (264 + 308) cm2 = 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.

Sol. Here, r = 3.5 cm
∴ h = (15.5 - 3.5) cm = 12.0 cm

Total surface area of the toy = CSA of the hemisphere + CSA of the cone
⇒ Surface area of the conical part = πrl
⇒ Surface area of the hemispherical part = 2πr2
Exercise 12.1

Slant height of the cone

Exercise 12.1

Total surface area of the toy = CSA of the hemisphere + CSA of the cone

Exercise 12.1

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.

Sol. Side of the block = 7 cm
⇒ The greatest diameter of the hemisphere = 7 cm
Surface area of the solid:
= [Total S.A. of the cubical block] + [S.A. of the hemisphere] - [Base area of the hemisphere]
= (6 × l2) + 2πr2 - πr2
Exercise 12.1

  Exercise 12.1

Exercise 12.1


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

Sol. Let 'l' be the side of the cube.
∴ The greatest diameter of the curved hemisphere = l
⇒ Radius of the curved hemisphere = l/2
∴ Surface area of hemisphere = 2πr2
Exercise 12.1

Exercise 12.1Base area of the hemisphere  
Exercise 12.1
Surface area of the cube = 6 × l2 = 6l2
∴ Surface area of the remaining solid
Exercise 12.1
Exercise 12.1

Q6. A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends (see Fig.). The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area.
Exercise 12.1

Sol. Radius of the hemispherical part
= 5/2 mm = 2.5 mm
Exercise 12.1∴ Surface area of one hemispherical part = 2πr2
⇒ Surface area of both hemispherical parts
Exercise 12.1
Area of cylindrical part = Exercise 12.1
∴ Total surface area
Exercise 12.1

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 m2. (Note that the base of the tent will not be covered with canvas.)

Sol. 

For cylindrical part:
 Radius (r) = 4/2 m = 2 m
 Height (h) = 2.1 m
∴ Curved surface area = 2πrh = Exercise 12.1
For conical part:
 Slant height (l) = 2.8 m
 Base radius (r)= 2m
∴ Curved surface area
Exercise 12.1
Total surface area:
= [Surface area of the cylindrical part] + [Surface area of conical part]
Exercise 12.1

Cost of the canvas used 
Cost of 1 m2 of canvas = Rs 500
∴ Cost of 44 m2 of canvas = Rs 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 cm2.

Sol. 

For cylindrical part:
 Height = 2.4 cm
 Diameter = 1.4 cm
⇒ Radius (r) = 0.7 cm
Exercise 12.1For conical part:
 Base area (r) = 0.7 cm
 Height (h) = 2.4 cm
∴ Slant height 
Exercise 12.1Exercise 12.1∴ Curved surface area of the conical part
Exercise 12.1Exercise 12.1
Base area of the conical part
Exercise 12.1

Total surface area of the remaining solid:
Exercise 12.1
Exercise 12.1

Exercise 12.1

Q9. A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in Fig. 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.
Exercise 12.1

Sol. Radius of the cylinder (r) = 3.5 cm
Height of the cylinder (h) = 10 cm
∴ Total surface area = 2πrh + 2πr2 = 2πr (h + r)
Exercise 12.1

Curved surface area of a hemisphere = 2πr2
∴ Curved surface area of both hemispheres
Exercise 12.1
Base area of a hemisphere = πr2

∴ Base area of both hemispheres = 2πr2
Exercise 12.1
∴ Total surface area of the remaining solid
= 297 cm2 + 154 cm2 - 77 cm2
= (451 - 77) cm2 = 374 cm2.

Exercise 12.2

[Unless stated otherwise, take π = 22/7]
Q1. A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1 cm and the height of the cone is equal to its radius. Find the volume of the solid in terms of π.

Sol. Here, r = 1 cm and h = 1 cm.
∵ Volume of the conical part = 1/3 πr2h
Volume of the hemispherical part =2/3 πr3
∴ Volume of the solid shape
Exercise 12.2
Exercise 12.2

Q2. Rachel, an engineering student, was asked to make a model shaped like a cylinder with two cones attached at its two ends by using a thin aluminium sheet. The diameter of the model is 3 cm and its length is 12 cm. If each cone has a height of 2 cm, find the volume of air contained in the model that Rachel made. (Assume the outer and inner dimensions of the model to be nearly the same.)

Sol. Here, diameter = 3 cm
⇒ Radius (r)= 3/2 cm
Total height = 12 cm
Height of a cone (h1) = 2 cm
∴ Height of both cones = 2 × 2 = 4 cm
⇒ Height of the cylinder (h2) = (12 - 4) cm 
= 8 cm.
Now, volume of the cylindrical part = πr2h2
Volume of both conical parts:
= Exercise 12.2
∴ Volume of the whole model
Exercise 12.2

Q3. 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 (see figure).
Exercise 12.2

Sol. Since a gulab jamun is like a cylinder with hemispherical ends.
Total height of the gulab jamun = 5 cm.
Diameter = 2.8 cm
⇒ Radius = 1.4 cm
∴ Length (height) of the cylindrical part 
= 5 cm - (1.4 + 1.4) cm
= 5 cm - 2.8 cm = 2.2 cm
Now, volume of the cylindrical part
= πr2h
Volume of a hemispherical end = 2/3 πr3
Volume of both the hemispherical ends Exercise 12.2
Exercise 12.2
∴ Volume of a gulab jamun
Exercise 12.2
Exercise 12.2
⇒ Volume of 45 gulab jamuns
Exercise 12.2
Since the quantity of syrup in gulab jamuns
Exercise 12.2
Exercise 12.2 = 338.184 cm3
= 338 cm3 (approx.)

Q4. A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm. The radius of each of the depressions is 0.5 cm and the depth is 1.4 cm. Find the volume of wood in the entire stand (see Fig.).
Exercise 12.2

Sol. Dimensions of the cuboid are 15 cm, 10 cm and 3.5 cm.
∴ Volume of the cuboid 

Exercise 12.2
Exercise 12.2
Since each depression is conical with base radius (r) = 0.5 cm and depth (h) = 1.4 cm,
∴ Volume of each depression (cone)
Exercise 12.2
Since there are 4 depressions,
∴ Total volume of 4 depressions
Exercise 12.2
Now, volume of the wood in entire stand
= [Volume of the wooden cuboid] - [Volume of 4 depressions]
Exercise 12.2

Q5. A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of its top, which is open, is 5 cm. It is filled with water up to the brim. When lead shots, each of which is a sphere of radius 0.5 cm are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped in the vessel.
Exercise 12.2

Sol. Height of the conical vessel (h) = 8 cm
Base radius (r) = 5 cm
∴ Volume of the cone = 1/3 πr2h
Exercise 12.2
Since Volume of the cone = [Volume of water in the cone]
Exercise 12.2
Now, Total volume of lead shots 
= 1/4 of [Volume of water in the cone]
Exercise 12.2
Since, radius of a lead shot (sphere) (r) = 0.5 cm
∴ Volume of 1 lead shot = Exercise 12.2

∴ Number of lead shots Exercise 12.2

Exercise 12.2
Thus, the required number of lead shots = 100.

Q6. A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, which is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the pole, given that 1 cm3 of iron has approximately 8 g mass. (Use π = 3.14) 
Exercise 12.2

Sol. Height of the big cylinder (h) = 220 cm
Base radius (r) = 24/2 cm = = 12 cm
∴ Volume of the big cylinder = πr2h = π (12)2 × 220 cm3
Also, height of smaller cylinder (h1) = 60 cm
Base radius (r1) = 8 cm
∴ Volume of the smaller cylinder πr12h1
= π (8)2 × 60 cm3
∴ Volume of iron
= [Volume of big cylinder] + [Volume of the smaller cylinder]
= π × 220 × 122 + π × 60 × 82 cm3
= 3.14 [220 × 12 × 12 + 60 × 8 × 8] cm3 = 314/100 [20 × 144 + 60 × 64] cm3
Exercise 12.2

Exercise 12.2

Q7. A solid consisting of a right circular cone of height 120 cm and radius 60 cm standing on a hemisphere of radius 60 cm is placed upright in a right circular cylinder full of water such that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is 60 cm and its height is 180 cm.

Sol. Height of the conical part = 120 cm.
Base radius of the conical part = 60 cm.
∴ Volume of the conical part = Exercise 12.2
Radius of the hemispherical part = 60 cm.
∴ Volume of the hemispherical part Exercise 12.2
∴ Volume of the solid
= [Volume of conical part] + [Volume of hemispherical part]
Exercise 12.2
Volume of the cylinder = πr2h
Exercise 12.2

Exercise 12.2

⇒ Volume of water in the cylinder   Exercise 12.2
∴ Volume of the water left in the cylinder
Exercise 12.2

Exercise 12.2 [∵ 1000000 cm3 = 1 m3]
= 1.13142857142 m3 = 1.131 m3 (approx).

Q8. A spherical glass vessel has a cylindrical neck 8 cm long, 2 cm in diameter; the diameter of the spherical part is 8.5 cm. By measuring the amount of water it holds, a child finds its volume to be 345 cm3. Check whether she is correct, taking the above as the inside measurements, and π = 3.14.

Sol. Volume of the cylindrical part
= πr2h = 3.14 × 12 × 8 cm3 [∵ Radius = 2/3 = 1 cm, height (h) = 8 cm]
Exercise 12.2
Volume of the spherical part
Exercise 12.2
Total volume of the glass-vessel
Exercise 12.2
⇒ Volume of water in the vessel = 346.51 cm3
Exercise 12.2
Since the child finds the volume as 345 cm3
∴ The child's answer is not correct
⇒ The correct answer is 346.51 cm3.

The document NCERT Solutions: Surface Areas & Volumes (Exercises 12.1, 12.2) is a part of the Class 10 Course Mathematics (Maths) Class 10.
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FAQs on NCERT Solutions: Surface Areas & Volumes (Exercises 12.1, 12.2)

1. How do I find the surface area of a cylinder in CBSE Class 10 exams?
Ans. Surface area of a cylinder equals 2πr(r + h), where r is the radius and h is the height. This formula combines the curved surface area (2πrh) and two circular bases (2πr²). Students must identify whether questions ask for total surface area or only curved surface area, as this affects which formula to apply during board exam calculations.
2. What's the difference between curved surface area and total surface area for cones?
Ans. Curved surface area of a cone is πrl (where l is slant height), covering only the lateral surface. Total surface area adds the circular base: πrl + πr². Understanding this distinction prevents calculation errors in Exercise 12.1 and 12.2 problems. Many students confuse these terms, leading to incorrect answers in NCERT solutions and exams.
3. How do I calculate the volume of a hemisphere and why is it half of a sphere's volume?
Ans. Volume of a hemisphere is (2/3)πr³, exactly half the sphere's volume (4/3)πr³. A hemisphere represents a sphere cut through its centre, dividing it into two equal parts. This relationship helps students verify calculations and understand three-dimensional geometry concepts tested in Class 10 surface areas and volumes chapters.
4. What formulas do I need to memorize for Exercise 12.1 and 12.2 problems?
Ans. Key formulas include: cylinder (TSA = 2πr² + 2πrh), cone (TSA = πr² + πrl), sphere (TSA = 4πr²), and cuboid (TSA = 2(lb + bh + hl)). Volume formulas follow: cylinder (πr²h), cone ((1/3)πr²h), and sphere ((4/3)πr³). Refer to mind maps and flashcards on EduRev to organise these formulas for quick revision before exams.
5. How do composite solids and real-world problems work in NCERT surface area and volume questions?
Ans. Composite solids combine two or more basic shapes-like a cylinder topped with a cone. Students must find total surface area by calculating individual surfaces and adding them, excluding overlapping regions. Real-world problems test application skills; examples include paint requirements for storage tanks or material needed for packaging, making these calculations practical and exam-relevant.
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