Chapter-12, SURFACE AREAS AND VOLUMES

1. A cylindrical pencil sharpened at one edge is the combination of (A) a cone and a cylinder (B) frustum of a cone and a cylinder (C) a hemisphere and a cylinder (D) two cylinders.

Explanation:

The right answer is C, which is a cone and a cylinder.

A cone and a cylinder make up a cylindrical sharpened pencil.

2. A Surabhi is the combination of (A) a sphere and a cylinder (B) a hemisphere and a cylinder (C) two hemispheres (D) a cylinder and a cone.

Explanation:

The correct response is option A, which describes a surabhi as the union of a sphere and a cylinder. It serves as a water storage container or a decorative element.

solution expand

3. A plumbline (Sahul) is the combination of (see Fig. 12.2) (A) a cone and a cylinder (B) a hemisphere and a cone (C) frustum of a cone and a cylinder (D) sphere and cylinder 

Explanation:

The right answer is B) From Cone: A solid thing with a circular base and a single vertex.

Hemisphere: A hemisphere is an exact half of a sphere in geometry.

4. The shape of a glass (tumbler) (see Fig. 12.3) is usually in the form of (A) a cone (B) the frustum of a cone (C) a cylinder (D) a sphere 

Explanation:

The right answer is B) According to the definition, a cone is removed by cutting across the plane parallel to its base.

A frustum of a cone is the new component that remains on the opposite side of the plane.

5. The shape of a gilli, in the gilli-danda game (see Fig. 12.4), is a combination of (A) two cylinders (B) a cone and a cylinder (C) two cones and a cylinder (D) two cylinders and a cone 

Explanation:

The fact that

That is: Two cones and a cylinder are equal to a cone plus a cylinder.

As a result, in the game of Gilli-Danda, Gilli's shape is made up of two cones and a cylinder.

6. A shuttle cock used for playing badminton has the shape of a combination of 

(A) a cylinder and a sphere (B) a cylinder and a hemisphere (C) a sphere and a cone (D) frustum of a cone and a hemisphere

Explanation:

The ideal selection is B.

A frustum cone with a hemisphere

The shape of a shuttlecock used in badminton is a hybrid of a hemisphere and a frustum cone.

7. A cone is cut through a plane parallel to its base and then the cone that is formed on one side of that plane is removed. The new part that is left over on the other side of the plane is called (A) a frustum of a cone (B) a cone (C) a cylinder (D) a sphere

Explanation:

The correct response is option A, which entails cutting a cone through a plane parallel to its base before removing the cone that results on one side of the plane. A frustum of a cone is the new component that remains on the opposite side of the plane.

8. A hollow cube of internal edge 22cm is filled with spherical marbles of diameter 0.5 cm and it is assumed that  ⅛  space of the cube remains unfilled. Then the number of marbles that the cube can accommodate is (A) 142296 (B) 142396 (C) 142496 (D) 142596

Explanation:

It follows that

A hollow cube's inside edge measures 22 cm.

Cube volume equals side³ + 22³ + 10648 cm³.

Special marbles have a 0.5 centimeter diameter.

Special marbles' radius: 0.5/2 = 1/4 cm

The fact that

One marble's volume is 4/3 r³.

Changing the values

One marble's volume is equal to 4/3 x 22/7 x (1/4)³ = 11/168 cm³.

Here

Cube volume - 1/8 volume of cube = 10648 - 1/8 10648 = 10648 - 7/8 volume of cube = 9317 cm³

The required number of marbles is calculated as follows: 9317/(11/168)

= 142296

 total space occupied by all the marbles in a cube; the volume of each marble.

The cube can hold 142296 marbles, which is the maximum number.

9. A metallic spherical shell of internal and external diameters 4 cm and 8 cm, respectively is melted and recast into the form of a cone of base diameter 8cm. The height of the cone is (A) 12cm (B) 14cm (C) 15cm (D) 18cm

Explanation:

It follows that

d1 = 4 cm for the internal diameter

The spherical shell's internal radius is r1 = 4/2 = 2 cm.

d2 = 8 cm external diameter

r² = 8/2 = 4 cm is the external radius of a spherical shell.

cone's base diameter is 8 cm.

The cone's base radius is 4 cm.

Think of h as the height of the cone.

We are aware that the volume of a sphere equals 4/3 (r2³ - r1³), where r1 and r2 are the interior and external radii, respectively.

Spherical shell volume = 4/3 (4³ - 2³) = 4/3 π(56) = (224/3)π

We are aware that the volume of a cone is (1/3)πr²h.

where r denotes the base radius and h the cone's height

Cone volume = (1/3)π 4²h = 16πh/3.

Here

Cone volume after melting equals the volume of the spherical shell times 224π/3 or 16πh/3.

By even more simplifying

16h = 224

16 divided by both sides yields 14 cm.

The cone is 14 cm tall as a result.

10. A solid piece of iron in the form of a cuboid of dimensions 49cm × 33cm × 24cm, is molded to form a solid sphere. The radius of the sphere is (A) 21cm (B) 23cm (C) 25cm (D) 19cm

Explanation:

We are aware that cuboid volume equals lbh.

where l = length, b = breadth, and h = height.

Since l = 49 cm

 b = 33 cm

 and h = 24 cm,

the values in the formula is replaced

Cuboid volume equals 49 cmx33 cmx and 24 cm= 38808 cm³.

Think of r as the cube's radius.

We are aware that spherical volume equals 4/3π r³.

Where r is the sphere's radius in this example, the volume of a cuboid equals the volume of a sphere molded in 38808 (4/3 r³)

Using more math, r³ = 29106 r³ = 29106/(22/7)

The result is r³ = 9261.

cube root on both sides yields r = 9261 = 21.

The sphere's radius is 21 cm as a result.

11. A mason constructs a wall of dimensions 270cm× 300cm × 350cm with the bricks each of size 22.5cm × 11.25cm × 8.75cm and it is assumed that 1 /8 8 space is covered by the mortar. Then the number of bricks used to construct the wall is (A) 11100 (B) 11200 (C) 11000 (D) 11300

Explanation:

The wall measures 270cm, 300cm, and 350cm.

We are aware that the wall's volume is 270, 300, 350, or 28350000 cm³.

It is assumed that the mortar covers 1/8 of the space.

The volume of the remaining wall minus the volume of the mortar

28350000 - 28350000 1/8 

= 28350000 - 3543750 

= 24806250 cm³ is the result of substituting the values.

Here

One brick has a volume of 22.5 x 11.25 x 8.75 or 2214.844 cm³.

Therefore, the necessary number of bricks is equal to 11200 (24806250/2214.844).

Consequently, 11200 bricks were utilized to build the wall.

12. Twelve solid spheres of the same size are made by melting a solid metallic cylinder with a base diameter of 2 cm and height of 16 cm. The diameter of each sphere is (A) 4 cm (B) 3 cm (C) 2 cm (D) 6 cm 

Explanation:

Because the cylinder's diameter is 2 cm

Cylinder radius = 2/2 = 1 cm

The cylinder's height is 16 cm.

The fact that

Cylinder volume = r2h

Changing the values

Cylinder volume = πx1² + 16 = 16 cm

Volume = 4/3 πr³ (using cm as the radius of a solid sphere).

From the inquiry

The volume of a cylinder 12 times 4/3 times πr³ equals the volume of twelve solid spheres.

Further computation yields r³= 1.

r = 1 cm after taking cube roots on both sides.

The diameter is thus 2 1 = 2 cm.

Thus, each sphere has a 2 cm diameter.

13. The radii of the top and bottom of a bucket of slant height 45 cm is 28 cm and 7 cm, respectively. The curved surface area of the bucket is (A) 4950 cm2 (B) 4951 cm2 (C) 4952 cm2 (D) 4953 cm2

Explanation:

Given that the bucket's top radius is 28 cm

7 cm is the radius of the bucket's bottom.

45 cm is the bucket's slant height.

Consequently, the bucket has a cone-shaped frustum.

We are aware that the bucket's curved surface area is equal to πx45x (28 + 7) (l (R + r)).

Further maths reveals that = πx45 x35 = 22/7 x45 x35 

= 4950 cm²

Consequently, the bucket's curved surface area is 4950 cm².

14. A medicine capsule is in the shape of a cylinder of diameter 0.5 cm with two hemispheres stuck to each of its ends. The length of the entire capsule is 2 cm. The capacity of the capsule is (A) 0.36 cm3 (B) 0.35 cm3 (C) 0.34 cm3 (D) 0.33 cm3

Explanation:

Given that cylinder diameter equals hemisphere diameter, which equals 0.5 cm

The cylinder radius equals the hemisphere radius, which is equal to 0.5/2 cm.

The capsule length overall is 2 cm.

Here, h is the length of the cylindrical portion of the capsule and is equal to the sum of the hemispheres' radii.

The capacity of the capsule equals the cylindrical component volume plus twice the hemisphere volume.

The equation is

Cylinder volume = πr²h

The volume of the hemisphere equals 2/3 πr².

Values replaced result in = 22/7 [(0.25)2 1.5 + 2 4/3 (0.25)²]

By adding the two numbers together, 22/7 [0.09375 + 0.0208] equals 22/7 [0.11455] = 0.36 cm².

Consequently, the capsule has a 0.36 cm² volume.

15. If two solid hemispheres of the same base radius r are joined together along their bases, then the curved surface area of this new solid is 

(A) 4πr2 (B) 6πr2 (C) 3πr2 (D) 8πr2

Explanation:

The answer is (A).

We unite solid hemispheres at their bases of radius r to create a solid sphere, which has a curved surface area of 2 πr². As a result, the new solid's curved surface area is equal to 2πr² + 2 πr²= 4πr².

16. A right circular cylinder of radius r cm and height h cm (h>2r) just encloses a sphere of diameter (A) r cm (B) 2r cm (C) h cm (D) 2h cm

Explanation:

It is proper to select (B).

The diameter of the sphere is equal to the diameter of the cylinder, which is 2 r cm, because the sphere is enclosed within the cylinder.

17. During the conversion of a solid from one shape to another, the volume of the new shape will (A) increase (B) decrease (C) remain unaltered (D) be doubled

Explanation:

It is proper to select (C).

The volume of the new shape won't change when a solid is transformed from one shape to another. It doesn't change.

18. The diameters of the two circular ends of the bucket are 44 cm and 24 cm. The height of the bucket is 35 cm. The capacity of the bucket is (A) 32.7 liters (B) 33.7 liters (C) 34.7 liters (D) 31.7 liters 

Explanation:

Given that h = 35 cm and D = 44 cm.

R = 44/2 = 22 cm

d = 24 cm

r/2 = 24/2 = 12 cm

We are aware that the volume of a cone's frustum equals πh/3(R² + Rr + r²).

Since 1 liter equals 1000 cubic centimeters

The volume of the cone's frustum is equal to [(22² + 22 + 12 + 12²)/3] or 32706.67 cubic centimeters, or 32.7 liters.

As a result, the bucket has a 32.7-liter capacity.

19. In a right circular cone, the cross-section made by a plane parallel to the base is an (A) circle (B) frustum of a cone (C) sphere (D) hemisphere

Explanation:

It is proper to select (B).

We are aware that the region between a plane and a cone's base that is sliced by the plane is known as the frustum of the cone.

20. Volumes of two spheres are in the ratio 64:27. The ratio of their surface areas is (A) 3 : 4 (B) 4 : 3 (C) 9 : 16 (D) 16 : 9

Explanation:

Think of r1 and r2 as the spheres' respective radii.

Volumes of two spheres are divided 64:27.

We are aware that spherical volume equals 4/3 r3.

V1/V2 here = 64/27

4/3 π r1³/ 4/3 π r2³ = 64/27

Through more simplification (r1/r2)³ = (4/3)³

Rooting like a cube on both sides

r1/r2 = 4/3

Think of S1 and S2 as the two spheres' surface areas.

4r12/4r22 = (r1/r2) where S1/S2 =²

Here, S1 = S2 = (4/3)² = 16/9

They have a 16: 9 surface area ratio as a result.

21. 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

Explanation:

Given, two identical solid hemispheres with bases that are separated by a distance of r cm are joined at their bases.

We must determine if the combination's total surface area is 6 πr².

We create a sphere by connecting two identical solid hemispheres with similar radii at their bases.

We are aware that the sphere's total surface area equals its curved surface area.

Since a sphere is a three-dimensional shape without a face or edge

The sphere's total surface area is equal to 4πr².

As a result, the assertion is untrue.

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

Explanation:

Both cylinders' radius and height are the same.

Now, how big is this shape's overall surface 

=(2πrh+πr+πr)+(2πrh+πr+πr)=(4πrh+4πr²)

23. A solid cone of radius r and height h is placed over a solid cylinder having the same base radius and height as that of a cone. The total surface area of the combined solid is 2 2   rrh r h   +++.

Explanation:

It follows that

r is the base radius of a solid cone.

h is the height of a solid cone.

A solid cylinder's base radius is equal to r.

The solid cylinder's height is equal to h.

The curved surface area of the shape is therefore equal to the sum of the curved surfaces of the cone and the cylinder, or rl + 2rh².

It can be expressed as = rh² + rh² + rh².

The assertion is accurate as a result.

24. A solid ball is exactly fitted inside the cubical box of side a. The volume of the ball is  4/3, a3

Explanation:

Assuming, a solid ball fits precisely inside the cubical box on side a.

We must determine if the ball's volume is (4/3)a3.

The diameter of the solid ball will be equal to the distance between the opposing cube sides since it fits exactly inside the cubical box.

hence, the ball diameter equals a

Radius = 2/a

Having said that The sphere's volume equals (4/3)πr³

 = (4/3)π(a/2)³ 

= (4/3)π(a³/8)

 = (1/3)π(a³/2) 

= πa³/6

The ball's volume is therefore  πa³/6.

25. The volume of the frustum of a cone is 1/3h r r rr +, where h is the vertical height of the frustum and r1, r2 are the radii of the ends.

Explanation:

volume of the frustum cone = πh/3 (R²+Rr+r²). False

26. The capacity of a cylindrical vessel with a hemispherical portion raised upward at the bottom as shown in the Fig.πh2/3 12.7 is [ ] 

Explanation:

A cylindrical vessel with a hemispherical part is provided.

The capacity of a cylindrical vessel with a raised hemispherical part at the bottom must be equal to (πr²/3)[3h - 2r].

The provided vessel's capacity is equal to the sum of its cylinder and hemisphere capacities.

Cylinder volume =πr²h

The hemisphere's volume equals (2/3)πr³.

The vessel's capacity is equal to πr²h - (2/3)πr³ equals πr²/3 (3h minus 2r) cubic units.

As a result, the vessel has a capacity of πr²/3 (3h - 2r) cubic units.

27. 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. Is the following statement true or false and justify your answer

Explanation:

 

Cone frustum curved surface area =l(r1+r2) where l is the slant height =h²+(r1+r2)², where h is the height and r1 and r2 are the radii of a cone's bottom and higher frusta.

28. 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 the curved surface area of the frustum of a cone + the area of the circular base + the curved surface area of the cylinder

Explanation:

The surface area of the metallic sheet in use is the sum of the curved surfaces of the frustum of a cone, the circular base, and the curved surfaces of the cylinder.

 So it is true.

29. Three metallic solid cubes whose edges are 3 cm, 4 cm, and 5 cm are melted and formed into a single cube. Find the edge of the cube so formed. 


Explanation:

Three solid cubes' edges are indicated as being 3 cm, 4 cm, and 5 cm, respectively.

Cube volume equals (side)³

The first cube's volume is (3)³=27 cm³.

The volume of the second cube is 64 cm³ (4)³.

And the third cube's volume is (5)³125cm³.

The three cubes' combined volume is (27+64+125)cm³.

Let R cm be the value for the cube's edge.

The volume of the resulting cube is then equal to R³=216 cm³,

R=³216 cm³,

and R=6 cm.

30. How many shots each having a diameter 3 cm can be made from a cuboidal lead solid of dimensions 9cm × 11cm × 12cm?

Explanation:

Considering that a cuboidal lead is 9 cm x 11 cm x12 cm

Cuboidal lead is used to create shots with a 3 cm diameter.

The number of shots must be determined.

The volume of a cuboid equals length, width, and depth 9 x11 x12 

= 99x  12

= 1188 cm³.

Sphere volume = (4/3)πr³.

Given that, a shot's diameter is 3 cm

Radius: 3/2 times 1.5 cm

Shot volume equals (4/3)(22/7)(1.5)3 = 14.14 cm³

Shot volume divided by cuboidal lead volume equals 1188/14.14, or 84 shots.

The result is that 84 shots were generated.

31. A bucket is in the form of a frustum of a cone and holds 28.490 liters of water. The radii of the top and bottom are 28 cm and 21 cm, respectively. Find the height of the bucket.

Explanation:

A bucket in the shape of a cone's frustum is provided.

The bucket has a water capacity of 28.490 liters.

The top and bottom radii are 28 cm and 21 cm, respectively.

We must determine the bucket's height.

The volume of a cone's frustum is equal to h/3 (R²+ r² + Rr).

Assuming R = 28 cm

r = 21 cm

Water volume equals 28.490 liters or 28.490 x 1000 cm³.

The bucket's volume is 28490 cm³.

Now, 28490 = πh/3 ((28)² + (21)² + 28(21))

28490 = (22/7)(h/3)(784 + 441 + 588)

28490 = 22h/21 (1813)

When h is figured out, 28490 = (39886/21)h 28490 = 1899.33h, and h = 28490/1899.33 h = 15 cm

Consequently, the bucket has a 15 cm height.= 2π(16) 

= 32π cm³

The volume of a smaller cone divided by the volume of a cone's frustum equals 32π/224π, or 1/7.


Consequently, the volume ratio between the two components is 1:7.

32. A cone of radius 8 cm and height 12 cm is divided into two parts by a plane through the mid-point of its axis parallel to its base. Find the ratio of the volumes of the two parts.

Explanation:

The cone has a given 8 cm diameter and 12 cm height.

A plane passing through the center of the cone's axis and running parallel to its base divides the object into two pieces.

The ratio of the two components' volumes must be determined.

Considering that the cone's height is h = 12 cm

The cone's radius is R = 8 cm.

We obtain a cone's frustum and a smaller cone by cutting the cone through the midway parallel to the base.

The respective sides of the comparable triangles OMN and OPD are proportionate.

OM/OP = MN/PD

MN = 8 cm for the cone's radius in the illustration.

PD equals rcm (smaller cone radius).

OM equals the cone's 12 cm height.

Since the smaller cone is divided in half at the middle, OP = 12/2 

= 6 cm.

So, 8/r = 12/6

8/r = 2

 2r = 8

 r = 4 cm

Volume of a cone's frustum = h/3 [R² + r² + Rr] = ( 6 / 3) [8² + 4² + 8 (4)] 

= (2) [64 + 16 + 32]

= 2( 112) 

= 224 cm³

(1/3) is the volume of the smaller cone.πr²h = π(1/3)(4)²(6)

33. Two identical cubes each of volume 64 cm3 is the surface area of the resulting cuboid?

Explanation:

Using the equation volume of a cube = a3, where the length of the edge is denoted by the letter "a," we can determine the length of each cube's edge.

The cubes will appear as follows when linked end to end:

Using the equation 2(lb + bh + lh), where l, b, and h are the cuboid's length, width, and height, respectively.

Each cube's edge length should be 'a'.

The volume of the cube is thus equal to a3.

A3 = 64 cm3, which is the cube's volume. A =∛ (64 cm3)

a = 4 cm

Therefore,

L = a = 4 cm is the length of the resulting cuboid.

The resulting cuboid's width, b = a = 4 cm

The resulting cuboid's height is given by h = 2a = 2 4 cm = 8 cm.

The resulting cuboid's surface area is equal to 2 (lb + bh + lh) = 2 (4 cm + 4 cm 8 cm + 4 cm 8 cm) 

= 2 (16 cm2 + 32 cm2 + 32 cm2)

= 2 80 cm2 = 160 cm2.

so that The resulting cuboid has a surface area of 160 cm2.

34. From a solid cube of side 7 cm, a conical cavity of height 7 cm and radius 3 cm is hollowed out. Find the volume of the remaining solid.

Explanation:

Considering that the (a) side of a solid cube is 7 cm

Conical cavity height, or cone h, is 7 cm.

The conical cavity fits vertically inside the cube because the height of the conical cavity and the side of the cube are equal.


The conical cavity's radius, or cone's, dimensions are 3 cm, 2 cm, and 3 cm, respectively. A solid 7 cm side cube is provided.

The cube has a conical cavity that is hollowed out that is 7 cm high and 3 cm in radius.

The volume of the remaining solid must be determined.

Cube volume = (Side)³ = (7)³ = 343 cm³.

Cone volume equals (1/3)πr²h where r = 3 cm and h = 7 cm.

Cone volume: (1/3)(22/7)(3)2(7)

= (22/3)

= 66 cm³.

The volume of the cube minus volume yields the volume of the remaining solid.

35. Two cones with the same base radius 8 cm and height 15 cm are joined together along their bases. Find the surface area of the shape so formed.

Explanation:

Given, the bases of two similar cones are attached to one another.

The base's radius is 8 cm.

The cone is 15 cm tall.

We need to determine the shape's surface area.

The cone's curved surface area equals πrl.

l = √r² + h²

provided, r = 8 cm

h = 15 cm

l = √8² + 15²

= √64+225

= √289

l = 17 cm

The new shape's curved surface area is equal to the sum of the curved surfaces of cones one and two.

The curved surface area will be same because the cones are the same in size.

Therefore, the new shape's curved surface equals double the cone's curved surface area.

Cone's curved surface area is (22/7)(8)(17).

= 2992/7

= 427.43 cm²

The new shape's curved surface area is equal to 2 (427.43).

= 854.858

= 855 cm²

As a result, the new shape's surface area is 855 cm2.

36. Two solid cones A and B are placed in a cylindrical tube as shown in Fig.12.9. The ratio of their capacities are 2:1. Find the heights and capacities of cones. Also, find the volume of the remaining portion of the cylinder.

Explanation:

Provided, a cylindrical tube is filled with two solid cones A and B.

Cones A and B have a volume ratio of 2:1.

We need to calculate the volume of the remaining part of the cylinder as well as the height and capacity of the cones.

Cone volume = (1/3)r2h

The figure shows,

Cone A's radius is 6/2, or 3 cm.

Cone B's radius is 6/2, or 3 cm.

Cone A's height should be h₁.

Therefore, cone B's height is equal to 21 minus h₁.

Cone A's volume is (1/3)(3)2h₁.

= 3πh₁ cm³

Cone B's volume is equal to (1/3)(3)2(21 - h1).

= 3π(21 - h₁)

= 63π - 3πh₁ cm³

Given that cone, A's volume is 2:1 larger than cone B's,

Therefore, the volume of cone A is equal to twice the volume of cone B. ------------------ (1)

3πh₁ = 2(63π - 3πh₁)

 3πh₁ = 126π - 6πh₁

3πh₁ + 6πh₁ = 126π

9πh₁ = 126π

9h₁ = 126

h₁ = 126/9

h₁ = 42/3 cm

h₁ = 14 cm

Cone B's height is equal to 21 - 14 cm.

Cone A's volume equals 3(22/7)(42/3)

= (22)(6)

= 132 cm³

The volume of cone B equals the volume of cone A/2 according to (1).

= 132/2

= 66 cm³

Cylinder volume = πr²h

provided, r = 6/2 = 3 cm

h = 21 cm

Cylinder volume = (22/7)(3)²(21)

= (22)(9)(3)

= 22(27)

= 594 cm³

The volume of the remaining section is equal to the sum of the volumes of the cones A and B.

= 594 - 132 - 66

= 594 - 198

= 396 cm³

Consequently, the remaining portion's volume is 396 cm3.

37. An ice cream cone full of ice cream having a radius 5 cm and height 10 cm as shown in Fig.12.10. Calculate the volume of ice cream, provided that its  ⅙ part is left unfilled with ice cream. 

Explanation:

Provided an ice cream cone with a radius of 5 cm and a height of 10 cm, we must determine the volume of ice cream when only 1/6 of the cone is filled with ice cream.

The volume of the hemisphere equals (2/3)r³ with r set to 5 cm.

Hemisphere volume is (2/3)(22/7)(5)³ = (2/3)(22/7)(125) 

= 261.90 cm³.

The cone's height is equal to 10 - 5 cm.

Cone volume = (1/3)r2h = (1/3)(22/7)(5)²(5) 

= (1/3)(22/7)(25)(5) 

= 130.95 cm³

Ice cream cone volume equals hemisphere plus cone volume (261.9 + 130.9 = 392.85 cm³).

When a portion of ice cream is left empty, its volume is 392.85 divided by six to equal 65.475 cm³.

327.375 cm³ of ice cream is equal to 392.85 minus 65.475 cm³.

Consequently, the ice cream's volume is 327.375 cm³.

38. Marbles of diameter 1.4 cm are dropped into a cylindrical beaker of diameter 7 cm containing some water. Find the number of marbles that should be dropped into the beaker so that the water level rises by 5.6 cm.

Explanation:

Provided, a cylindrical beaker with a diameter of 7 cm and some water is filled with marbles that are 1.4 cm in diameter.

We need to determine how many marbles should be poured into the beaker in order to raise the water level by 5.6 cm.

Marble radius = 1.4/2 = 0.7 cm

Marble has a volume of (4/3)πr³ = (4/3)(0.7)3 = 0.457 cm³.

Given that the cylindrical beaker's radius is 7/2, or 3.5 cm

The beaker's water level needs to increase by 5.6 cm.

So, h = 5.6 cm

Cylinder volume = πr²h = (3.5)²(5.6) = 68.6 cm³

The number of marbles equals cylindrical beaker volume divided by marble volume, or 68.6/0.457, which equals 150.

There should be 150 marbles dropped as a result.

39. How many spherical lead shots each of diameter 4.2 cm can be obtained from a solid rectangular lead piece with dimensions 66 cm, 42 cm, and 21 cm.

Explanation:

A solid lead piece measuring 66 cm, 42 cm, and 21 cm is provided.

We need to determine how many solid rectangular lead pieces can be converted into spherical lead shots, each measuring 4.2 cm in diameter.

Cuboid volume equals length, breadth, and height

The volume of a solid rectangle is equal to 66, 42, and 21 cm³

Sphere volume = (4/3)r³.

Given that the spherical lead shot diameter is 4.2 cm

Radius equals 4.2/2, or 2.1 cm.

Spherical lead shot volume is equal to (4/3)(22/7)(2.1)3 (4/3)(22)(0.3)(2.1)2 (4/3)(22)(0.1)(2.1)2 = 38.808 cm³).

The volume of a rectangular lead piece divided by the volume of spherical lead shots yields the number of spherical lead shots that can be obtained: 58212/38.808 

= 1500

As a result, the rectangular lead piece may produce 1500 spherical lead shots.

40. How many spherical lead shots of diameter 4 cm can be made out of a solid cube of lead whose edge measures 44 cm.

Explanation:

A solid lead cube with a 44-centimeter edge is provided.

We need to determine how many spherical lead shots with a diameter of 4 cm can be constructed from the solid cube.

Cube volume equals a³.

edge = 44 cm given

Solid cube volume equals (44)³ = 85184 cm³.

Sphere volume = (4/3)r³.

Given that the lead shot's diameter is 4 cm

Radius = 4/2 equals 2 cm.

Lead shot volume equals (4/3)(22/7)(2)³ 

= (4/3)(22/7)(8) 

= 33.52 cm³.

The volume of solid cube/volume of lead shot = 85184/33.52

= 2541, which is the maximum number of spherical lead shots that can be generated.

As a result, the solid lead cube can be used to create 2541 lead shots.

41. A wall 24 m long, 0.4 m thick, and 6 m high is constructed with the bricks each of dimensions 25 cm × 16 cm × 10 cm. If the mortar occupies  1/10 th of the volume of the wall, then find the number of bricks used in constructing the wall.

Explanation:

Given, a wall 24 m long, 0.4 m thick, and 6 m high is built with bricks that are each 25 cm 16 cm 10 cm in size.

One-tenth of the volume of the wall is taken up by a mortar.

We need to figure out how many bricks were used to build the wall.

Wall volume equals length, breadth, and height

Wall volume, including bricks, is equal to 24 x 0.4 x 6 = 57.6 m³.

Given that the bricks are 25 cm, 16 cm, and 10 cm in size

Brick length is 25 cm, or 25/100, or 0.25 m.

The bricks' width is 16 cm, or 16/100, or 0.16 meters.

The bricks' height is 10 cm, or 10/100, or 0.1 m.

The brick volume is equal to 0.25 x 0.16 x 0.10, or 0.004 m³.

Mortar volume equals one-tenth of the wall's volume (1/10 (57.6) = 5.76 m³).

how many bricks were utilized in the building = (volume of the wall with bricks and mortar - the volume of bricks)/volume of bricks = (57.6 - 5.76)/0.004

= 51.84/0.004 

= 12960

As a result, 12960 bricks were utilized to build the wall.

42. Find the number of metallic circular discs with 1.5 cm base diameter and of height 0.2 cm to be melted to form a right circular cylinder of height 10 cm and diameter 4.5 cm.

Explanation:

Provided is a metallic disc with a base diameter of 1.5 cm and a height of 0.2 cm.

A right circular cylinder of 10 cm in height and 4.5 cm in diameter is created by melting the disc.

The number of metallic discs that can form must be determined.

The disc diameter is assumed to be 1.5 cm.

Radius equals 1.5/2, or 0.75 cm

The disc's height is 0.2 cm.

The volume of the disc is equal to πr²h = (0.75)²(0.2) = 0.1125 cm³.

Given that the cylinder's diameter is 4.5 cm

Radius: 4.5/2=2.25 cm

10 cm is the cylinder's height.

Cylinder volume = πr²h 

= (2.25)2(10) 

= 50.625 cm³.

The volume of the cylinder divided by the volume of the produced circular disc gives us 50.625/0.1125, or 450.

450 round discs can be created as a result.

43. A solid metallic hemisphere of radius 8 cm is melted and recast into a right circular cone of base radius 6 cm. Determine the height of the cone.

Explanation:

A right circular cone is created by melting a solid metallic hemisphere.

The radius of a sphere is 8 cm.

The base radius of the right circular cone is 6 cm.

We must ascertain the cone's height.

Hemisphere's volume equals (2/3)πr³ = (2/3)π(8)³ = 1024π/3 cm³

Cone volume equals (1/3)"πr²h" = (1/3)"(6)²h" = "12h cm3

Given that volume of the hemisphere equals the volume of the cone 1024/3, 12πh

1024/36 h = 28.44 cm 

1024/3 = 12hπ 

h = 1024/3(12) h

The cone measures 28.44 cm tall as a result.

44. A rectangular water tank of base 11 m × 6 m contains water up to a height of 5 m. If the water in the tank is transferred to a cylindrical tank of radius 3.5 m, find the height of the water level in the tank.

Explanation:

Provided, a rectangular water tank with a base measuring 11 m by 6 m and a height of 5 m can hold water.

The water in the tank is transported to a 3.5 m-diameter cylindrical tank.

We must ascertain the tank's height.

 The rectangular tank's water volume is calculated as follows: 11 6 5 = 11 30 = 330 m3

Cylinder volume =  πr²h

= (22/7)(3.5)²h 

= 38.5h m³.

Considering, water is moved from a rectangular tank to a cylindrical tank.

As a result, both tanks contain the same amount of water.

38.5h= 330 

h = 330/38.5

 h = 8.57 m

As a result, the cylindrical tank's water level is 8.57 m high.

45. How many cubic centimeters of iron are required to construct an open box whose external dimensions are 36 cm, 25 cm and 16.5 cm provided the thickness of the iron is 1.5 cm. If one cubic cm of iron weighs 7.5 g, find the weight of the box.

Explanation:

Because an open box has exterior measurements of 36, 25, and 16.5 cm.

Iron makes up the open box.

The iron has a 1.5 centimeter thickness.

We need to determine how much iron is needed to build the open box.

If one cubic centimeter of iron weighs 7.5 g, we must determine the weight of the box.

Let x be the iron's thickness.

Assuming x = 1.5 cm

An open box's external volume is equal to its length, width, and height.

= 36 × 25 × 16.5

= 14850 cm³

Box internal length equals length minus two.

= 36 - 2(1.5)

= 36 - 3

= 33 cm

Box internal breath equals breadth minus two.

= 25 - 2(1.5)

= 25 - 3

= 22 cm

Box height (internal) = height - x

= 16.5 - 1.5

= 15 cm

Box interior volume = (length - 2x) (Breadth = 2x, Height = x)

= 33 × 22 × 15

= 10890 cm³

The amount of iron needed to build an open box is equal to the sum of its external and inside volumes.

= 14850 - 10890

= 3960 cm³

Consequently, 3960 cm³ of iron is needed to create an iron box. Given that 1 cm3 of iron weighs 7.5 g

Iron weighing 3960 cm³ equals 7.5(3960) pounds.

= 29700 g

1 kg = 1000 g

Iron weighs 3960 cm³ at a weight of 29700/1000.

= 29.7 kg

Consequently, 29.7 kg of iron is needed to build an open box.

46. The barrel of a fountain pen, cylindrical in shape, is 7 cm long and 5 mm in diameter. A full barrel of ink in the pen is used up on writing 3300 words on average. How many words can be written in a bottle of ink containing one-fifth of a liter?

Explanation:

Dimensions of a barrel =22/7x0.25x0.25x7=1.375 cm³

Bottle's ink content in percentage  =1/5 litre = 1000/5=200 cm³

As a result, the total number of barrels that the specified amount of ink may fill is equal to  200/1.375

So, the necessary word count is  200/1.3756x330=48000.

47. Water flows at the rate of 10m/minute through a cylindrical pipe 5 mm in diameter. How long would it take to fill a conical vessel whose diameter at the base is 40 cm and depth is 24 cm?

Explanation:

Given that a cylindrical pipe with a diameter of 5 mm moves water at a rate of 10 m/min.

A conical vase is 24 cm deep and 40 cm in diameter at the base.

We need to determine how long it will take to fill the conical vessel.

Cylinder volume = πr² h

assuming the diameter is 5 mm

Radius: 5/2 = 2.5 mm

1 cm = 10 mm

So, r = 2.5/10 = 0.25 cm

flow rate equals 10 m/min

1 m = 100 cm

The rate of flow is therefore 10(100) cm/min.

The volume of water emitted in a minute is equal to (0.25)2(10)(100) = 62.5 cm³.

Conical vessel volume equals (1/3) πr²h

assuming a diameter of 40 cm

Radius: (20 cm) = 40/2

Depth h = 24 centimetres = (1/3)(20)²(24)

 = 400π(8)

 = 3200π cm³

Given that the volume of the conical vessel divided by the amount of water that flows through a cylinder pipe in a minute is 3200π/62.5, which equals 51.2 minutes.

As a result, 51.2 minutes are needed to fill the conical vessel.

48. A heap of rice is in the form of a cone of diameter 9 m and height 3.5 m. Find the volume of the rice. How much canvas cloth is required to just cover the heap?

Explanation:

Given, a pile of rice is in the shape of a cone with a 9 m circumference and 3.5 m height.

We must determine the rice's volume.

We need to determine how much canvas fabric is needed to just cover the pile.

Cone volume = (1/3)πr²h

Given that diameter is 9 metres

Radius: 9/2, or 4.5 metres

Size = 3.5 metres

Rice has a volume of (1/3)(22/7)(4.5)2(3.5) = 74.25 m³

As a result, 74.25 m³ of rice is present.

Curved surface area results from using canvas material to simply cover the heap.

The cone's curved surface area equals rl.

We are aware that l = r2 + h2 l = (4.5)2 + (3.5).² 

l = √20.25 + 12.25

 l = √32.5

 l = 5.7 m

Surface area of a curve: (22/7)(4.5)(5.7) = 80.61 m²

Therefore, 80.61 m² of canvas cloth is needed to just cover the heap.

49. A factory manufactures 120000 pencils daily. The pencils are cylindrical in shape a of length 25 cm and circumference of the base as 1.5 cm. Determine the cost of coloring the curved surfaces of the pencils manufactured in one day at Rs 0.05 per dm²

Explanation:

Assume that a factory produces 120000 pencils every day.

The pencils are cylinder-shaped, measuring 25 cm in length and 1.5 cm around the base.

At Rs. 0.05 per dm², we must calculate the cost of coloring the curved surfaces of the pencils produced in a single day.

Given that the circle's circumference is 2 πr 2π r, or 1.5 cm

Given that a pencil is 25 cm long

The cylinder's curved surface area is equal to 2πrh 

= 1.5(25) 

= 37.5 cm².

1 dm = 10 cm

The surface area of the curve is 37.5/100, or 0.375 dm².

120000 pencils have a curved surface area of 120000(0.375) = 45000 dm².

1 dm² of pencil coloring paper costs Rs. 0.05.

The cost of coloring a pencil surface area of 45000 dm² is 45000(0.05) = Rs. 2250.

Consequently, it costs Rs. 2250 to color 45000 dm² of a pencil.

50. Water is flowing at the rate of 15 km/h through a pipe of diameter 14 cm into a cuboidal pond which is 50 m long and 44 m wide. In what time will the level of water in the pond rise by 21 cm?

Explanation:

assuming a cylinder with a 14 cm diameter

Consequently, the cylinder's radius is 7 cm.

Cylinder volume = πr²h cubic units

Water volume flowing through the cylinder at a pace of 15 km/hr in an hour equals (22/7) x (7/100) x (7/100) x 15000 = 231 m³.

We are aware that cuboid volume equals lbs cubic units.

Water volume in the tank is therefore equal to 54 x 44 x (21/100) = 462 m³.

Time taken is equal to (462/231)  = 2 hours.

51. A solid iron cuboidal block of dimensions 4.4 m × 2.6 m × 1m is recast into a hollow cylindrical pipe of internal radius 30 cm and thickness 5 cm. Find the length of the pipe.

Explanation:

Given: The cuboid's dimensions are 4.4 m, 2.6 m, and 1 m.

L = 4.4 m = 440 cm for the length.

Inches (B) = 2.6 metres = 260 cm

Height (H) equals 1 m and 100 cm

Let r be the hollow cylindrical pipe's inner radius.

Hence, r = 30 cm.

Allow 'x' to represent the pipe's thickness.

⇒ x = 5 cm

Assume that the hollow cylindrical pipe is 'h' cm long.

Let 'R' cm be the pipe's outer radius.

Inner radius plus thickness equals outer radius.

R = r + x = 30 + 5 = 35 cm, then.

The volume will not change because the cuboidal block was cast into a hollow cylinder.

As a result, the cuboid's volume is equal to the hollow cylinder's volume, L, B, and H = π(R2- r2) h, 440, and 260×100 = (22/7) × (352 - 302) × h ⇒ h = (440 × 260 ×100 × 7) / (22 × 325) (By switching the terms)

Since 100 cm = 1 m, the answer is that h = 11200 cm or 112 m.

52. 500 persons are taking a dip into a cuboidal pond which is 80 m long and 50 m broad. What is the rise of water level in the pond, if the average displacement of the water by a person is 0.04m3?

Explanation:

When 500 people dip their toes into a cuboidal pond, let the rise in water level be h m.

Due to that,

The cuboidal pond is 80 meters long.

The cuboidal pond's width is 50 meters.

Now, the volume needed to raise the water level in the pond is given by: length x breadth x height = 80 x 50 x h = 4000 h m3.

And one individual displaces on average 0.04 m3 of water.

Therefore, 500 people displace an average of 0.04 m3 of water.

Now, by the provided condition, the average amount of water displaced by 500 people will cause the water level in the pond to rise.

⇒4000 h=500×0.04

∴h=5004000

=204000

=1200 m

=0.005 m   (1m=100cm)

=0.005×100 cm

=0.5 cm

Therefore, a 0.5 cm elevation in the pond's water level is necessary.

53. 16 glass spheres each of radius 2 cm are packed into a cuboidal box of internal dimensions 16 cm × 8 cm × 8 cm and then the box is filled with water. Find the volume of water filled in the box.

Explanation:

Given, a cuboidal box is filled with water before 16 glass spheres are placed within the box.

Each sphere has a 2 cm radius.

The cuboidal box's internal dimensions are 16 cm by 8 cm by 8 cm.

We must determine how much water is in the box.

Sphere volume = (4/3)πr³.

= (4/3)(22/7)(2)³

= (4/3)(22/7)8

= 33.52 cm³

16 spheres' volume equals 16 (33.52)

= 536.32 cm³

The dimensions of a cuboidal box are length, breadth, and height.

= 16 × 8 × 8

= 16 × 64

= 1024 cm³

The volume of the box minus the volume of the 16 spheres equals the volume of water within.

= 1024 - 536.32

= 487.68 cm³

Consequently, 487.68 cm³ of water is contained in the box.

54. A milk container of height 16 cm is made of a metal sheet in the form of a frustum of a cone with radii of its lower and upper ends as 8 cm and 20 cm respectively. Find the cost of milk at the rate of Rs. 22 per liter which the container can hold.

Explanation:

Provided, a metal milk container shaped like the frustum of a cone.

The container is 16 cm tall and has a bottom and upper-end radii of 8 cm and 20 cm, respectively.

We need to determine how much milk the container can store at the price of Rs. 22 per liter.

The area of a cone's frustum equals h/3 [R²+ r² + Rr].

Provided , R = 20 cm

r = 8 cm

h = 16 cm

Container volume equals (22/7)(16/3)[(20)² + (8)² + 20(8)]

= (22/7)(16/3)[400 + 64 + 160]

= (22/7)(16/3)[624]

= 10459.428 cm³

1 cm³ = 1000 liters

Thus, the container's volume is 10459.428/1000, or 10.459 liters.

One liter of milk costs Rs. 22.

10.459 liters of milk cost 22(10.459), or Rs. 230.10, to buy.

As a result, 10.459 liters of milk cost Rs. 230.10/-.

55. A cylindrical bucket of height 32 cm and base radius 18 cm is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm, find the radius and slant height of the heap.

Explanation:

Given, sand is placed within a cylindrical bucket.

The height of the cylindrical bucket is 32 cm, and the base radius is 18 cm.

Sand is dumped from the bucket onto the ground, creating a conical mound.

The 24 cm-tall sand conical heap is conical in shape.

The heap's radius and slant height must be determined.

The volume of a cylinder equals πr²h

Given, h = 32 cm and r = 18 cm

Sand volume in a cylindrical bucket is equal to (22/7)(18)²(32) = 32585.14 cm³.

Cone volume = (1/3)πr²h

Assuming h = 24 cm

= (1/3)(22/7)r²(24)

= 25.14r² cm³

Provided that, the volume of a conical sand heap equals the volume of a cylindrical bucket.

25.14r² = 32585.14

r² = 32585.14/25.14

r² = 1296.14

Using the square root,

r = √1296.14

r = 36 cm

Consequently, the sand conical heap has a 36 cm radius.

Angle height, l = √r² + h²

= √(36)² + (24)²

= √1296 + 576

= √1872

= 43.27 cm

Consequently, the conical sand heap's slant height is 43.27 cm.

56. A rocket is in the form of a right circular cylinder closed at the lower end and surmounted by a cone with the same radius as that of the cylinder. The diameter and height of the cylinder are 6 cm and 12 cm, respectively. If the slant height of the conical portion is 5 cm, find the total surface area and volume of the rocket [Use π = 3.14].

Explanation:

Considering that a rocket is made up of a cone and a right circular cylinder.

Given that the cylinder's diameter is 6 cm Cylinder's radius = 62 + 3 cm

And the cylinder's height is 12 cm.

∴ Cylinder volume =π r²h=3.14(3)212 = 339.12 cm³.

And the curved surface area is equal to 2rh =23.14312=226.08.

Currently, in the right-angled AOC, h=l2r2 

h=5232=259=16=4

∴ Cone's height, 

h = 4 cm

The cone's radius is 3 cm.

The cone's volume is now 13r²h=13r³.

14(3)²(4)=113.043=37.68 cm³, and its curved surface area is rl=3.14(3)5(4)=47.

Therefore, the rocket's entire volume is 339.12 cm3 plus 37.68 cm³ and its total surface area is 47.1 cm² plus 226.08 cm² plus 28.26 cm2 (or 301.44 cm²).

57. A building is in the form of a cylinder surmounted by a hemispherical vaulted dome and contains  41 19/21  m3  of air. If the internal diameter of the dome is equal to its total height above the floor, find the height of the building.

Explanation:

Given, a structure has the shape of a cylinder with a hemispherical vaulted dome on top.

Air volume in the building is 41 19/21 m3.

The internal diameter of the hemispherical dome equals the building's overall height.

We must ascertain the building's height.

The volume of a building equals the sum of its cylinder and hemisphere volumes.

Let the building's overall height be h.

Given, Dimension = h

Radius = 2h

Cylinder height equals h - h/2 = h/2

Cylinder volume = πr²h

Volume equals (h/2)²(h/2) = h3/8 m³.

The hemisphere's volume equals (2/3)r³.

= (2/3)π(h/2)³

= 2πh³/24 m³

Because the building's air volume is 41 19/21 m³,

= [41(21) + 19]/21

= [861 + 19]/21

= 880/21 m³

Now, 880/21 = πh³/8 + 2πh³/24

880/21 = (3πh³ + 2πh³)/24

880/21 = 5πh³/24

by finding h,

h³ = 880(24)/21(5π)

h³ = 64

Using the cube root,

h = 4 m

As a result, the building has a 4 m height.

58. A hemispherical bowl of internal radius 9 cm is full of liquid. The liquid is to be filled into cylindrical shaped bottles each of radius 1.5 cm and height 4 cm. How many bottles are needed to empty the bowl? 

Explanation:

Provided that the hemispherical bowl's radius is 9 cm, the proponent states that the bowl's volume is 2/3π r2

suggests that =⅔ x 22/7 x9x9x9

suggests that =1527.42cm3

Since the cylindrical bottles' diameter (d) is 3 cm and the bottle's height (h) is 4 cm

suggests that the radius is 1.5 cm

Therefore, the cylindrical ball's volume is equal to πr2h.

=22/7 x1.5x1.5x4cm3

Let n be the number of bottles, which implies 

=22/7 x9x9x9

suggests that 

=nx 22/7 x9x9x9

suggests that n=5

Therefore, there are 54 bottles in total.

59. A solid right circular cone of height 120 cm and radius 60 cm is placed in a right circular cylinder full of water of height 180 cm such that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is equal to the radius of the cone.

Explanation:

Given, a solid right circular cone with a height of 120 cm and a radius of 60 cm is positioned within a 180 cm tall right circular water cylinder so that it touches the bottom.

The cylinder's radius equals the cone's radius

We need to determine how much water is still in the cylinder.

Cone volume = (1/3)πr²h

Assuming r = 60 cm and h = 120 cm

Cone volume is equal to 1/3 (22/7/60)2(120) 

= 1/3 (22/7/3600)(120)

= 452571.429 cm³.

Cylinder volume = πr²h

Given that 60 cm is the distance between the cylinder and the cone.

h = 180 cm

= (22/7)(60)²(180)

= (22/7)(3600)(180)

= 2036571.429 cm³

The formula for the amount of water still in the cylinder is: cylinder volume - cone volume.

= 2036571.429 - 452571.429

= 1583999.999

= 1584000 cm³

Consequently, there are 1584000 cm³ of water remaining in the cylinder.

60. Water flows through a cylindrical pipe, whose inner radius is 1 cm, at the rate of 80 cm/sec in an empty cylindrical tank, the radius of whose base is 40 cm. What is the rise of the water level in the tank in half an hour?

Explanation:

Assume that a cylindrical tank with a base radius of 40 cm is empty and water is flowing through a pipe with an inner radius of 1 cm at a speed of 80 cm/sec.

In a half-hour, we need to determine how much the water level in the tank has risen.

Cylinder volume =  πr²h

Assuming r = 1 cm

flow rate is 80 cm/second.

Water flowing through a cylindrical pipe has a volume of (22/7)(1)2(80) = 251.429 cm²/sec.

60 minutes in an hour.

60 seconds make up a minute.

452571.429 cm³ of water runs through a pipe in 30 minutes (251.429 x 30 x 60).

Volume of a cylinder equals  πr²h

Assuming that r = 40 cm = (22/7)(40)²h 

= 5028.571h cm³

Given that the water volume in the cylindrical tank is equal to the volume of water flowing through the cylindrical pipe every 30 minutes, 5028.571 hours, or 452571.429 hours,

h = 452571.429/5028.571

h = 90 cm

As a result, the tank's water level has risen by 90 cm.

61. The rainwater from a roof of dimensions 22 m × 20 m drains into a cylindrical vessel having a diameter of base 2 m and a height 3.5 m. If the rainwater collected from the roof just fills the cylindrical vessel, then find the rainfall in cm.

Explanation:

It is assumed that rainwater from a roof enters a cylindrical container.

The roof is 22 by 20 metres.

3.5 m in height and 2 m in base diameter for the cylindrical tank.

If the cylindrical container is only partially filled with rainwater collected from the roof, we must calculate the rainfall in cm.

Cylinder volume = πr²h

assuming a 2 m base diameter

Radius = 1 metre

1 m = 100 cm

Radius equals 100 cm

h = 3.5 m

= 3.5(100)

h = 350 cm

A cylindrical container filled to capacity with rainwater has a volume equal to π(100)²(350).

= 3500000π cm³

Provided that length is 22 metres, the formula for the amount of rainwater gathered on the roof is length x breadth x height.


1 m = 100 cm

hence, the length is 2200 cm.

2000 cm wide at 20 metres.

= 2200 × 2000 × h

= 4400000h cm³

Now, 4400000h = 3500000π

h = 3500000(22/7) / 4400000

= 35(22/7) / 44

= 5(22) / 44

= 5/2

= 2.5 cm

the rainfall is 2.5 cm as a result.

62. A pen stand made of wood is in the shape of a cuboid with four conical depressions and a cubical depression to hold the pens and pins, respectively. The dimension of the cuboid are 10 cm, 5 cm and 4 cm. The radius of each of the conical depressions is 0.5 cm and the depth is 2.1 cm. The edge of the cubical depression is 3 cm. Find the volume of the wood in the entire stand.

Explanation:

We can see from the above illustration that there is no wood inside the conical depressions. The volumes of all four conical depressions will be the same because their diameters are the same.

The volume of the conical depression divided by the volume of the wooden cuboid equals the total volume of wood in the pen stand.

Using formulas, we shall determine the solid's volume;

Cuboid volume is equal to lbh, where l, b, and h are the cuboid's length, width, and height, respectively.

The cone's volume is equal to 1/3 of πr2h1, where r and h1 are the cone's radius and height, respectively.

Each conical depression's depth, h₁ = 1.4 cm

Each conical depression's radius is equal to 0.5 cm.

The cuboid is 15 cm by 10 cm by 3.5 cm.

The volume of the wooden cuboid minus 4 times the volume of the conical depression is the total amount of wood in the pen stand.

= l × b × h - 4 × 1/3 πr2h₁

= (15 cm × 10 cm × 3.5 cm) - (4 × 1/3 × 22/7 × 0.5 cm × 0.5 cm × 1.4 cm)

= 525 cm3 - 1.47 cm3

= 523.53 cm3

523.53 cm3 of wood makes up the complete stand.



Chapter-12, SURFACE AREAS AND VOLUMES