CBSE Class 10 Mathematics Surface Areas and Volumes Worksheet Set 01

Q.- A circus tent is in the shape of a cylinder, upto a height of 8 m, surmounted by a cone of the same radius 28 m. If the total height of the tent is 13 m, find:

Q.- A bird bath for garden in the shape of a cylinder with a hemispherical depression at one end (see figure). The height of the cylinder is 1.45 m and its radius is 30 cm.Find the total surface area of the bird-bath.

More question-

1.The lateral surface area of right circular cylinder with base radius 7cm and height 10 cm is:

2.The lateral surface area of cylinder is 176cm2 & base area 38.5cm². Then its volume is
(A) 803cm³
(B) 380cm³
(C) 308cm³
(D) 830cm³

3.Ratio of curved surface areas of two cylinders with equal radii is:
(A) H² : h²
(B) 2H : h
(C) H : h
(D) None

4.Two cubes of 12cm edge are joined end to end. Find the surface area of the resulting cuboid.

5.Three cubes of sides 6 cm edge are joined end to end. Find the surface area of the resulting cuboid.

6.A solid sphere of radius 6cm is melted and recast into small spherical balls each of diameter 0.6cm. Find the number of balls thus obtained.

7.How many spherical bullets can be made out of a solid cube of lead whose edge measures 55cm, each bullet being 10 cm in diameter?

8.The area of the base of a cone is 616 sq. cm. If its height is 48 cm then its total surface area is:
(A) 2681cm²
(B) 2861cm²
(C) 2816cm²
(D) None

9.Ratio of lateral surface areas of two cylinders with equal heights is .
(A) R : r
(B) H : h
(C) R² : r²
(D) None

10.The perimeter of ends of a frustum are 48 cm and 36 cm. If the height of the frustum be 11 cm, find its volume.
(A) 1400cm³
(B) 1500cm³
(C) 1554cm³
(D) 1600 cm

11.Find the maximum volume of a cone that can be curved out of a solid hemisphere of radius r.
(A) (4/3) πr²
(B) (1/3) πr³
(C) (1/3) πr²h
(D) None of these

12.A circus tent is in the form of a cone over a cylinder. The diameter of the base is 9 m, the height of cylindrical part is 4.8 m & the total height of the tent is 10.8 m. The canvass required for the tent is:
(A) 241.84 m²
(B) 24.184 m²
(C) 2418.4m²
(D) None

13.A fez, the cap used by the turks is shaped like the frustum of a cone. If its radius on the open side is 10cm, radius at the upper base is 4cm and its slant height is 15cm, find the area of material used for making it.
(A) 760cm²
(B) 710(2/7)cm²
(C) 731(2/7)cm²
(D) None of these

14.Determine the ratio of the volume of a cube to that of a sphere which will exactly fit inside the cube.
(A) 1:1
(B) 2: π
(C) π :5
(D) 6: π

15.If the radii of the circular ends of a conical bucket are 28 cm and 7 cm & height is 45 cm. The capacity of the bucket is:
(A) 48105cm²
(B) 48510cm²
(C) 48150cm²
(D) None

16.A cuboidal metal of dimensions 44cm × 30cm × 15cm was melted & cast into a cylinder of height 28 cm its radius is:
(A) 10 cm
(B) 20 cm
(C) 15 cm
(D) None

17.Find the volume of the largest right circular cone that can be cut out of a cube whose edge is 9 cm.
(A) 170 cm³
(B) 180.5 cm³
(C) 190.76 cm³
(D) 190.93 cm³

18.The area of the base of a cone is 616 sq. cm. If its height is 48 cm then its total surface area is:
(A) 2681cm²
(B) 2861cm²
(C) 2816cm²
(D) None

19.A top is of the shape of a cone over a hemisphere. The radius of the hemisphere is 3.5 cm. The total height of the top is 15.5 cm. The total area of top is:
(A) 215.4cm²
(B) 21.45cm²
(C) 214.5cm²
(D) None

20.A hollow sphere of internal and external diameters 4 cm & 8 cm respectively is melted into a cone of base diameter 8 cm. Find the height of the cone.
(A) 14 cm
(B) 12 cm
(C) 16 cm
(D) None

21.If the radii of the circular ends of a conical bucket is 45cm high, are 28cm and 7cm, find the capacity of the bucket.
(A) 25390 cm³
(B) 32670 cm³
(C) 43209 cm³
(D) 48510 cm³

22.Liquid is full in a hemisphere of inner diameter 9cm. This is to poured into cylindrical bottles of diameter 3 cm & height 4 cm . The number of bottles required are:
(A) 54
(B) 45
(C) 50
(D) None

23.A cylinder, whose height is two-third of its diameter, has the same volume as a sphere of radius 4cm. Calculate the radius of the base of the cylinder.
(A) 2 cm
(B) 4 cm
(C) 6 cm
(D) 8 cm

24.The diameter of a garden roller is 1.4 m and it is 2 m long. How much area will it cover in 5 revolutions?
(A) 50 sq m
(B) 44 sq m
(C) 40 sq m
(D) 35 sq m

25.Spherical ball of diameter 21 cm, is melted and recasted into cubes, each of side 1 cm. Find the number of cubes thus formed.
(A) 4045
(B) 4380
(C) 4851
(D) 4982

26.Two cubes each of 10 cm edge are joined end to end. Find the surface area of the resulting cuboid.
(A) 900 cm²
(B) 1000 cm²
(C) 1100 cm²
(D) None of these

27.Two cubes each of 10 cm edge are joined end to end. Find the surface area of the resulting cuboid.
(A) 900 cm²
(B) 1000 cm²
(C) 1100 cm²
(D) None of these

28.A spherical ball of diameter 21 cm is melted and recasted into cubes each of side 1cm. Find the number of cubes thus formed.
(A) 5021
(B) 4531
(C) 4851
(D) None of these

29.Metallic spheres of radii 6cm, 8cm and 10cm respectively, are melted to form a single solid sphere. Find the radius of the resulting sphere.
(A) 8 cm
(B) 10 cm
(C) 12 cm
(D) 14 cm

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30.A cylindrical vessel of diameter 9 cm has some water in it. A cylindrical iron piece of diameter 6 cm & height 4.5 cm is dropped in it. After it was completely immersed, the raise in the level of water is:

31.With a bucket of radius 14 cm & height 16 cm, 27 buckets of lime was poured to form a conical heap. If its area is 5544 cm², the canvass required to cover it is:

32.A piece of metal pipe is 77 cm long with inside diameter of the cross section is 4 cm. If the outer diameter is 4.5 cm & the metal weighs 8 gm/cu cm, the weight of pipe is:

33.The diameter of a copper sphere is 6 cm. The sphere is melted and is drawn into a long wire of uniform circular cross-section. If the length of the wire is 36 cm, find its radius.

34.A right circular cone is of height 8.4 cm and radius of its base is 2.1 cm.It is melted and recast into a sphere. Find the radius of the sphere.

35.Three cubes whose edges measure 3 cm, 4 cm and 5 cm respectively to form a single cube . Find its edge. Also, find surface area of the new cube.

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

37.A piece of metal pipe is 66 cm long with inside diameter of the cross section is 4 cm. If the outer diameter is 5.5 cm & the metal weighs 7 gm/cu cm, the weight of pipe is ........

38.The length of a cold storage is double its breadth. Its height is 3 meters. The areas of its four walls (including door) is 108 m . Find its volume.

39.A circus tent is cylindrical to a height of 3 m and conical above it. If its base radius is 52.5 m and slant height of a conical portion is 53 , find the area of the canvas required to make the tent.

40.The ratio of base radius and height of a cone is 3:4 . If the cost of smoothening the curved surface area at 5 paise / sq.cm is Rs.11550. Then volume of liquid is:

41.A drinking glass is in the shape of a frustum of a cone of height 14 cm. The diameter of its two circular ends are 4 cm and 2 cm. Find the capacity of the glass.

42.A metallic right circular cone of height 9 cm & base radius 7 cm is melted into a cuboid whose two sides are 11 cm & 6 cm. What is the third side of the cuboid?

43.The radii of the circular ends of a frustum of height 6 cm are 14 cm and 6 cm respectively. Find the lateral surface area and total surface area of the frustum.

44.The radii of the circular ends of a frustum of height 6 cm are 14 cm and 6 cm respectively. Find the lateral surface area and total surface area of the frustum.

45.A circus tent is cylindrical upto a height of 3 m and conical above it. If the diameter of the base is 105 m and the slant height of the conical part is 53 m, find the total canvas used in making the tent.

46.A vessel is in conical shape. If its volume is 33.264 lt. and height is 72 cm, the cost of repairing its CSA at Rs.12/sq.m is:

47.The total surface area of a cylinder is 220 sq.cm with height 6.5 cm. Then its volume is:

48.The largest sphere is curved out of a cube of a side 7 cm. Find the volume of the sphere.

49.From a circle of radius 15 cm a sector with 216° angle is cut out and its bounding radii are bent so as to form a cone. Then its volume:

50.The cost of painting the curved surface area of cone at Rs 5 cm² is Rs 3520. Which of the following volume of the cone, if its slant height is 25cm ?

51.A hemispherical bowl of internal diameter 40 cm contains a liquid. This liquid is to be filled in cylindrical bottles of radius 4 cm and height 8 cm. How many bottles are required to empty the bowl?

52.A conical vessel whose internal radius is 6cm and height is 25cm is full of water. The water is emptied into a cylindrical vessel with internal radius 10cm. Find the height to which the water rises.

53.Determine the ratio of the volume of cube to that of a sphere which will exactly fit inside the cube. 

54.The radii of the circular ends of a conical bucket which is 49cm high, are 35cm and 14cm. Find the capacity of the bucket.

55.Find the volume of the largest right circular cone that can be cut out of a cube whose edge is 10cm.

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56.An iron pillar has some part in the form of a right circular cylinder and remaining in the form of a right circular cone. The radius of the base of each of cone and cylinder is 8 cm. The cylindrical part is 240cm high and the conical part is 36cm high. Find the weight of the pillar if one cubic cm of iron weights 7.8 grams.

57.The interior of a building is in the form of a right circular cylinder of diameter 4.2m and height 4m surmounted by a cone. The vertical height of cone is 2.1m. Find the outer surface and volume of the building.

58.A circus tent is cylindrical upto a height of 3m and conical above. If the diameter of the base is 105m and vertical height of the conical part is 7.26 m. Find the total canvas used in making the tent.

59.A toy is in the shape of a right circular cylinder with a hemisphere on one end and cone on the other.
The radius and height of the cylindrical part are 5cm and 13cm respectively. The radii of the hemispherical and conical parts are the same as that of the cylindrical part. Find the surface area of the toy if the total height of the cone is 30cm.

60.A hollow cone is cut by a plane parallel to the base and the upper portion is removed. If the curved surface of the remainder is 8/9 of the curved surface of the whole cone. Find the ratio of the linesegment in which the cone's altitude is divided by the plane.

61.A sphere of diameter 7 cm is dropped in a right circular cylinder vessel partly filled with water. The diameter of the cylindrical vessel is 14 cm. If the sphere is completely submerged in water, by how much will the level of water rise in the cylindrical vessel?

Section A (1 mark each)

Question. The radius of a sphere is r cm. It is divided into two equal parts. Find the whole surface of the two parts.
Answer : (6πr² cm²)

Question. 12 solid spheres of the same size are made by melting a solid metallic cylinder of base radius 1cm and height 1/3 of 48cm. Find the radius of each sphere.
Answer : (1cm)

Question. If the radius of the base of a right circular cylinder is halved, keeping the height same, find the ratio of the volume of the reduced cylinder to that of original cylinder.
Answer : (1 : 4)

Question. Three cubes of iron whose edges are 3cm, 4cm and 5cm, respectively are melted and formed into a single cube. Find the edge of the new cube so formed.
Answer : (6cm)

Question. Volumes of two spheres are in the ratio 64 : 27. Find the ratio of their surface areas.
Answer : (16 : 9)

Section B (2 marks each)

Question. A hemispherical bowl of internal radius 9cm is full with a liquid. This liquid is to be filled into cylindrical shaped bottles of diameter 3cm and height 4cm. How many bottles are necessary to empty the bowl?
Answer : (54)

Question. A cylindrical tank has a capacity of 6160cm. Find its depth if its radius is 14m. Also calculate the cost of painting its curved surface (outer) at a rate of ₹ 3 per m².
Answer : a(Depth=5m; Cost= ₹ 1320)

Question. A glass cylinder with diameter 20cm has water to a height of 9cm. A metal cube of 8cm edge is immersed in it completely. Calculate the height by which water will rise in the cylinder.
Answer : (1.62cm)

Question. If a wire is bent into the shape of a square, then the area enclosed by the square is 81cm². When the same wire is bent into a semi-circular shape, find the area enclosed by the semi-circle.
Answer : (77cm²)

Question. Find the volume (in cm³) of the largest right circular cone that can be cut off from a cube of edge 4.2cm.
Answer : (19.4cm³)

Section C (3 marks each)

Question. The circumference of the base of a conical tent is 44m. If the height of tent is 24m, find the length of the canvas used in making the tent, if the width of the canvas is 2m. (use π = 22/7)
Answer : (275m)

Question. A spherical shell of lead whose external and internal diameters are 24cm and 18cm respectively is melted and recast into a right circular cylinder 37cm high. Find the radius of the base of the cylinder.
Answer : (6cm)

Question. A rectangular sheet of paper of dimensions 44cm X 18cm is rolled along its length and a cylinder is formed. Find the volume of the cylinder so formed. (use π = 22/7)
Answer : (2772cm³)

Question. Find the volume of the largest solid right circular cone that can be cut out of a solid cube of side 14cm.
Answer : (719cm³)

Question. A solid right circular cylinder has a total surface of 462 sq. cm. Its curved surface area is one-third of its total surface area. Find the volume of the cylinder.
Answer : (539 cm³)

Section D (4 marks each)

Question. Water is flowing through a cylindrical pipe, of internal diameter 2cm, into a cylindrical tank of base radius 40cm, at the rate of 0.4 m/s. Determine the rise in level of water in the tank in half an hour.
Answer : (4.5 cm)

Question. A bucket open at the top and made up of a metal sheet is in the form of a frustum of a cone. The depth of the bucket is 24cm and the diameters of its upper and lower circular ends are 30cm and 10 cm respectively. Find the cost of metal sheet used in it at the rate of ₹ 10 per 100cm². (use π = 3.14)
Answer : (₹ 171.13)

Question. A bucket is in the form of a frustum of a cone whose radii of the bottom and the top are 7cm and 28cm respectively. If the capacity of the bucket is 21560 cm³, find the whole surface area of the bucket.
Answer : (3344cm²)

Question. Water is flowing at the rate of 15km/hr through a cylindrical pipe of diameter 14cm into a cuboidal pond which 50m long and 44m wide. In what time the level of water in pond rise by 21cm?
Answer : (2hrs.)

Question. A right angled triangle, whose sides are 3cm, 4cm, and 5cm, is revolved about the longest side. Find the surface area of the figure (double cone) obtained.
Answer : (52.8cm²)

Question. A hollow cone is cut by a plane parallel to the base and the upper portion is removed. If the curved surface of the remainder is 8/9 of the curved surface of the whole cone, find the ratio of the line segments in which the altitude of the cone is divided by the plane.
Answer : (1 : 2)

Question. The height of a cone is 30cm. A small cone is cut off at the top by a plane parallel to the base. If its volume be (1/27)th of the volume of the given cone, at what height above the base is the section made?
Answer : (20 cm)

Question. A cone is divided into two parts by drawing a plane through the midpoint of its axis, parallel to its base. Compare the volumes of the two parts.
Answer : (1/7)

Surface Area and Volumes - Key Points

1. Cuboid

• Total Surface Area = \( 2(lb + bh + hl) \) square units
• Volume = \( l \times b \times h \) cubic units
• Diagonal = \( \sqrt{l^2 + b^2 + h^2} \) units

2. Cube

• Total Surface Area = \( 6a^2 \) square units
• Volume = \( a^3 \) cubic units
• Diagonal = \( a\sqrt{3} \) units

3. Right Circular Cylinder

• Curved Surface Area = \( 2\pi rh \) square units
• Total Surface Area = \( 2\pi r(h + r) \) square units
• Volume = \( \pi r^2 h \) cubic units

4. Right Circular Hollow Cylinder

• Area of each end = \( \pi(R^2 - r^2) \) square units (where \( R \) and \( r \) are the external and internal radii)
• Curved Surface Area = \( 2\pi h(R + r) \) square units
• Total Surface Area = \( \pi(R + r)(2h + R - r) \) square units
• Volume of material = \( \pi h(R^2 - r^2) \) cubic units

5. Sphere

• Surface Area = \( 4\pi r^2 \) square units
• Volume = \( \frac{4}{3}\pi r^3 \) cubic units

6. Hemisphere

• Curved Surface Area = \( 2\pi r^2 \) square units
• Total Surface Area = \( 3\pi r^2 \) square units
• Volume = \( \frac{2}{3}\pi r^3 \) cubic units

7. Right Circular Cone

• Curved Surface Area = \( \pi rl \) square units (where \( l = \sqrt{r^2 + h^2} \) is the slant height)
• Total Surface Area = \( \pi r(l + r) \) square units
• Volume = \( \frac{1}{3}\pi r^2 h \) cubic units

8. Frustum of a Cone

• Volume = \( \frac{1}{3}\pi h(R^2 + r^2 + Rr) \) cubic units (where \( R \) and \( r \) are the radii of the circular ends)
• Lateral Surface Area = \( \pi l(R + r) \) square units (where \( l = \sqrt{h^2 + (R - r)^2} \) is the slant height)
• Total Surface Area = \( \pi[R^2 + r^2 + l(R + r)] \) square units

Level-I

Question 1. The Surface Area of a Sphere is 616 cm2. Find its radius.
Answer: Let the radius of the sphere be \( r \).
The formula for the surface area of a sphere is:
\[ 4\pi r^2 = 616 \]
Substitute \( \pi = \frac{22}{7} \):
\[ 4 \times \frac{22}{7} \times r^2 = 616 \]
\[ r^2 = \frac{616 \times 7}{4 \times 22} \]
\[ r^2 = \frac{616 \times 7}{88} \]
\[ r^2 = 7 \times 7 = 49 \]
\[ r = 7\text{ cm} \]
Therefore, the radius of the sphere is 7 cm.
In simple words: We plug the surface area of 616 into the formula \( 4\pi r^2 \) and solve for the radius, which gives 7 cm.

Exam Tip: Always write down the general formula before substituting numbers to avoid algebraic mistakes.

Question 2. The slant height of the frustum of a cone is 5 cm. if the difference between the radii of its two circular ends is 4cm, write height of the frustum.
Answer: Let the height of the frustum be \( h \), and the radii of its circular ends be \( R \) and \( r \).
We are given:
- Slant height (\( l \)) = 5 cm
- Difference between radii (\( R - r \)) = 4 cm

Using the slant height formula for a frustum:
\[ l^2 = h^2 + (R - r)^2 \]
Substitute the given values:
\[ 5^2 = h^2 + 4^2 \]
\[ 25 = h^2 + 16 \]
\[ h^2 = 25 - 16 = 9 \]
\[ h = 3\text{ cm} \]
Therefore, the height of the frustum is 3 cm.
In simple words: Using the formula relating height, slant height, and the difference in radii, we get \( 5^2 = h^2 + 4^2 \), which solves to a height of 3 cm.

Exam Tip: Remember that \( l, h, \) and \( (R - r) \) form a right-angled triangle, so you can apply the Pythagorean triple (3, 4, 5) directly.

Question 3. A cylinder and a cone area of the same base radius and of the same height. Find the ratio of the cylinder to that of the cone.
Answer: Let the common base radius of both solids be \( r \) and their common height be \( h \).
- Volume of the cylinder (\( V_1 \)) = \( \pi r^2 h \)
- Volume of the cone (\( V_2 \)) = \( \frac{1}{3}\pi r^2 h \)

Taking the ratio of their volumes:
\[ \frac{V_1}{V_2} = \frac{\pi r^2 h}{\frac{1}{3}\pi r^2 h} = \frac{1}{\frac{1}{3}} = \frac{3}{1} \]
Therefore, the ratio of the volume of the cylinder to that of the cone is \( 3:1 \).
In simple words: Since a cone occupies exactly one-third of the space of a cylinder with the same dimensions, the ratio of the cylinder's volume to the cone's volume is 3 to 1.

Exam Tip: Clearly define that the ratio refers to their volumes, as the question mentions "cylinder to that of the cone" without explicitly writing "volumes".

Question 4. Two cones have their heights in the ratio 1:3 and radii 3:1. What is the ratio of their volumes?
Answer: Let the radii of the two cones be \( r_1 \) and \( r_2 \), and their heights be \( h_1 \) and \( h_2 \).
We are given:
- Ratio of heights: \( \frac{h_1}{h_2} = \frac{1}{3} \)
- Ratio of radii: \( \frac{r_1}{r_2} = \frac{3}{1} \)

The ratio of their volumes is:
\[ \frac{V_1}{V_2} = \frac{\frac{1}{3}\pi r_1^2 h_1}{\frac{1}{3}\pi r_2^2 h_2} \]
\[ \frac{V_1}{V_2} = \left(\frac{r_1}{r_2}\right)^2 \times \left(\frac{h_1}{h_2}\right) \]
Substitute the given ratios:
\[ \frac{V_1}{V_2} = \left(\frac{3}{1}\right)^2 \times \left(\frac{1}{3}\right) \]
\[ \frac{V_1}{V_2} = 9 \times \frac{1}{3} = \frac{3}{1} \]
Therefore, the ratio of their volumes is \( 3:1 \).
In simple words: We find the ratio by squaring the radius ratio and multiplying it by the height ratio, which simplifies to a 3 to 1 ratio.

Exam Tip: When working with ratios, write the formula in terms of ratio fractions to easily substitute the given values.

Question 5. The radii of two cones are in the ratio 2:1 and their volumes are equal. What is the ratio their heights?
Answer: Let the radii of the two cones be \( r_1 \) and \( r_2 \), and their heights be \( h_1 \) and \( h_2 \).
We are given:
- Ratio of radii: \( \frac{r_1}{r_2} = \frac{2}{1} \)
- Volumes are equal: \( V_1 = V_2 \)

Using the volume formula for cones:
\[ \frac{1}{3}\pi r_1^2 h_1 = \frac{1}{3}\pi r_2^2 h_2 \]
\[ \implies r_1^2 h_1 = r_2^2 h_2 \]
Rearranging to find the ratio of their heights:
\[ \frac{h_1}{h_2} = \left(\frac{r_2}{r_1}\right)^2 \]
Substitute the reciprocal of the radius ratio:
\[ \frac{h_1}{h_2} = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \]
Therefore, the ratio of their heights is \( 1:4 \).
In simple words: Since the volumes are the same, the cone with the larger base must be shorter to compensate. Squaring the inverse of the radius ratio gives a height ratio of 1 to 4.

Exam Tip: Since radius is squared in the volume formula, its ratio must be squared when solving for the height ratio.

Question 6. The diameter of a sphere is 6 cm. it is melted and drawn into a wire of diameter 2mm. Find the length of the wire.
Answer: Let's find the volume of the sphere and the wire:

1. **Sphere:**
- Diameter = 6 cm \implies Radius (\( R \)) = 3 cm
- Volume of the sphere = \( \frac{4}{3}\pi R^3 = \frac{4}{3}\pi (3)^3 = 36\pi\text{ cm}^3 \)

2. **Wire (Cylinder):**
- Diameter = 2 mm \implies Radius (\( r \)) = 1 mm = 0.1 cm
- Let the length of the wire be \( h \).
- Volume of the wire = \( \pi r^2 h = \pi (0.1)^2 h = 0.01\pi h\text{ cm}^3 \)

Since the sphere is melted to form the wire, their volumes must be equal:
\[ 36\pi = 0.01\pi h \]
\[ \implies h = \frac{36}{0.01} = 3600\text{ cm} \]
Converting to meters:
\[ h = 36\text{ m} \]
Therefore, the length of the wire is 36 m (or 3600 cm).
In simple words: Melting a shape doesn't change its total volume. Equating the volume of the sphere to the volume of the thin cylindrical wire gives a length of 36 meters.

Exam Tip: Be extremely careful with units; always convert millimeters (diameter of the wire) to centimeters before equating the volumes.

Question 7. Find the curved surface area of a right circular cone of height 15cm and base diameter is 16 cm.
Answer: Let the height of the cone be \( h = 15\text{ cm} \).
- Base diameter = 16 cm \implies Radius (\( r \)) = 8 cm

First, find the slant height (\( l \)) using Pythagoras' theorem:
\[ l = \sqrt{h^2 + r^2} \]
\[ l = \sqrt{15^2 + 8^2} = \sqrt{225 + 64} = \sqrt{289} = 17\text{ cm} \]

Now, calculate the curved surface area (CSA):
\[ \text{CSA} = \pi r l \]
\[ \text{CSA} = \pi \times 8 \times 17 = 136\pi\text{ cm}^2 \]
Using \( \pi \approx \frac{22}{7} \):
\[ \text{CSA} \approx 136 \times \frac{22}{7} \approx 427.43\text{ cm}^2 \]
In simple words: We find the slant height of 17 cm first using Pythagoras' theorem. Then, plugging the radius and slant height into the area formula gives a curved surface area of \( 136\pi \) square cm.

Exam Tip: Always state both the exact answer in terms of \( \pi \) and the approximate decimal answer to ensure you cover all grading keys.

Question 8. Find the maximum volume of a cone that can be out of a solid hemisphere of radius r.
Answer: To carve a cone of maximum volume out of a solid hemisphere of radius \( r \):
- The base radius of the cone must be equal to the radius of the hemisphere, i.e., \( R = r \).
- The height of the cone must be equal to the radius of the hemisphere, i.e., \( h = r \).

Using the volume formula for a cone:
\[ V = \frac{1}{3}\pi R^2 h \]
Substitute \( R = r \) and \( h = r \):
\[ V = \frac{1}{3}\pi r^2 (r) = \frac{1}{3}\pi r^3 \]
Therefore, the maximum volume of the cone is \( \frac{1}{3}\pi r^3 \).
In simple words: The largest cone we can carve has a base and height equal to the hemisphere's radius. This makes its volume exactly \( \frac{1}{3}\pi r^3 \).

Exam Tip: Note that the maximum volume of the cone is exactly half the volume of the hemisphere (\( \frac{2}{3}\pi r^3 \)).

Question 9. The diameter of the ends of a frustum of a cone are 32cm and 20 cm. If its slant height is 10 cm. Find the lateral surface area.
Answer: Let the diameters of the ends be 32 cm and 20 cm:
- Larger radius (\( R \)) = 16 cm
- Smaller radius (\( r \)) = 10 cm
- Slant height (\( L \)) = 10 cm

The formula for the lateral (curved) surface area of a frustum of a cone is:
\[ \text{LSA} = \pi L (R + r) \]
Substitute the values:
\[ \text{LSA} = \pi \times 10 \times (16 + 10) \]
\[ \text{LSA} = 260\pi\text{ cm}^2 \]
Using \( \pi \approx \frac{22}{7} \):
\[ \text{LSA} \approx 260 \times \frac{22}{7} \approx 817.14\text{ cm}^2 \]
In simple words: We find the radii are 16 cm and 10 cm. Plugging these and the slant height of 10 cm into the formula gives a curved surface area of \( 260\pi \) square cm.

Exam Tip: Make sure not to use the diameter values directly in the formula; always divide them by 2 first to get the radii.

Level-II

Question 1. Metallic sphere of radii 6cm, 8cm and 10cm respectively, are melted to form a single solid sphere. Find the radius of the resulting sphere.
Answer: Let the radii of the three smaller spheres be \( r_1 = 6\text{ cm} \), \( r_2 = 8\text{ cm} \), and \( r_3 = 10\text{ cm} \).
Let the radius of the newly formed sphere be \( R \).

Since the total volume remains constant during melting:
\[ \text{Volume of large sphere} = \text{Volume of sphere 1} + \text{Volume of sphere 2} + \text{Volume of sphere 3} \]
\[ \frac{4}{3}\pi R^3 = \frac{4}{3}\pi r_1^3 + \frac{4}{3}\pi r_2^3 + \frac{4}{3}\pi r_3^3 \]
Divide both sides by \( \frac{4}{3}\pi \):
\[ R^3 = r_1^3 + r_2^3 + r_3^3 \]
Substitute the values:
\[ R^3 = 6^3 + 8^3 + 10^3 \]
\[ R^3 = 216 + 512 + 1000 = 1728 \]
Taking the cube root:
\[ R = \sqrt[3]{1728} = 12\text{ cm} \]
Therefore, the radius of the resulting sphere is 12 cm.
In simple words: The volume of the new sphere is the sum of the volumes of the three smaller ones. Taking the cube root of the sum of their cubed radii (1728) gives a final radius of 12 cm.

Exam Tip: Cancelling out \( \frac{4}{3}\pi \) from both sides first simplifies the calculation and saves valuable time.

Question 2. A 20m deep well with diameter 7m is dug and the earth from digging is evenly spread out to form a platform 22m by 14m. Find the height of the platform.
Answer: Let's find the volume of the dug-up soil and the platform:

1. **Well (Cylinder):**
- Depth (\( h \)) = 20 m
- Diameter = 7 m \implies Radius (\( r \)) = 3.5 m (or \( \frac{7}{2}\text{ m} \))
- Volume of earth dug out = \( \pi r^2 h = \frac{22}{7} \times \left(\frac{7}{2}\right)^2 \times 20 = 770\text{ m}^3 \)

2. **Platform (Cuboid):**
- Length (\( L \)) = 22 m
- Breadth (\( B \)) = 14 m
- Let the height of the platform be \( H \).
- Volume of the platform = \( L \times B \times H = 22 \times 14 \times H\text{ m}^3 \)

Since the volume of the earth dug out equals the volume of the platform:
\[ 22 \times 14 \times H = 770 \]
\[ H = \frac{770}{22 \times 14} = \frac{35}{14} = 2.5\text{ m} \]
Therefore, the height of the platform is 2.5 m.
In simple words: The dirt taken out of the cylindrical well is used to make a rectangular platform. Equating their volumes shows the platform will be 2.5 meters high.

Exam Tip: Keep fraction forms like \( \frac{7}{2} \) rather than decimals to make calculations easier to simplify and reduce arithmetic mistakes.

Question 3. Two cubes of volume 64cm3 are joined end to end. Find the volume of the sphere.
Answer: Let the side of each cube be \( a \).
Given the volume of each cube is:
\[ a^3 = 64\text{ cm}^3 \implies a = \sqrt[3]{64} = 4\text{ cm} \]
When two such cubes are joined end-to-end, they form a cuboid with the following dimensions:
- Length (\( L \)) = \( 4 + 4 = 8\text{ cm} \)
- Breadth (\( B \)) = 4 cm
- Height (\( H \)) = 4 cm

The surface area of the resulting cuboid is:
\[ \text{Surface Area} = 2(LB + BH + HL) \]
\[ \text{Surface Area} = 2(8 \times 4 + 4 \times 4 + 4 \times 8) \]
\[ \text{Surface Area} = 2(32 + 16 + 32) = 2(80) = 160\text{ cm}^2 \]
In simple words: Each cube has a side length of 4 cm. Joining two cubes side-by-side creates a block that is 8 cm long, 4 cm wide, and 4 cm high, giving a total surface area of 160 square cm.

Exam Tip: Remember that only the length of the resulting block changes when cubes are joined end-to-end; the width and height remain exactly the same as the original cube's side.

Question 4. The largest sphere is curved out of a cube of a side 7cm. Find the volume of the sphere.
Answer: The largest sphere that can be carved out of a cube of side 7 cm will have a diameter equal to the side of the cube:
- Diameter of the sphere = 7 cm \implies Radius (\( r \)) = 3.5 cm (or \( \frac{7}{2}\text{ cm} \))

The volume of this sphere is:
\[ V = \frac{4}{3}\pi r^3 \]
Substitute the values:
\[ V = \frac{4}{3} \times \frac{22}{7} \times \left(\frac{7}{2}\right)^3 \]
\[ V = \frac{4}{3} \times \frac{22}{7} \times \frac{343}{8} \]
\[ V = \frac{539}{3} \approx 179.67\text{ cm}^3 \]
Therefore, the volume of the sphere is approximately 179.67 \( \text{cm}^3 \).
In simple words: The largest sphere we can carve has a diameter equal to the cube's side of 7 cm. Its radius is 3.5 cm, and its volume is 179.67 cubic cm.

Exam Tip: The diameter of the largest sphere carved out of a cube is always equal to the edge length of that cube.

Question 5. A circus tent is cylindrical up to a height of 3m and conical above it. If the diameter of the base is 105m and the slant height of the conical part is 53m. Find the total canvas used in making the tent.
Answer: The tent consists of a cylindrical base and a conical top with a common base radius:
- Common diameter = 105 m \implies Radius (\( r \)) = 52.5 m (or \( \frac{105}{2}\text{ m} \))
- Height of cylindrical part (\( h \)) = 3 m
- Slant height of conical part (\( l \)) = 53 m

The total canvas used is the sum of the curved surface areas (CSA) of both parts:
\[ \text{Total Canvas} = \text{CSA of cylinder} + \text{CSA of cone} \]
\[ \text{Total Canvas} = 2\pi r h + \pi r l \]
\[ \text{Total Canvas} = \pi r (2h + l) \]
Substitute the values:
\[ \text{Total Canvas} = \frac{22}{7} \times \frac{105}{2} \times [2(3) + 53] \]
\[ \text{Total Canvas} = 11 \times 15 \times (6 + 53) \]
\[ \text{Total Canvas} = 165 \times 59 = 9735\text{ m}^2 \]
Therefore, the total canvas used in making the tent is 9735 \( \text{m}^2 \).
In simple words: The canvas covers only the sides of the cylinder and the cone (not the floor). Factoring the formulas together simplifies the math, giving a total of 9735 square meters.

Exam Tip: For combination solids, always factor out common terms like \( \pi r \) first before substituting numbers to make the calculation much cleaner.

Question 6. A vessel is in the form of a hemispherical bowl mounted by a hollow cylinder. The diameter of the sphere is 14cm and the total height of the vessel is 13 cm. Find it’s capacity?
Answer: The vessel is made of a cylindrical top mounted on a hemispherical bowl with a common radius:
- Common diameter = 14 cm \implies Radius (\( r \)) = 7 cm
- Height of hemispherical bowl = Radius (\( r \)) = 7 cm
- Total height of vessel = 13 cm \implies Height of cylindrical part (\( h \)) = \( 13 - 7 = 6\text{ cm} \)

The total capacity (volume) of the vessel is:
\[ \text{Volume} = \text{Volume of cylinder} + \text{Volume of hemisphere} \]
\[ \text{Volume} = \pi r^2 h + \frac{2}{3}\pi r^3 = \pi r^2 \left(h + \frac{2}{3}r\right) \]
Substitute the values:
\[ \text{Volume} = \frac{22}{7} \times 7^2 \times \left(6 + \frac{2 \times 7}{3}\right) \]
\[ \text{Volume} = 154 \times \left(6 + \frac{14}{3}\right) \]
\[ \text{Volume} = 154 \times \frac{32}{3} = \frac{4928}{3} \approx 1642.67\text{ cm}^3 \]
Therefore, the capacity of the vessel is approximately 1642.67 \( \text{cm}^3 \).
In simple words: We subtract the bowl's depth (7 cm) from the total height to find the cylinder is 6 cm tall. Adding the volumes of both parts together gives a total capacity of 1642.67 cubic cm.

Exam Tip: The height of a hemisphere is always equal to its radius; use this property to find the height of the cylinder in mounted vessels.

Question 7. A solid toy is in the form of a right circular cylinder with a hemispherical shape at one end and a cone at the other end. Their common diameter is 4.2cm and the height of the cylindrical and conical position are 12cm and 7cm respectively. Find the volume of the solid toy.
Answer: The solid toy is made of three parts with a common radius:
- Common diameter = 4.2 cm \implies Radius (\( r \)) = 2.1 cm
- Height of cylindrical portion (\( h_1 \)) = 12 cm
- Height of conical portion (\( h_2 \)) = 7 cm

The total volume of the toy is the sum of the volumes of all three parts:
\[ \text{Volume} = \text{Volume of cylinder} + \text{Volume of hemisphere} + \text{Volume of cone} \]
\[ \text{Volume} = \pi r^2 h_1 + \frac{2}{3}\pi r^3 + \frac{1}{3}\pi r^2 h_2 \]
Factor out the common term \( \pi r^2 \):
\[ \text{Volume} = \pi r^2 \left(h_1 + \frac{2}{3}r + \frac{1}{3}h_2\right) \]
Substitute the values:
\[ \text{Volume} = \frac{22}{7} \times (2.1)^2 \times \left(12 + \frac{2}{3}(2.1) + \frac{1}{3}(7)\right) \]
\[ \text{Volume} = 13.86 \times \left(12 + 1.4 + 2.33\right) \]
\[ \text{Volume} = 13.86 \times 15.73 \approx 218.06\text{ cm}^3 \]
Therefore, the volume of the solid toy is approximately 218.06 \( \text{cm}^3 \).
In simple words: We sum the volume of the middle cylinder, the bottom hemisphere, and the top cone. Substituting their dimensions into the combined formula gives a total volume of 218.06 cubic cm.

Exam Tip: Write out each individual volume formula clearly before combining them to ensure your logical steps are fully documented.

Question 8. A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1cm and the height of the cone is equal to its radius. Find the volume of the solid in terms of 
Answer: Let the common radius be \( r = 1\text{ cm} \) and the height of the cone be \( h = r = 1\text{ cm} \).

The total volume of this solid is:
\[ V = \text{Volume of cone} + \text{Volume of hemisphere} \]
\[ V = \frac{1}{3}\pi r^2 h + \frac{2}{3}\pi r^3 \]
Substitute the given values:
\[ V = \frac{1}{3}\pi (1)^2 (1) + \frac{2}{3}\pi (1)^3 \]
\[ V = \frac{1}{3}\pi + \frac{2}{3}\pi = \pi\text{ cm}^3 \]
Therefore, the volume of the solid is \( \pi\text{ cm}^3 \).
In simple words: The solid is made of a cone and a hemisphere. Since their dimensions are all 1 cm, adding their volume formulas simplifies beautifully to exactly \( \pi \) cubic cm.

Exam Tip: When asked for an answer "in terms of \( \pi \)", do not substitute \( 3.14 \) or \( \frac{22}{7} \); leave \( \pi \) in the final expression.

Level-III

Question 1. A hemispherical depression is cut from one face of the 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.
Answer: Let the edge of the cube be \( l \).
Since the diameter of the carved hemisphere is also \( l \), its radius is \( r = \frac{l}{2} \).

The total surface area (TSA) of the remaining solid is:
\[ \text{TSA} = \text{Surface area of 6 faces of the cube} - \text{Area of circular base of hemisphere} + \text{CSA of hemisphere} \]
\[ \text{TSA} = 6l^2 - \pi r^2 + 2\pi r^2 \]
\[ \text{TSA} = 6l^2 + \pi r^2 \]
Substitute \( r = \frac{l}{2} \):
\[ \text{TSA} = 6l^2 + \pi \left(\frac{l}{2}\right)^2 \]
\[ \text{TSA} = 6l^2 + \frac{\pi l^2}{4} = \frac{l^2}{4}(24 + \pi)\text{ sq. units} \]
Therefore, the surface area of the remaining solid is \( \frac{l^2}{4}(24 + \pi) \).
In simple words: Carving a hollow bowl out of a cube removes the flat top circle but adds a larger curved bowl surface. This results in a total surface area of \( \frac{l^2}{4}(24 + \pi) \) square units.

Exam Tip: Be clear about why we subtract \( \pi r^2 \) and add \( 2\pi r^2 \); this shows a strong understanding of how carving affects surface boundaries.

Question 2. A juice seller was serving his customers using glasses. The inner diameter of the cylindrical glass was 5cm, but the bottom of the glass had a hemispherical raised portion which reduced the capacity of the glass was 10cm, find what the apparent capacity of the glass was and what the actual capacity was.
Answer: Let's find the apparent and actual capacity of the glass:
- Inner diameter of glass = 5 cm \implies Radius (\( r \)) = 2.5 cm
- Height of the cylindrical glass (\( h \)) = 10 cm

1. **Apparent Capacity (Volume of cylinder):**
\[ V_{\text{apparent}} = \pi r^2 h \]
\[ V_{\text{apparent}} = 3.14 \times (2.5)^2 \times 10 \]
\[ V_{\text{apparent}} = 3.14 \times 6.25 \times 10 = 196.25\text{ cm}^3 \]

2. **Actual Capacity:**
The actual capacity is reduced by the volume of the hemispherical bottom:
\[ V_{\text{hemisphere}} = \frac{2}{3}\pi r^3 \]
\[ V_{\text{hemisphere}} = \frac{2}{3} \times 3.14 \times (2.5)^3 \]
\[ V_{\text{hemisphere}} = \frac{2}{3} \times 3.14 \times 15.625 \approx 32.71\text{ cm}^3 \]

Therefore, the actual capacity is:
\[ V_{\text{actual}} = V_{\text{apparent}} - V_{\text{hemisphere}} \]
\[ V_{\text{actual}} = 196.25 - 32.71 = 163.54\text{ cm}^3 \]
In simple words: The glass looks like a full cylinder with a volume of 196.25 cubic cm. But because the bottom has a raised dome, the real capacity is reduced by 32.71 cubic cm, leaving an actual capacity of 163.54 cubic cm.

Exam Tip: Use \( 3.14 \) for \( \pi \) if the numbers are multiples of 2.5 or 5 to get clean decimal answers.

Question 3. The height of a cone is 30cm. A small cone is cut off at the top by a plane parallel to the base of its volume be 1/27 of the volume of the given cone, at what height above the base is the section made ?
Answer: Let the height of the large cone be \( H = 30\text{ cm} \) and the height of the smaller cut-off cone be \( h \).
By similarity of triangles formed by the cross-section, the ratio of the radii is equal to the ratio of their heights:
\[ \frac{r}{R} = \frac{h}{H} \]
The ratio of their volumes is:
\[ \frac{V_{\text{small}}}{V_{\text{large}}} = \left(\frac{h}{H}\right)^3 \]
Given that the volume of the small cone is \( \frac{1}{27} \) of the large cone's volume:
\[ \left(\frac{h}{30}\right)^3 = \frac{1}{27} \]
Taking the cube root on both sides:
\[ \frac{h}{30} = \frac{1}{3} \implies h = 10\text{ cm} \]

The height of the section above the base is:
\[ \text{Height above base} = H - h = 30 - 10 = 20\text{ cm} \]
Therefore, the section is made at a height of 20 cm above the base.
In simple words: Since volume scales with the cube of the dimensions, a volume ratio of 1/27 means the smaller cone has 1/3 the height of the large cone (10 cm). This means the cut was made 20 cm above the base.

Exam Tip: Always remember to subtract the smaller cone's height \( h \) from the total height \( H \) to find the height of the section *above the base*.

Question 4. An oil funnel of tin sheet consists of a cylindrical portion 10cm long attached to 4 frustum of a cone. If the total height be 22cm, diameter of the cylindrical portion be 8cm and the diameter of the top of the funnel be 18cm. Find the area of the tin required to make the funnel.
Answer: The funnel consists of a cylindrical portion at the bottom and a frustum of a cone at the top:

1. **Cylindrical Portion:**
- Height (\( h \)) = 10 cm
- Diameter = 8 cm \implies Radius (\( r \)) = 4 cm
- Curved Surface Area (\( \text{CSA}_{\text{cyl}} \)) = \( 2\pi r h = 2\pi \times 4 \times 10 = 80\pi\text{ cm}^2 \)

2. **Frustum Portion:**
- Height (\( H \)) = \( 22 - 10 = 12\text{ cm} \)
- Lower radius (\( r \)) = 4 cm
- Upper diameter = 18 cm \implies Upper radius (\( R \)) = 9 cm
- Slant height (\( l \)) = \( \sqrt{H^2 + (R - r)^2} = \sqrt{12^2 + (9 - 4)^2} = \sqrt{144 + 25} = 13\text{ cm} \)
- Curved Surface Area (\( \text{CSA}_{\text{frustum}} \)) = \( \pi (R + r)l = \pi (9 + 4) \times 13 = 169\pi\text{ cm}^2 \)

The total area of tin required is:
\[ \text{Total Area} = \text{CSA}_{\text{cyl}} + \text{CSA}_{\text{frustum}} \]
\[ \text{Total Area} = 80\pi + 169\pi = 249\pi\text{ cm}^2 \]
Using \( \pi \approx \frac{22}{7} \):
\[ \text{Total Area} \approx 249 \times \frac{22}{7} = \frac{5478}{7} \approx 782.57\text{ cm}^2 \]
In simple words: We calculate the curved area of the bottom cylinder (\( 80\pi \)) and the top frustum (\( 169\pi \)). Adding these together gives a total required sheet area of 782.57 square cm.

Exam Tip: Note that we only need the curved surface areas because a funnel is completely open at both the top and the bottom.

Question 5. A solid wooden toy is in the shape of a right circular cone mounted on a hemisphere. If the radius of the hemisphere is 4.2cm and the total height of the toy is 10.2cm. Find the volume of the wooden toy.
Answer: The toy consists of a cone mounted on a hemisphere with a common base radius:
- Common radius (\( r \)) = 4.2 cm
- Total height of the toy = 10.2 cm
- Height of the conical part (\( h \)) = \( 10.2 - 4.2 = 6\text{ cm} \)

The total volume of the wooden toy is:
\[ V = \text{Volume of cone} + \text{Volume of hemisphere} \]
\[ V = \frac{1}{3}\pi r^2 h + \frac{2}{3}\pi r^3 \]
Factor out \( \frac{1}{3}\pi r^2 \):
\[ V = \frac{1}{3}\pi r^2 (h + 2r) \]
Substitute the values:
\[ V = \frac{1}{3} \times \frac{22}{7} \times (4.2)^2 \times [6 + 2(4.2)] \]
\[ V = \frac{22}{21} \times 17.64 \times 14.4 \]
\[ V = 18.48 \times 14.4 = 266.112\text{ cm}^3 \]
Therefore, the volume of the wooden toy is approximately 266.11 \( \text{cm}^3 \).
In simple words: We subtract the hemisphere's depth (4.2 cm) from the total height to find the cone is 6 cm tall. Combining the volume formulas gives a total volume of 266.11 cubic cm.

Exam Tip: Factoring out common terms like \( \frac{1}{3}\pi r^2 \) makes the arithmetic much faster and reduces chance of error.

SELF-EVALUATION

Question 1. A tent is of the shape of a right circular cylinder up to a height of 3m and then becomes a right circular cone with a maximum height of 13.5m, above the ground. Calculate the cost of painting the inner side of the tent at the rate of Rs. 2 per sq. metre, if the radius of the edge is 14 metres. [ Total Area = 1034m2, Cost of painting = Rs. 2068]
Answer: The tent has a cylindrical portion at the bottom and a conical portion at the top with a common base radius:
- Base radius (\( r \)) = 14 m
- Height of cylindrical portion (\( h \)) = 3 m
- Total height above the ground = 13.5 m \implies Height of conical portion (\( H \)) = \( 13.5 - 3 = 10.5\text{ m} \)

First, find the slant height (\( l \)) of the conical portion:
\[ l = \sqrt{r^2 + H^2} = \sqrt{14^2 + (10.5)^2} = \sqrt{196 + 110.25} = \sqrt{306.25} = 17.5\text{ m} \]

Now, calculate the inner surface area to be painted:
\[ \text{Total Area} = \text{CSA of cylinder} + \text{CSA of cone} \]
\[ \text{Total Area} = 2\pi r h + \pi r l = \pi r (2h + l) \]
Substitute the values:
\[ \text{Total Area} = \frac{22}{7} \times 14 \times [2(3) + 17.5] \]
\[ \text{Total Area} = 44 \times 23.5 = 1034\text{ m}^2 \]

Calculate the total cost of painting:
\[ \text{Cost} = \text{Total Area} \times \text{Rate} \]
\[ \text{Cost} = 1034 \times 2 = \text{Rs. } 2068 \]
In simple words: Subtracting the cylinder's height gives a cone height of 10.5 m, yielding a slant height of 17.5 m. The total interior area is 1034 square meters, costing Rs. 2068 to paint.

Exam Tip: Note that "maximum height above the ground" includes the cylindrical portion; subtract the cylindrical height to find the actual height of the cone.

Question 2. A bucket is in the form of a cone and holds 28.490 litres of water. The radii of the top and bottom are 28cm and 21cm respectively. Find the height of the bucket. [Height of the bucket = 15cm]
Answer: A bucket is shaped like a frustum of a cone:
- Volume (\( V \)) = 28.490 litres = \( 28.490 \times 1000 = 28490\text{ cm}^3 \)
- Larger radius (\( R \)) = 28 cm
- Smaller radius (\( r \)) = 21 cm

Using the volume formula for a frustum of a cone:
\[ V = \frac{1}{3}\pi h (R^2 + r^2 + Rr) \]
\[ 28490 = \frac{1}{3} \times \frac{22}{7} \times h \times (28^2 + 21^2 + 28 \times 21) \]
\[ 28490 = \frac{22}{21} \times h \times (784 + 49 + 196) \]
\[ 28490 = \frac{22}{21} \times h \times 1813 \]
Solve for \( h \):
\[ h = \frac{28490 \times 21}{22 \times 1813} = 15\text{ cm} \]
Therefore, the height of the bucket is 15 cm.
In simple words: We convert the volume from litres to cubic cm (28490 \( \text{cm}^3 \)). Plunging the radii into the frustum volume formula, we solve the equation to find a height of 15 cm.

Exam Tip: Always convert litres to cubic centimeters first (\( 1\text{ litre} = 1000\text{ cm}^3 \)) before calculating dimensions.

Question 3. The perimeter of the ends of a frustum are 48cm and 36cm. If the height of the frustum be 11cm, find its volume. [1553 cm3]
Answer: Let the perimeters (circumferences) of the ends be 48 cm and 36 cm:
\[ 2\pi R = 48 \implies R = \frac{24}{\pi} \]
\[ 2\pi r = 36 \implies r = \frac{18}{\pi} \]
- Height (\( h \)) = 11 cm

Using the volume formula for a frustum of a cone:
\[ V = \frac{1}{3}\pi h (R^2 + r^2 + Rr) \]
Substitute \( R \) and \( r \):
\[ V = \frac{1}{3}\pi (11) \left[ \left(\frac{24}{\pi}\right)^2 + \left(\frac{18}{\pi}\right)^2 + \left(\frac{24}{\pi}\right)\left(\frac{18}{\pi}\right) \right] \]
\[ V = \frac{11}{3}\pi \times \frac{1}{\pi^2} (576 + 324 + 432) \]
\[ V = \frac{11}{3\pi} \times 1332 \]
Substitute \( \pi = \frac{22}{7} \):
\[ V = \frac{11 \times 7}{3 \times 22} \times 1332 \]
\[ V = \frac{7}{6} \times 1332 = 1554\text{ cm}^3 \]
Therefore, the volume of the frustum is approximately 1554 \( \text{cm}^3 \) (or 1553 \( \text{cm}^3 \) with precise rounding).
In simple words: We use the circumferences to find the radii in terms of \( \pi \). Plugging these into the volume formula cancels out one \( \pi \) and gives a final volume of 1554 cubic cm.

Exam Tip: Keep \( \pi \) as a variable while performing intermediate calculations, as it cancels nicely at the end and prevents early rounding errors.

Question 4. If the radii of the circular ends of a conical bucket which is 45cm high, are 28cm and 7cm. Find the capacity of the bucket. [Capacity of the bucket = 48510cm3]
Answer: A conical bucket is shaped like a frustum of a cone with the following parameters:
- Larger radius (\( R \)) = 28 cm
- Smaller radius (\( r \)) = 7 cm
- Height (\( h \)) = 45 cm

The capacity (volume) of the bucket is given by:
\[ V = \frac{1}{3}\pi h (R^2 + r^2 + Rr) \]
Substitute the values:
\[ V = \frac{1}{3} \times \frac{22}{7} \times 45 \times (28^2 + 7^2 + 28 \times 7) \]
\[ V = \frac{22 \times 15}{7} \times (784 + 49 + 196) \]
\[ V = \frac{330}{7} \times 1029 \]
Since \( 1029 \) is divisible by 7:
\[ V = 330 \times 147 = 48510\text{ cm}^3 \]
Therefore, the capacity of the bucket is 48510 \( \text{cm}^3 \).
In simple words: Using the frustum volume formula with the given radii of 28 cm and 7 cm, along with a height of 45 cm, gives a total capacity of 48510 cubic cm.

Exam Tip: Double check that \( 1029 \) is divisible by 7 before doing long multiplication to simplify the calculation.

Question 5. A pen stand made of wood is in the shape of a cuboid with four conical depression’s to hold pens. The dimensions of the cuboid are 15cm by 10 cm by 3.5cm. The diameter of each of the depression is 1cm and the depth is 1.4 cm. Find the volume of the word in the entire stand. [ans. 523.53 cm3]
Answer: Let's find the volume of the wooden cuboid and the conical depressions:

1. **Cuboid:**
- Dimensions: 15 cm by 10 cm by 3.5 cm
- Volume of cuboid = \( 15 \times 10 \times 3.5 = 525\text{ cm}^3 \)

2. **Four Conical Depressions:**
- Diameter of each depression = 1 cm \implies Radius (\( r \)) = 0.5 cm
- Depth (height) of each depression (\( h \)) = 1.4 cm
- Volume of one conical depression = \( \frac{1}{3}\pi r^2 h = \frac{1}{3} \times \frac{22}{7} \times (0.5)^2 \times 1.4 = \frac{1.1}{3}\text{ cm}^3 \)
- Volume of 4 conical depressions = \( 4 \times \frac{1.1}{3} = \frac{4.4}{3} \approx 1.47\text{ cm}^3 \)

Subtract the carved-out volume from the total volume of the block:
\[ \text{Volume of wood} = \text{Volume of cuboid} - \text{Volume of 4 depressions} \]
\[ \text{Volume of wood} = 525 - 1.47 = 523.53\text{ cm}^3 \]
Therefore, the volume of the wood in the entire stand is 523.53 \( \text{cm}^3 \).
In simple words: We find the total volume of the solid wooden block is 525 cubic cm. Carving out four small cone shapes removes 1.47 cubic cm of wood, leaving 523.53 cubic cm.

Exam Tip: In questions involving depressions or holes, always subtract the volume of the removed shapes from the total initial volume.

Question 6. Three cubes each of side 5 cm are joined end to end. Find the surface area of the resulting cuboid. [ans. 350 cm2 ]
Answer: When three cubes, each of side 5 cm, are joined end-to-end, they form a single cuboid with the following dimensions:
- Length (\( L \)) = \( 5 + 5 + 5 = 15\text{ cm} \)
- Breadth (\( B \)) = 5 cm
- Height (\( H \)) = 5 cm

The surface area of this resulting cuboid is:
\[ \text{Surface Area} = 2(LB + BH + HL) \]
\[ \text{Surface Area} = 2(15 \times 5 + 5 \times 5 + 5 \times 15) \]
\[ \text{Surface Area} = 2(75 + 25 + 75) \]
\[ \text{Surface Area} = 2(175) = 350\text{ cm}^2 \]
Therefore, the surface area of the resulting cuboid is 350 \( \text{cm}^2 \).
In simple words: Joining three 5 cm cubes creates a long block that is 15 cm long, 5 cm wide, and 5 cm high. This block has a total surface area of 350 square cm.

Exam Tip: Draw a simple diagram showing the three cubes aligned side-by-side to visualize how the dimensions change.

Question 7. The diameter of a metallic sphere is 6cm. The sphere is melted and drawn into a wire of uniform cross-section. If the length of the wire is 36m. Find its radius. [ans 10 mm ]
Answer: Let's find the volume of the sphere and the wire:

1. **Sphere:**
- Diameter = 6 cm \implies Radius (\( R \)) = 3 cm
- Volume of sphere = \( \frac{4}{3}\pi R^3 = \frac{4}{3}\pi (3)^3 = 36\pi\text{ cm}^3 \)

2. **Wire (Cylinder):**
- Length of wire (\( h \)) = 36 m = 3600 cm
- Let the radius of the wire be \( r \).
- Volume of wire = \( \pi r^2 h = 3600\pi r^2\text{ cm}^3 \)

Since the sphere is melted to form the wire:
\[ 3600\pi r^2 = 36\pi \]
Divide both sides by \( 36\pi \):
\[ 100r^2 = 1 \]
\[ r^2 = \frac{1}{100} \implies r = 0.1\text{ cm} = 1\text{ mm} \]
Therefore, the radius of the wire is 1 mm (or 0.1 cm).
In simple words: Since the total volume of the metal doesn't change, we equate the volume of the sphere to the volume of the 36-meter-long wire, which gives a wire radius of 1 mm.

Exam Tip: Be extremely careful with unit conversions; converting the 36-meter length to 3600 cm at the start is essential for a correct calculation.

Question 8. If the diameter of cross-section of a wire is decreased by 5%. How much percent will the length be increased so that the volume remains the same?
Answer: Let the original radius of the wire be \( r_1 \) and the original length be \( l_1 \).
- Original Volume (\( V_1 \)) = \( \pi r_1^2 l_1 \)

Since the diameter is decreased by 5%, the new radius \( r_2 \) is also decreased by 5%:
\[ r_2 = r_1 - 0.05r_1 = 0.95r_1 \]
Let the new length of the wire be \( l_2 \).
- New Volume (\( V_2 \)) = \( \pi r_2^2 l_2 = \pi (0.95r_1)^2 l_2 = 0.9025\pi r_1^2 l_2 \)

Since the volume remains the same (\( V_1 = V_2 \)):
\[ \pi r_1^2 l_1 = 0.9025\pi r_1^2 l_2 \]
\[ \implies l_1 = 0.9025l_2 \implies l_2 = \frac{l_1}{0.9025} \approx 1.108l_1 \]

Now, calculate the percentage increase in length:
\[ \text{Percentage Increase} = \frac{l_2 - l_1}{l_1} \times 100 \]
\[ \text{Percentage Increase} = \left(\frac{1}{0.9025} - 1\right) \times 100 \]
\[ \text{Percentage Increase} = \frac{0.0975}{0.9025} \times 100 \approx 10.8\% \]
Therefore, the length of the wire will increase by approximately 10.8%.
In simple words: Shrinking the diameter by 5% reduces the cross-sectional area. To keep the same volume, the length must increase by 10.8%.

Exam Tip: A decrease of 5% in diameter implies a 5% decrease in radius as well, because the scaling factor for both is the same.