Solids [Surface Area & volume] — Exercise 20

The area of a square on the diagonal of a cube is 48 cm². Each edge of the cube is :

1. 2 cm

2. 4 cm

3. 6 cm

4. 8 cm

Answer

Given:

Area of the square = 48 cm²

Let a be the Side of square.

Area of square = side²

⇒ a² = 48 cm²

⇒ a = $\sqrt{48}$ cm

⇒ a = $\sqrt{16 \times 3}$ cm

⇒ a = 4 $\sqrt{3}$ cm

Let s be the side of cube.

Diagonal of cube = side $\sqrt{3}$

⇒ Side of square = Diagonal of cube

⇒ s $\sqrt{3}$ = 4 $\sqrt{3}$ cm

⇒ s $\cancel{\sqrt{3}}$ = 4 $\cancel{\sqrt{3}}$ cm

⇒ side = 4 cm

Each edge of the cube is 4 cm.

Hence, option 2 is the correct option.

A cuboid has length = 4 cm, breadth = 3 cm and diagonal = 13 cm. The volume of the cuboid is :

1. 156 cm³

2. 288 cm³

3. 144 cm³

4. 78 cm³

Answer

Given:

Length of cuboid = 4 cm

Breadth of cuboid = 3 cm

Diagonal of cuboid = 13 cm

Let h be the height of cuboid.

Length of diagonal of cuboid = $\sqrt{l^2 + b^2 + h^2}$

⇒ $\sqrt{4^2 + 3^2 + h^2}$ = 13

⇒ $\sqrt{16 + 9 + h^2}$ = 13

Squaring both sides, we get

⇒ $({\sqrt{16 + 9 + h^2}})^2$ = (13)²

⇒ 16 + 9 + h² = 169

⇒ 25 + h² = 169

⇒ h² = 169 - 25

⇒ h² = 144

⇒ h = $\sqrt{144}$

⇒ h = 12 cm

Volume of cuboid = l x b x h

= 4 x 3 x 12 cm²

= 144 cm²

Volume of cuboid is 144 cm².

Hence, option 3 is the correct option.

A cuboid with dimensions 12 cm x 9 cm x 2 cm is made of a metal. It is melted and recast into a solid cube. The edge of the cube is :

1. 6 cm

2. 12 cm

3. 15 cm

4. 8 cm

Answer

Given:

Dimensions of cuboid = 12 cm x 9 cm x 2 cm

Let s be the side of cube.

Volume of cuboid = Volume of cube

⇒ side³ = l x b x h

⇒ s³ = 12 x 9 x 2 cm³

⇒ s³ = 216 cm³

⇒ s = $\sqrt[3]{216}$ cm

⇒ s = 6 cm

The edge of the cube is 6 cm.

Hence, option 1 is the correct option.

The dimensional ratio of the sides of a cuboid is 3 : 2 : 1. If its volume is 1296 cm³; the actual dimensions of the cuboid are :

1. 12 cm, 12 cm and 8 cm

2. 8 cm, 8 cm and 12 cm

3. 12 cm, 12 cm and 12 cm

4. 18 cm, 12 cm and 6 cm

Answer

Given:

Volume of cuboid = 1296 cm³

Dimensions of cuboid = 3 : 2 : 1

Let the sides of cuboid be 3a, 2a and a.

Volume of cuboid = l x b x h

⇒ 3a x 2a x a = 1296 cm³

⇒ 6a³ = 1296 cm³

⇒ a³ = $\dfrac{1296}{6}$ cm³

⇒ a³ = 216 cm³

⇒ a = $\sqrt[3]{216}$ cm

⇒ a = 6 cm

Sides are 3a, 2a and a = 3 x 6 cm, 2 x 6 cm and 6 cm

= 18 cm, 12 cm and 6 cm

Hence, option 4 is the correct option.

A tank with dimensions 12 m, 10 m and 8 m is dug and the soil taken out of it is spread uniformly on a field of length = 40 m and breadth = 32 m. The rise in level of the field is :

1. 7.5 m

2. 7.5 cm

3. 75 cm

4. 75 m

Answer

Given:

Dimensions of tank = 12 m, 10 m and 8 m

Dimensions of field = 40 m, 32 m

Let h be the height of field.

Volume of tank = Volume of soil spread on the field

⇒ l_tank x b_tank x h_tank = l_field x b_field x h_field

⇒ 12 x 10 x 8 = 40 x 32 x h

⇒ 960 = 1,280 x h

⇒ h = $\dfrac{960}{1,280}$ m

⇒ h = 0.75 m = 75 cm

The rise in level of the field is 75 cm.

Hence, option 3 is the correct option.

Through the pipe of uniform cross-section (12 cm²) water flows with the speed of 20 cm/s. The volume of water that flows in 1 minute is :

1. 0.0144 m³

2. 0.144 m³

3. 144 cm³

4. 240 cm³

Answer

Given:

Area of cross section = 12 cm²

Speed of water = 20 cm/s

Volume of water flowing out per second = Area of cross section x speed of water

= 12 x 20 cm³

= 240 cm³/s

Volume of water flowing out per minute = 240 x 60 cm³

= 14,400 cm³

= 0.0144 m³

The total volume of water flowing out in one minute is 0.0144 m³.

Hence, option 1 is the correct option.

In the given figure, all the dimensions are given in cm. The volume of the solid is :

1. 7 cm x 13 cm x 20 cm

2. (7 + 13) x 20 x 3 cm³

3. $\dfrac{1}{4}(7 + 13)$ x 3 x 20 cm³

4. $\dfrac{1}{2}(7 + 13)$ x 3 x 20 cm³

Answer

Volume of solid = Area of base x height

Area of base = Area of trapezium = $\dfrac{1}{2}$ x (sum of parallel side) x height

= $\dfrac{1}{2}$ x (7 + 13) x 3

Volume of solid = $\dfrac{1}{2}$ x (7 + 13) x 3 x 20

Hence, option 4 is the correct option.

The length, breadth and height of a rectangular solid are in the ratio 5 : 4 : 2. If the total surface area is 1216 cm², find the length, the breadth and the height of the solid.

Answer

Given:

Total surface area = 1216 cm²

Ratio of dimensions = 5 : 4 : 2

Let the sides of rectangular solid be 5a, 4a and 2a.

Total surface area = 2(lb + bh + hl)

⇒ 2(5a x 4a + 4a x 2a + 2a x 5a) = 1216 cm²

⇒ 2(20a² + 8a² + 10a²) = 1216 cm²

⇒ 2 x 38a² = 1216 cm²

⇒ 76a² = 1216 cm²

⇒ a² = $\dfrac{1216}{76}$ cm²

⇒ a² = 16 cm²

⇒ a = $\sqrt{16}$ cm

⇒ a = 4 cm

Dimensions of rectangular solid are:

Length = 5a = 5 x 4 cm = 20 cm

Breadth = 4a = 4 x 4 cm = 16 cm

Height = 2a = 2 x 4 cm = 8 cm

Hence, length, breadth and height of rectangular solid are 20 cm, 16 cm and 8 cm, respectively.

The volume of a cube is 729 cm³. Find its total surface area.

Answer

Given:

Volume of a cube = 729 cm³

Let a be the side of cube.

Volume of a cube = side³

⇒ a³ = 729 cm³

⇒ a = $\sqrt{729}$ cm

⇒ a = 9 cm

Total surface area of a cube = 6a²

= 6 x 9² cm²

= 6 x 81 cm²

= 486 cm²

Hence, the total surface area of a cube is 486 cm².

The dimensions of a Cinema Hall are 100 m, 60 m and 15 m. How many persons can sit in the hall, if each requires 150 m³ of air for healthy breathing ?

Answer

Given:

Dimensions of hall = 100 m, 60 m and 15 m

Volume of hall = l x b x h

= 100 x 60 x 15 m³

= 90,000 m³

Let n be the number of persons.

Volume of air = Number of person x Air required by each person

⇒ 90,000 m³ = n x 150 m³

⇒ n = $\dfrac{90,000}{150}$

⇒ n = 600

Hence, the number of people that can sit in the Cinema Hall is 600.

75 persons can sleep in a room 25 m by 9.6 m. If each person requires 16 m³ of air; find the height of the room.

Answer

Given:

Number of persons that can sleep in the room = 75

Air required by each person = 16 m³

Let h be the height of the room.

Dimensions of room = 25 m x 9.6 m x h m

Volume of air in the room = Number of persons x Air required by each person

⇒ 25 x 9.6 x h = 75 x 16

⇒ 240 x h = 75 x 16

⇒ 240 x h = 1,200

⇒ h = $\dfrac{1,200}{240}$

⇒ h = 5 m

Hence, the height of the room is 5 m.

The edges of three cubes of metal are 3 cm, 4 cm and 5 cm. They are melted and formed into a single cube. Find the edge of the new cube.

Answer

Given:

Edges of three cubes:

Cube 1: s₁ = 3 cm
Cube 2: s₂ = 4 cm
Cube 3: s₃ = 5 cm

Volume of new cube formed = Total volume of the 3 melted cubes

Let S be the edge of the new cube formed from the melted material.

Total Volume of the Three Cubes = V₁ + V₂ + V₃

⇒ s₁³ + s₂³ + s₃³ = S³

⇒ 3³ + 4³ + 5³ = S³

⇒ 27 + 64 + 125 = S³

⇒ S³ = 216

⇒ S = $\sqrt[3]{216}$

⇒ S = 6 cm

Hence, the edge of the new cube is 6 cm.

Three cubes, whose edges are x cm, 8 cm and 10 cm respectively, are melted and recast into a single cube of edge 12 cm. Find 'x'.

Answer

Given:

Edges of three cubes:

Cube 1: s₁ = x cm
Cube 2: s₂ = 8 cm
Cube 3: s₃ = 10 cm

Edge of the new cube formed from the melted material = 12 cm

Total Volume of the Three Cubes = V₁ + V₂ + V₃

⇒ s₁³ + s₂³ + s₃³ = S³

⇒ x³ + 8³ + 10³ = 12³

⇒ x³ + 512 + 1,000 = 1,728

⇒ x³ + 1,512 = 1,728

⇒ x³ = 1,728 - 1,512

⇒ x³ = 216

⇒ x = $\sqrt[3]{216}$

⇒ x = 6 cm

Hence, x = 6 cm.

Three equal cubes are placed adjacently in a row. Find the ratio of the total surface area of the resulting cuboid to that of the sum of the total surface areas of the three cubes.

Answer

Given:

Let a be the side of each cube.

When the cubes are placed adjacently, they form a cuboid with the following dimensions:

Length of cuboid = a + a + a = 3a

Breadth of cuboid = a

Height of cuboid = a

The ratio = $\dfrac{\text{Tot. surface area of cuboid}}{\text{Sum of tot. surface areas of 3 cubes}}$

$= \dfrac{2(lb + bh + hl)}{6 \times side^2 + 6 \times side^2 + 6 \times side^2}\\[1em] = \dfrac{2(3a \times a + a \times a + a \times 3a)}{3 \times 6a^2}\\[1em] = \dfrac{2(3a^2 + a^2 + 3a^2)}{18a^2}\\[1em] = \dfrac{2 \times 7a^2}{18a^2}\\[1em] = \dfrac{14a^2}{18a^2}\\[1em] = \dfrac{14 \cancel{a^2}}{18 \cancel{a^2}}\\[1em] = \dfrac{14}{18}\\[1em] = \dfrac{7}{9}\\[1em]$

Hence, the ratio of the total surface area of the resulting cuboid to that of the sum of the total surface areas of the three cubes is 7:9.

The cost of papering the four walls of a room at 75 paise per square metre is ₹ 240. The height of the room is 5 metres. Find the length and the breadth of the room, if they are in the ratio 5 : 3.

Answer

Given:

Total cost = ₹ 240

Cost of papering = 75 paise per sq. m. = ₹ 0.75 per sq. m.

Height of the room = 5 m

Ratio of length to breadth = 5 : 3

Let the length and breadth of room be 5a and 3a.

Area of the four walls = 2(l + b)h

= 2(5a + 3a)5

= 2 x 8a x 5 sq. m.

= 80a sq. m.

Total cost = Area of the four walls x Cost of papering

⇒ 240 = 80a x 0.75

⇒ 240 = 60a

⇒ a = $\dfrac{240}{60}$

⇒ a = 4 m

The length and breadth of room are 5a and 3a = 5 x 4 m and 3 x 4 m

= 20 m and 12 m

Hence, the length and breadth of the room are 20 m and 12 m, respectively.

The area of a playground is 3650 m². Find the cost of covering it with gravel 1.2 cm deep, if the gravel costs ₹ 6.40 per cubic metre.

Answer

Given:

Area of playground = 3650 m²

Depth of gravel = 1.2 cm = 0.012 m

Cost of gravel = ₹ 6.40 per cubic metre

Volume of Gravel Needed = Area of playground x Depth

= 3,650 x 0.012 m³

= 43.8 m³

Total cost of the gravel = Volume of Gravel Needed x Cost per cubic metre

= 43.8 x 6.40

= ₹ 280.32

Hence, the cost of covering the playground with gravel is ₹ 280.32.

A square plate of side 'x' cm is 8 mm thick. If its volume is 2880 cm³; find the value of x.

Answer

Given:

Volume of the square plate = 2880 cm³

Thickness of the square plate = 8 mm = 0.8 cm

Side of the square plate = x cm

Volume of the square plate = side x side x thickness

⇒ 2,880 = x $\times$ x $\times$ 0.8

⇒ 2,880 = x² $\times$ 0.8

⇒ x² = $\dfrac{2,880}{0.8}$

⇒ x² = 3600

⇒ x = $\sqrt{3600}$

⇒ x = 60 cm

Hence, the value of x is 60 cm.

The external dimensions of a closed wooden box are 27 cm, 19 cm and 11 cm. If the thickness of the wood in the box is 1.5 cm; find :

(i) volume of the wood in the box;

(ii) the cost of the box, if wood costs ₹ 1.20 per cm³;

(iii) number of 4 cm cubes that could be placed into the box.

Answer

(i) Given:

External dimensions of wooden box = 27 cm, 19 cm and 11 cm.

Thickness of the wood = 1.5 cm

External volume of box = l x b x h

= 27 x 19 x 11

= 5,643 cm ³

Internal volume of box = l x b x h

= (27 - 2 x 1.5) x (19 - 2 x 1.5) x (11 - 2 x 1.5)

= (27 - 3) x (19 - 3) x (11 - 3)

= 24 x 16 x 8

= 3,072 cm ³

Volume of wood = External volume of box - Internal volume of box

= 5,643 cm ³ - 3,072 cm ³

= 2,571 cm ³

Hence, volume of wood in the box is 2,571 cm ³.

(ii) Cost of wood = ₹ 1.20 per cm³

Total cost = Volume of wood x Cost of wood

= ₹ 2,571 x 1.20

= ₹ 3,085.20

Hence, total cost of the box is ₹ 3,085.20.

(iii) Side of cube = 4 cm

Let n be the number of cubes.

Internal volume of box = Number of cubes x Volume of cube

⇒ 3,072 = n x side³

⇒ 3,072 = n x 4³

⇒ 3,072 = n x 64

⇒ n = $\dfrac{3,072}{64}$

⇒ n = 48

Hence, the number of 4 cm cubes that could be placed into the box is 48.

A tank 20 m long, 12 m wide and 8 m deep is to be made of iron sheet. It is open at the top. Determine the cost of iron-sheet, at the rate of ₹ 12.50 per metre, if the sheet is 2.5 m wide.

Answer

Given:

Dimensions of the tank = 20 m x 12 m x 8 m

Width of the iron sheet = 2.5 m

Rate of iron sheet = ₹ 12.50 per metre

Area of sheet = Surface area of the tank

⇒ Length_sheet x Width_sheet = Area of 4 walls of the tank + Area of base

⇒ Length_sheet x 2.5 m = 2(l + b)h + l x b

⇒ Length_sheet x 2.5 m = 2(20 + 12)8 + 20 x 12 m²

⇒ Length_sheet x 2.5 m = 2 x 32 x 8 + 240 m²

⇒ Length_sheet x 2.5 m = 512 + 240 m²

⇒ Length_sheet x 2.5 m = 752 m²

⇒ Length_sheet = $\dfrac{752}{2.5}$ m

⇒ Length_sheet = 300.8 m

Cost of the sheet = Length of the sheet x Rate of iron sheet

= 300.8 x 12.50

= ₹ 3,760

Hence, the total cost of the iron sheet is ₹ 3,760.

A closed rectangular box is made of wood of 1.5 cm thickness. The exterior length and breadth are respectively 78 cm and 19 cm, and the capacity of the box is 15 cubic decimetres. Calculate the exterior height of the box.

Answer

Given:

External length of the box = 78 cm

External breadth of the box = 19 cm

Thickness of the wood = 1.5 cm

Volume of the box = 15 dm³ = 15,000 cm³

Let h be the external height of the box.

Volume of the box = l x b x h

⇒ (78 - 2 x 1.5) x (19 - 2 x 1.5) x (h - 2 x 1.5) cm³ = 15,000 cm³

⇒ (78 - 3) x (19 - 3) x (h - 3) cm³ = 15,000 cm³

⇒ 75 x 16 x (h - 3) cm³ = 15,000 cm³

⇒ 1,200 x (h - 3) cm³ = 15,000 cm³

⇒ (h - 3) = $\dfrac{15,000}{1,200}$ cm

⇒ h - 3 = 12.5 cm

⇒ h = 12.5 + 3 cm

⇒ h = 15.5 cm

Hence, the exterior height of the box is 15.5 cm.

The square on the diagonal of a cube has an area of 1875 sq. cm. Calculate :

(i) the side of the cube.

(ii) the total surface area of the cube.

Answer

(i) Given:

Area of the square = 1875 cm²

Let a be the side of the square.

Let s be the side of cube.

Area of the square = side²

⇒ a² = 1875 cm²

⇒ a = $\sqrt{1875}$ cm

⇒ a = 25 $\sqrt{3}$ cm

Side of the square = Diagonal of the cube

Diagonal of the cube = side $\sqrt{3}$

⇒ s $\sqrt{3}$ = 25 $\sqrt{3}$ cm

⇒ side $\cancel{\sqrt{3}}$ = 25 $\cancel{\sqrt{3}}$ cm

⇒ side = 25 cm

Hence, the side of the cube is 25cm.

(ii) Total surface area of the cube = 6 x side²

= 6 x (25)² cm²

= 6 x 625 cm²

= 3,750 cm²

Hence, the total surface area of the cube is 3,750 cm².

The following figure shows a solid of uniform cross-section. Find the volume of the solid. All measurements are in centimetres. Assume that all angles in the figure are right angles.

Answer

The solid can be divided into two cuboids.

Dimensions of cuboid 1:

Length(l₁) = 9 cm Breadth(b₁) = 4 cm Height(h₁) = 3 cm

Dimension of cuboid 2:

Length(l₂) = 6 cm Breadth(b₂) = 4 cm Height(h₂) = 3 cm

Volume of solid = V₁ + V₂

= l₁b₁h₁ + l₂b₂h₂

= 9 x 4 x 3 + 6 x 4 x 3 cm³

= 108 + 72 cm³

= 180 cm³

Hence, the volume of the solid is 180 cm³.

A swimming pool is 40 m long and 15 m wide. Its shallow and deep ends are 1.5 m and 3 m deep respectively. If the bottom of the pool slopes uniformly, find the amount of water in litres required to fill the pool.

Answer

Length of the pool (L) = 40 m

Width of the pool (W) = 15 m

Depth at the shallow end = 1.5 m

Depth at the deep end = 3 m

The cross-section of the pool is a trapezium with:

Parallel sides (depths) as 1.5 m and 3 m.

Height of the trapezium equal to the width of the pool (15 m).

Area of the trapezium = $\dfrac{1}{2}$ x (sum of parallel lines) x height

= $\dfrac{1}{2}$ x (1.5 + 3) x 15 m²

= $\dfrac{1}{2}$ x 4.5 x 15 m²

= $\dfrac{1}{2}$ x 67.5 m²

= 33.75 m²

Volume of the pool = Area of the trapezium x length

= 33.75 x 40 m³

= 1350 m³ = 13,50,000 litres

Hence, the amount of water required to fill the swimming pool is 13,50,000 litres.

The cross-section of a tunnel perpendicular to its length is a trapezium ABCD as shown in the following figure; also given that : AM = BN; AB = 7 m; CD = 5 m. The height of the tunnel is 2.4 m. The tunnel is 40 m long. Calculate :

(i) the cost of painting the internal surface of the tunnel (excluding the floor) at the rate of ₹ 5 per m² (sq. metre).

(ii) the cost of paving the floor at the rate of ₹ 18 per m².

Answer

(i) Length of the tunnel = 40 m

Height of the tunnel = 2.4 m

Cross-section dimensions: AB = 7 m, CD = 5 m

AM = BN = $\dfrac{AB - CD}{2}$ = $\dfrac{7 - 5}{2}$ = $\dfrac{2}{2}$ = 1 m

In Δ ADM,

Using the pythagorean theorem in triangle ADM,

Base² + Height² = Hypotenuse²

⇒ AM² + DM² = AD²

⇒ 1² + (2.4)² = AD²

⇒ 1 + 5.76 = AD²

⇒ 6.76 = AD²

⇒ AD = $\sqrt{6.76}$

⇒ AD = 2.6 m

Area of the internal surface of the tunnel (excluding the floor) = Area of vertical walls - Area of floor

= 2.6 x 40 + 5 x 40 + 2.6 x 40 m²

= 104 + 200 + 104 m²

= 408 m²

Total cost of painting = Area of vertical walls x Cost of painting

= ₹ 408 x 5

= ₹ 2,040

Hence, total cost of painting the internal surface = ₹ 2,040.

(ii) Area of floor = 7 x 40 m²

= 280 m²

Total cost of paving = Area of floor x rate of paving

= ₹ (280 x 18)

= ₹ 5,040

Hence, total cost of paving = ₹ 5,040.

Water is discharged from a pipe of cross-section area 3.2 cm² at the speed of 5m/s. Calculate the volume of water discharged :

(i) in cm³ per sec.

(ii) in litres per minute.

Answer

(i) Cross-sectional area = 3.2 cm²

Speed of water = 5 m/s = 500 cm/s

Volume of water discharged per second = Cross-sectional area x Speed

= 3.2 x 500

= 1,600

Hence, the volume of water discharged per second is 1600 cm^3.

(ii) Volume of water discharged per min = 1600 x 60

= 96,000 cm³

= $\dfrac{96,000}{1,000}$ = 96 litres

Hence, the volume of water discharged per minute is 96 litres.

A hose-pipe of cross-section area 2 cm² delivers 1500 litres of water in 5 minutes. What is the speed of water in m/s through the pipe ?

Answer

Given:

Cross-sectional area of the hose-pipe = 2 cm²

Volume of water = 1,500 liters = 1,500,000 cm³

Time = 5 minutes = 5 x 60 seconds = 300 seconds

Volume of water flowing per second = $\dfrac{1,500,000}{300}$

= 5,000 cm³/ sec

Volume of water flowing per second = Area of cross-section x Speed of water

⇒ 5,000 = 2 x Speed of water

⇒ Speed of water = $\dfrac{5,000}{2}$

⇒ Speed of water = 2,500 cm/s = 25 m/s

Hence, the speed of water through the pipe is 25 m/s.

The cross-section of a piece of metal 4 m in length is shown below. Calculate :

(i) the area of the cross-section;

(ii) the volume of the piece of metal in cubic centimetres.

If 1 cubic centimetre of the metal weighs 6.6 g, calculate the weight of the piece of metal to the nearest kg.

Answer

(i) Area of cross-section = Area of rectangle ABCE + Area of Δ DEF

= length x breadth + $\dfrac{1}{2}$ x base x height

= 12 x 10 + $\dfrac{1}{2}$ x (16 - 10) x (12 - 7.5) m²

= 120 + $\dfrac{1}{2}$ x 6 x 4.5 m²

= 120 + 3 x 4.5 m²

= 120 + 13.5 m²

= 133.5 m²

Hence, the area of cross-section is 133.5 m².

(ii) Length of the piece of metal = 4 m = 400 cm

Volume of the piece of metal = Area of cross-section x length

= 133.5 x 400 cm³

= 53,400 cm³

Weight of 1 cubic cm of metal = 6.6 g

Weight of 53,400 cubic cm metal = 6.6 x 53,400 g

= $\dfrac{352,440}{1000}$ kg = 352.44 ≈ 352 kg.

Hence, the volume of the piece of metal is 53,400 cm³ and weight of the piece of metal is 352 kg.

A rectangular water-tank measuring 80 cm x 60 cm x 60 cm is filled from a pipe of cross-sectional area 1.5 cm², the water emerging at 3.2 m/s. How long does it take to fill the tank?

Answer

Given:

Dimensions of the rectangular tank = 80 cm x 60 cm x 60 cm

Cross-sectional area of the pipe = 1.5 cm²

Speed of water = 3.2 m/s = 3.2 × 100 cm/s = 320 cm/s

Volume of the rectangular tank = 80 x 60 x 60 cm³

= 2,88,000 cm³

Volume of water flowing per second = Cross-sectional area x Speed of water

= 1.5 x 320 cm³/s

= 480 cm³/s

Volume of water flowing per minute = 480 x 60 cm³

= 28,800 cm³/min

Total time required to fill the tank = $\dfrac{\text{Volume of tank}}{\text{Volume per minute}}$

= $\dfrac{288,000}{28,800}$ min

= 10 minutes

Hence, the time taken by the pipe to fill the tank is 10 minutes.