SURFACE AREA, VOLUME AND CAPACITY (Cuboid, Cube and Cylinder) - Q&A

1. Multiple Choice Type:

(i) The volume of a cuboid is 4800 cm³. If its length is 24 cm and breadth is 20 cm; its height is:
(a) 5 cm
(b) 10 cm
(c) 12 cm
(d) 15 cm
Answer: (b) 10 cm
Volume = length × breadth × height
4800 = 24 × 20 × h
4800 = 480 × h
h = 4800 / 480 = 10 cm

(ii) The length, breadth and height of cuboid are in the ratio 3 : 2 : 1 and its volume is 162 cm³; the longest side of the cuboids is:
(a) 6 cm
(b) 9 cm
(c) 18 cm
(d) 54 cm
Answer: (b) 9 cm
Let dimensions be 3x, 2x, and 1x.
Volume = 3x × 2x × 1x = 6x³
6x³ = 162
x³ = 27 ⇒ x = 3
Longest side = 3x = 3 × 3 = 9 cm

(iii) The length of a cuboid is doubled, breadth is halved and height is tripled, the volume of the cuboid will become :
(a) 1.5 times
(b) 2 times
(c) three times
(d) six times
Answer: (c) three times
Original Volume V = l × b × h
New length = 2l, New breadth = b/2, New height = 3h
New Volume = (2l) × (b/2) × (3h) = 3 × (l × b × h) = 3V

(iv) Each side of a cube is tripled, its surface area will become :
(a) 3 times
(b) six times
(c) nine times
(d) 27 times
Answer: (c) nine times
Original Surface Area = 6a²
New side = 3a
New Surface Area = 6(3a)² = 6(9a²) = 9 × (6a²)

(v) A cuboid has a total surface area of 80 m² and the lateral surface area of 50 m², the area of its base is:
(a) 30 m²
(b) 60 m²
(c) 15 m²
(d) 10 m²
Answer: (c) 15 m²
Total Surface Area = Lateral Surface Area + 2 × (Area of Base)
80 = 50 + 2 × Base Area
30 = 2 × Base Area
Base Area = 15 m²

2. The length, the breadth and the height of a cuboid are in the ratio 5 : 3 : 2. If its volume is 240 cm³, find its dimensions. Also, find the total surface area of the cuboid.
Solution:
Let the dimensions be 5x, 3x, and 2x.
Volume = 5x × 3x × 2x = 30x³
30x³ = 240
x³ = 8 ⇒ x = 2
Dimensions are:
Length = 5(2) = 10 cm
Breadth = 3(2) = 6 cm
Height = 2(2) = 4 cm
Total Surface Area = 2(lb + bh + hl)
= 2(10×6 + 6×4 + 4×10)
= 2(60 + 24 + 40)
= 2(124) = 248 cm²

3. The length, breadth and height of a cuboid are in the ratio 6 : 5 : 3. If its total surface area is 504 cm², find its dimensions. Also, find the volume of the cuboid.
Solution:
Let dimensions be 6x, 5x, and 3x.
Total Surface Area = 2(lb + bh + hl) = 504
2(6x⋅5x + 5x⋅3x + 3x⋅6x) = 504
2(30x² + 15x² + 18x²) = 504
2(63x²) = 504
126x² = 504
x² = 4 ⇒ x = 2
Dimensions are: L = 12 cm, B = 10 cm, H = 6 cm
Volume = l × b × h = 12 × 10 × 6 = 720 cm³

4. Find the length of each edge of a cube, if its volume is:
(i) 216 cm³
(ii) 1.728 m³ (Note: Book text "4.728" is likely a typo for "1.728" as 1.728 is a perfect cube)
Solution:
(i) Volume = a³ = 216
a = ∛216 = 6 cm
(ii) Volume = a³ = 1.728 m³
a = ∛1.728 = 1.2 m

5. The total surface area of a cube is 216 cm². Find its volume.
Solution:
Total Surface Area = 6a² = 216
a² = 36 ⇒ a = 6 cm
Volume = a³ = 6³ = 216 cm³

6. A wall 9 m long, 6 m high and 20 cm thick, is to be constructed using bricks of dimensions 30 cm, 15 cm and 10 cm. How many bricks will be required?
Solution:
Convert all wall dimensions to cm:
Length (L) = 9 m = 900 cm
Height (H) = 6 m = 600 cm
Thickness (B) = 20 cm
Volume of Wall = 900 × 600 × 20 = 10,800,000 cm³
Volume of 1 Brick = 30 × 15 × 10 = 4,500 cm³
Number of bricks = Volume of Wall / Volume of Brick
= 10,800,000 / 4,500
= 2,400 bricks

7. A solid cube of edge 14 cm is melted down and recast into smaller and equal cubes each of edge 2 cm. Find the number of smaller cubes obtained.
Solution:
Volume of big cube = 14 × 14 × 14 = 2744 cm³
Volume of one small cube = 2 × 2 × 2 = 8 cm³
Number of cubes = 2744 / 8 = 343

8. A closed box is a cuboid in shape with length = 40 cm, breadth = 30 cm and height = 50 cm. It is made of thin metal sheet. Find the cost of metal sheets required to make 20 such boxes, if 1 m² of metal sheet costs ₹ 45.
Solution:
Surface Area of 1 box = 2(lb + bh + hl)
= 2(40×30 + 30×50 + 50×40)
= 2(1200 + 1500 + 2000)
= 2(4700) = 9400 cm²
Convert to m²: 9400 / 10000 = 0.94 m²
Area for 20 boxes = 20 × 0.94 = 18.8 m²
Cost = 18.8 × 45 = ₹ 846

9. Four cubes, each of edge 9 cm, are joined as shown below (in a line). Write the dimensions of the cuboid obtained. Also, find total surface area and volume.
Solution:
When 4 cubes are joined end to end:
Length = 9 + 9 + 9 + 9 = 36 cm
Breadth = 9 cm
Height = 9 cm
Total Surface Area = 2(36×9 + 9×9 + 9×36)
= 2(324 + 81 + 324)
= 2(729) = 1458 cm²
Volume = 36 × 9 × 9 = 2916 cm³

10. What is the maximum length of a rod which can be kept in a rectangular box with internal dimensions 32 cm × 24 cm × 8 cm.
Solution:
Max length = Diagonal of cuboid = √(l² + b² + h²)
= √(32² + 24² + 8²)
= √(1024 + 576 + 64)
= √1664
= √(64 × 26) = 8√26 cm (approx 40.79 cm)

11. The diagonal of a cube is 25√3 m. Find its surface area.
Solution:
Diagonal of cube = a√3
a√3 = 25√3 ⇒ a = 25 m
Total Surface Area = 6a² = 6 × (25)² = 6 × 625 = 3750 m²

12. A rectangular room is 4.5 m long, 4 m wide and 3 m high. Find the cost of white washing its walls and the roof at ₹ 15 per square metre.
Solution:
Area to whitewash = Area of 4 walls + Area of roof
= 2h(l + b) + (l × b)
= 2×3(4.5 + 4) + (4.5 × 4)
= 6(8.5) + 18
= 51 + 18 = 69 m²
Cost = 69 × 15 = ₹ 1035

1. Multiple Choice Type

(i) The dimensions of a hall are 40 m × 25 m × 5 m. The number of persons which can be accommodated in the hall is (each person requires 5 m³ of air) :
(a) 1000
(b) 2000
(c) 5000
(d) 2500
Answer: (a) 1000
Volume of hall = 40 × 25 × 5 = 5000 m³
Number of persons = 5000 / 5 = 1000

(ii) The external dimensions of a closed rectangular box are 82 cm × 47 cm × 60 cm. If it is made of wood of 1 cm thickness, the internal dimensions of the box are:
(a) 80 cm × 47 cm × 60 cm
(b) 82 cm × 45 cm × 60 cm
(c) 82 cm × 47 cm × 58 cm
(d) 80 cm × 45 cm × 58 cm
Answer: (d) 80 cm × 45 cm × 58 cm
Internal L = 82 - 2(1) = 80
Internal B = 47 - 2(1) = 45
Internal H = 60 - 2(1) = 58

(iii) The outer dimensions of a closed small box are 12 cm × 12 cm × 10 cm. If it is made of 1 cm thick walls, its capacity is:
(a) 1440 cm³
(b) 1000 cm³
(c) 640 cm³
(d) 800 cm³
Answer: (d) 800 cm³
Internal L = 12 - 2 = 10
Internal B = 12 - 2 = 10
Internal H = 10 - 2 = 8
Capacity = 10 × 10 × 8 = 800 cm³

(iv) A room is 3 m long, 2 m broad and 2 m high. It has one door 2 m × 1 m; two windows each 1 m × 0.5 m; the remaining area of the walls is:
(a) 16 m²
(b) 17 m²
(c) 20 m²
(d) 5 m²
Answer: (b) 17 m²
Total Area of walls = 2h(l + b) = 2×2(3 + 2) = 20 m²
Area of door = 2 × 1 = 2 m²
Area of 2 windows = 2 × (1 × 0.5) = 1 m²
Remaining area = 20 - (2 + 1) = 17 m²

2. A room 5 m long, 4.5 m wide and 3.6 m high has one door 1.5 m by 2.4 m and two windows, each 1 m by 0.75 m. Find:
(i) the area of its walls, excluding doors and windows.
(ii) the cost of distempering its walls at the rate of ₹ 4.50 per m²
(iii) the cost of painting its roof at the rate of ₹ 9 per m²
Solution:
(i) Area of 4 walls = 2h(l+b) = 2(3.6)(5 + 4.5) = 7.2(9.5) = 68.4 m²
Area of door = 1.5 × 2.4 = 3.6 m²
Area of 2 windows = 2 × (1 × 0.75) = 1.5 m²
Excluding area = 68.4 - (3.6 + 1.5) = 68.4 - 5.1 = 63.3 m²
(ii) Cost of distempering = 63.3 × 4.50 = ₹ 284.85
(iii) Area of roof = l × b = 5 × 4.5 = 22.5 m²
Cost of painting roof = 22.5 × 9 = ₹ 202.50

3. The dining hall of a hotel is 75 m long, 60 m broad and 16 m high. It has five doors 4 m by 3 m each and four windows 3 m by 1.6 m each. Find the cost of :
(i) papering its walls at the rate of ₹ 12 per m²
(ii) carpeting its floor at the rate of ₹ 25 per m²
Solution:
(i) Total Wall Area = 2h(l+b) = 2(16)(75+60) = 32(135) = 4320 m²
Area of 5 doors = 5 × (4×3) = 60 m²
Area of 4 windows = 4 × (3×1.6) = 19.2 m²
Net Wall Area = 4320 - (60 + 19.2) = 4240.8 m²
Cost of papering = 4240.8 × 12 = ₹ 50,889.60
(ii) Floor Area = 75 × 60 = 4500 m²
Cost of carpeting = 4500 × 25 = ₹ 112,500

4. Find the volume of wood required to make a closed box of external dimensions 80 cm, 75 cm and 60 cm, the thickness of walls of the box being 2 cm throughout.
Solution:
External Volume = 80 × 75 × 60 = 360,000 cm³
Internal Dimensions:
L = 80 - 4 = 76 cm
B = 75 - 4 = 71 cm
H = 60 - 4 = 56 cm
Internal Volume = 76 × 71 × 56 = 302,176 cm³
Volume of Wood = External - Internal = 360,000 - 302,176 = 57,824 cm³

5. A closed box measures 66 cm, 36 cm and 21 cm from outside. If its walls are made of metal sheet, 0.5 cm thick, find:
(i) the capacity of the box;
(ii) volume of metal sheet used and
(iii) weight of the box, if 1 cm³ of metal weighs 3.6 g.
Solution:
(i) Internal Dimensions (subtract 2 × 0.5 = 1 cm):
L = 65 cm, B = 35 cm, H = 20 cm
Capacity = 65 × 35 × 20 = 45,500 cm³
(ii) External Volume = 66 × 36 × 21 = 49,896 cm³
Volume of metal = 49,896 - 45,500 = 4,396 cm³
(iii) Weight = 4,396 × 3.6 = 15,825.6 g = 15.8256 kg

6. The internal length, breadth and height of a closed box are 1 m, 80 cm and 25 cm respectively. If its sides are made of 2.5 cm thick wood, find:
(i) the capacity of the box
(ii) the volume of wood used to make the box.
Solution:
(i) Capacity = Internal Volume
L = 100 cm, B = 80 cm, H = 25 cm
Capacity = 100 × 80 × 25 = 200,000 cm³
(ii) External Dimensions (add 2 × 2.5 = 5 cm):
Ext L = 105 cm, Ext B = 85 cm, Ext H = 30 cm
External Volume = 105 × 85 × 30 = 267,750 cm³
Volume of wood = 267,750 - 200,000 = 67,750 cm³

7. Find the area of metal sheet required to make an open tank of length = 10 m, breadth = 7.5 m and depth = 3.8 m.
Solution:
Since it is an open tank (no lid):
Area = Area of Base + Area of 4 Walls
= (l × b) + 2h(l + b)
= (10 × 7.5) + 2(3.8)(10 + 7.5)
= 75 + 7.6(17.5)
= 75 + 133 = 208 m²

8. A tank 30 m long, 24 m wide and 4.5 m deep is to be made. It is open from the top. Find the cost of iron sheet required, at the rate of ₹ 65 per m² to make the tank.
Solution:
Area = (l × b) + 2h(l + b)
= (30 × 24) + 2(4.5)(30 + 24)
= 720 + 9(54)
= 720 + 486 = 1206 m²
Cost = 1206 × 65 = ₹ 78,390

9. The edges of three solid cubes are 6 cm, 8 cm and 10 cm. These cubes are melted and recasted into a single cube. Find the edge of the resulting cube.
Solution:
Total Volume = 6³ + 8³ + 10³
= 216 + 512 + 1000
= 1728 cm³
Edge of new cube = ∛1728 = 12 cm

10. The ratio between the lengths of the edges of two cubes are in the ratio 3 : 2. Find the ratio between their :
(i) total surface area
(ii) volume.
Solution:
Let edges be 3x and 2x.
(i) Ratio of TSA = 6(3x)² : 6(2x)² = 9x² : 4x² = 9 : 4
(ii) Ratio of Volume = (3x)³ : (2x)³ = 27x³ : 8x³ = 27 : 8

11. The length, breadth and height of a cuboid (rectangular solid) are 4 : 3 : 2.
(i) If its surface area is 2548 cm², find its volume.
(ii) If its volume is 3000 m³, find its surface area.
Solution:
(i) Let dims be 4x, 3x, 2x.
TSA = 2(12x² + 6x² + 8x²) = 2(26x²) = 52x²
52x² = 2548 ⇒ x² = 49 ⇒ x = 7
Dims: 28, 21, 14.
Volume = 28 × 21 × 14 = 8,232 cm³
(ii) Volume = 4x × 3x × 2x = 24x³
24x³ = 3000 ⇒ x³ = 125 ⇒ x = 5
Dims: 20, 15, 10.
Surface Area = 52x² = 52(25) = 1,300 m²

1. Multiple Choice Type:

(i) The curved surface area of a cylinder, with length 10 cm and radius 7 cm, is:
(a) 22 cm²
(b) 308 cm²
(c) 44 cm²
(d) 440 cm²
Answer: (d) 440 cm²
CSA = 2πrh = 2 × (22/7) × 7 × 10 = 440 cm²

(ii) Radius of the base of a solid cylinder is 1 cm and its length is 7 cm. Its volume is :
(a) 11 cm³
(b) 22 cm³
(c) 33 cm³
(d) 44 cm³
Answer: (b) 22 cm³
Volume = πr²h = (22/7) × 1² × 7 = 22 cm³

(iii) The radius and the length of a cylindrical rod are 5 cm and 10 cm respectively. Its lateral surface area is :
(a) 10 cm²
(b) 50 cm²
(c) 75 cm²
(d) none of these
Answer: (d) none of these
LSA = 2πrh = 2π(5)(10) = 100π ≈ 314 cm². 50 and 75 are incorrect.

(iv) The formula for the volume of the given cylindrical pipe (hollow cylinder) is :
(a) π(R + r) h
(b) π(R - r)h
(c) π(R² + r²) h
(d) π(R² - r²) h
Answer: (d) π(R² - r²) h

(v) A rectangular piece of paper 10 cm by 8 cm is rolled along its width to get the cylinder of largest size. The curved surface area of the cylinder formed is:
(a) 44 cm²
(b) 20 cm²
(c) 80 cm²
(d) 80 cm²
Answer: (c) 80 cm²
When rolled along width, the width (8 cm) becomes the circumference (or the paper covers the curved area directly). The area of the paper becomes the curved surface area.
Area = 10 × 8 = 80 cm²

2. Find the height of the cylinder whose radius is 7 cm and the total surface area is 1100 cm².
Solution:
Total Surface Area = 2πr(h + r)
1100 = 2 × (22/7) × 7 × (h + 7)
1100 = 44(h + 7)
h + 7 = 1100 / 44 = 25
h = 25 - 7 = 18 cm

3. The ratio between the curved surface area and the total surface area of a cylinder is 1 : 2. Find the ratio between the height and the radius of the cylinder.
Solution:
CSA / TSA = 1 / 2
2πrh / [2πr(h + r)] = 1 / 2
h / (h + r) = 1 / 2
2h = h + r
h = r
Ratio h : r = 1 : 1

4. The total surface area of a cylinder is 6512 cm² and the circumference of its base is 88 cm. Find :
(i) its radius
(ii) its volume
Solution:
(i) Circumference = 2πr = 88
2 × (22/7) × r = 88 ⇒ r = (88 × 7) / 44 = 14 cm
(ii) TSA = 2πrh + 2πr² = 6512
(Circumference × h) + 2πr² = 6512
88h + 2(22/7)(14)² = 6512
88h + 1232 = 6512
88h = 5280 ⇒ h = 60 cm
Volume = πr²h = (22/7) × 14 × 14 × 60
= 616 × 60 = 36,960 cm³

5. The sum of the radius and the height of a cylinder is 37 cm and the total surface area of the cylinder is 1628 cm². Find the height and the volume of the cylinder.
Solution:
Given: r + h = 37
TSA = 2πr(h + r) = 1628
2 × (22/7) × r × (37) = 1628
(44/7) × 37 × r = 1628
r = (1628 × 7) / (44 × 37)
r = 11396 / 1628 = 7 cm
Since r + h = 37 ⇒ 7 + h = 37 ⇒ h = 30 cm
Volume = πr²h = (22/7) × 7² × 30 = 154 × 30 = 4,620 cm³

6. A cylindrical pillar has radius 21 cm and height 4 m. Find:
(i) the curved surface area of the pillar
(ii) the cost of polishing 36 such cylindrical pillars at the rate of ₹ 12 per m².
Solution:
Convert units to meters: r = 21 cm = 0.21 m, h = 4 m
(i) CSA = 2πrh = 2 × (22/7) × 0.21 × 4
= 44 × 0.03 × 4 = 5.28 m²
(ii) Total Area for 36 pillars = 36 × 5.28 = 190.08 m²
Cost = 190.08 × 12 = ₹ 2,280.96

7. If the radii of two cylinders are in the ratio 4 : 3 and their heights are in the ratio 5 : 6, find the ratio of their curved surfaces.
Solution:
r1/r2 = 4/3; h1/h2 = 5/6
Ratio CSA = (2π r1 h1) / (2π r2 h2)
= (r1/r2) × (h1/h2)
= (4/3) × (5/6) = 20 / 18 = 10 : 9

8. A thin rectangular card board has dimensions 44 cm × 22 cm. It is rolled along its length to get a hollow cylinder of largest size. Find the volume of the cylinder formed.
Solution:
Rolled along length means circumference of base = length = 44 cm.
Height = breadth = 22 cm.
2πr = 44 ⇒ 2(22/7)r = 44 ⇒ r = 7 cm
Volume = πr²h = (22/7) × 7² × 22
= 154 × 22 = 3,388 cm³

9. The ratio between the curved surface area and the total surface area of a right circular cylinder is 3 : 5. Find the ratio between the height and the radius of the cylinder.
Solution:
CSA / TSA = 3 / 5
2πrh / 2πr(h+r) = 3 / 5
h / (h+r) = 3 / 5
5h = 3h + 3r
2h = 3r
h / r = 3 / 2
Ratio h : r = 3 : 2

10. If radii of two circular cylinders are in the ratio 3 : 4 and their heights are in the ratio 6 : 5, find the ratio of their curved surface areas.
Solution:
Ratio CSA = (r1/r2) × (h1/h2)
= (3/4) × (6/5)
= 18 / 20 = 9 : 10

1. Choose the correct answer from the options given below.

(i) The lateral surface area of a cube is 100 cm² then its volume is:
(a) 25 cm³
(b) 125 cm³
(c) 625 cm³
(d) none of these
Answer: (b) 125 cm³
Lateral Area = 4a² = 100 ⇒ a² = 25 ⇒ a = 5
Volume = a³ = 125

(ii) The length, the breadth and the height of a cuboid are doubled, the ratio between the volumes of the new cuboid and the original cuboid is:
(a) 4 : 1
(b) 1 : 4
(c) 8 : 1
(d) 1 : 8
Answer: (c) 8 : 1
V_new = (2l)(2b)(2h) = 8lbh = 8 × V_original

(iii) The radius of a cylinder is doubled and its height is halved; then the new volume is :
(a) same
(b) 2 times
(c) 4 times
(d) 8 times
Answer: (b) 2 times
V = πr²h. V_new = π(2r)²(h/2) = π(4r²)(h/2) = 2πr²h

(iv) The dimensions of a solid metallic cuboid are 9 cm × 8 cm × 3 cm. It is melted and recast into a cube. The edge of the cube so formed is :
(a) 6 cm
(b) 12 cm
(c) 18 cm
(d) 24 cm
Answer: (a) 6 cm
Volume = 9×8×3 = 216. a³ = 216 ⇒ a = 6

(v) The volume of a cuboid is 448 cm³. Its height is 7 cm and the base is square. Then each side of the base is:
(a) 64 cm
(b) 16 cm
(c) 4 cm
(d) 8 cm
Answer: (d) 8 cm
Vol = Area_base × h ⇒ 448 = x² × 7 ⇒ x² = 64 ⇒ x = 8

(vi) Statement 1: The volume of a right circular cylinder = Area of cross section (area of circle) × Distance between two circular parallel ends.
Statement 2: Lateral surface area of a closed cylinder is 2πr(h + r) cubic units.
Which of the following options is correct?
(a) Both the statements are true.
(b) Both the statements are false.
(c) Statement 1 is true, and statement 2 is false.
(d) Statement 1 is false, and statement 2 is true.
Answer: (c) Statement 1 is true, and statement 2 is false.
Statement 2 is false because LSA is 2πrh (not total area) and unit is square, not cubic.

Assertion-Reason based questions:

(vii) Assertion (A): The length, breadth and height of an open cuboid are 10 cm, 12 cm and 6 cm respectively. If the thickness is 1 cm, then internal dimensions are 8 cm, 10 cm and 5 cm respectively.
Reason (R): If l, b and h are the external dimensions of an open cuboid of thickness 'x', then its internal dimensions are (l-2x), (b-2x) and (h-x) respectively.
(a) (1)
(b) (2)
(c) (3)
(d) (4)
Answer: (a) (1) Both A and R are correct, and R is the correct explanation for A.
L_in = 10 - 2(1) = 8. B_in = 12 - 2(1) = 10. H_in (open) = 6 - 1 = 5. Logic holds.

(viii) Assertion (A): Three solid silver cubes of sides 6 cm, 8 cm and 10 cm are melted and recasted into a single solid cube. The side of the new cube = 2 times the side of the smallest cube.
Reason (R): Volume of a cuboid = (l × b × h) cubic units.
(a) (1)
(b) (2)
(c) (3)
(d) (4)
Answer: (b) (2) Both A and R are correct, but R is not the correct explanation for A.
Vol = 216+512+1000 = 1728. New side = 12. Smallest side = 6. 12 = 2×6. A is true. R is true formula but volume of cube (a³) is the specific logic needed, though R is a general truth about volume. However, usually, if R explains the *method* (conservation of volume), it's explanation. Here R defines volume of *cuboid*. It's related but slightly distinct. Given standard testing patterns, B is likely correct because the formula for a cube is specific, or simply because defining volume doesn't explain the ratio '2 times'.

(ix) Assertion (A): The length, breadth and height of a cuboid are 15 cm, 12 cm and 9 cm respectively. Lateral surface area of the cuboid = 846 cm².
Reason (R): Lateral surface area of cuboid = 2 × h × (l + b) square units.
(a) (1)
(b) (2)
(c) (3)
(d) (4)
Answer: (d) (4) A is false, but R is true.
LSA = 2×9(15+12) = 18(27) = 486 cm². A says 846. A is false.

(x) Assertion (A): If radius of a right circular cylinder is doubled and the height is reduced to 1/2 of the original, the ratio of the volume of the new cylinder thus formed to the volume of the original cylinder is 1:1.
Reason (R): Volume of a cylinder = πr²h, where r is the radius of circular base and h is height.
(a) (1)
(b) (2)
(c) (3)
(d) (4)
Answer: (d) (4) A is false, but R is true.
Check A: New Vol = π(2r)²(h/2) = 2πr²h. Ratio is 2:1, not 1:1. A is false.

2. A cuboid is 8 m long, 12 m broad and 3.5 m high. Find its
(i) total surface area
(ii) lateral surface area
Solution:
(i) TSA = 2(8×12 + 12×3.5 + 3.5×8) = 2(96 + 42 + 28) = 2(166) = 332 m²
(ii) LSA = 2h(l + b) = 2(3.5)(8 + 12) = 7(20) = 140 m²

3. How many bricks will be required for constructing a wall which is 16 m long, 3 m high and 22.5 cm thick, if each brick measures 25 cm × 11.25 cm × 6 cm?
Solution:
Volume Wall = 1600 × 300 × 22.5 = 10,800,000 cm³
Volume Brick = 25 × 11.25 × 6 = 1,687.5 cm³
Number of bricks = 10,800,000 / 1,687.5 = 6,400

4. The length, breadth and height of a cuboid are in the ratio 6 : 5 : 3. If its total surface area is 504 cm², find its volume.
Solution:
TSA = 2(6x⋅5x + 5x⋅3x + 3x⋅6x) = 126x² = 504 ⇒ x² = 4 ⇒ x = 2
Dims: 12, 10, 6.
Volume = 12 × 10 × 6 = 720 cm³

5. The external dimensions of an open wooden box are 65 cm, 34 cm and 25 cm. If the box is made up of wood 2 cm thick, find the capacity of the box and the volume of wood used to make it.
Solution:
Open box (no lid):
Internal L = 65 - 4 = 61 cm
Internal B = 34 - 4 = 30 cm
Internal H = 25 - 2 = 23 cm (only bottom thickness deducted)
Capacity = 61 × 30 × 23 = 42,090 cm³
External Vol = 65 × 34 × 25 = 55,250 cm³
Volume of wood = 55,250 - 42,090 = 13,160 cm³

6. The curved surface area and the volume of a toy, cylindrical in shape, are 132 cm² and 462 cm³ respectively. Find its diameter and its length.
Solution:
CSA = 2πrh = 132
Volume = πr²h = 462
Divide Vol by CSA: (πr²h) / (2πrh) = 462 / 132
r / 2 = 3.5 ⇒ r = 7 cm
Diameter = 2r = 14 cm
From CSA: 2(22/7)(7)h = 132 ⇒ 44h = 132 ⇒ h = 3 cm
Length = 3 cm

7. The floor of a rectangular hall has a perimeter 250 m. If the cost of painting the four walls at the rate of ₹ 10 per m² is ₹ 15,000, find the height of the hall.
Solution:
Cost = Area × Rate ⇒ 15000 = Area × 10 ⇒ Area of walls = 1500 m²
Area of walls = Perimeter × Height
1500 = 250 × h
h = 1500 / 250 = 6 m

8. The length of a hall is double its breadth. Its height is 3 m. The area of its four walls (including doors and windows) is 108 m², find its volume.
Solution:
l = 2b, h = 3
Area of walls = 2h(l + b) = 108
2(3)(2b + b) = 108
6(3b) = 108
18b = 108 ⇒ b = 6 m
l = 12 m
Volume = l × b × h = 12 × 6 × 3 = 216 m³

9. A solid cube of side 12 cm is cut into 8 identical cubes. What will be the side of the new cube? Also, find the ratio between the surface area of the original cube and the total surface area of all the small cubes formed.
Solution:
Vol original = 12³ = 1728
Vol small = 1728 / 8 = 216
Side small = ∛216 = 6 cm
SA Original = 6(12)² = 864
SA of 8 small cubes = 8 × 6(6)² = 8 × 216 = 1728
Ratio = 864 : 1728 = 1 : 2

10. The diameter of a garden roller is 1.4 m and it is 2 m long. Find the maximum area covered by it in 50 revolutions?
Solution:
d = 1.4 ⇒ r = 0.7 m. h = 2 m.
Area in 1 revolution = CSA = 2πrh
= 2 × (22/7) × 0.7 × 2 = 8.8 m²
Area in 50 revolutions = 50 × 8.8 = 440 m²

11. In a building, there are 24 cylindrical pillars. For each pillar, radius is 28 cm and height is 4 m. Find the total cost of painting the curved surface area of the pillars at the rate of ₹ 8 per m².
Solution:
r = 28 cm = 0.28 m, h = 4 m
CSA of 1 pillar = 2πrh = 2 × (22/7) × 0.28 × 4 = 7.04 m²
Total Area = 24 × 7.04 = 168.96 m²
Cost = 168.96 × 8 = ₹ 1351.68

12. The ratio between the curved surface area and the total surface area of a cylinder is 1 : 2. Find the volume of the cylinder, given its total surface area is 616 cm².
Solution:
CSA/TSA = 1/2 ⇒ CSA = 308, TSA = 616
Ratio 1:2 implies h = r (as proved in Ex 23C Q3).
TSA = 2πr(r+r) = 4πr² = 616
4 × (22/7) × r² = 616
r² = (616 × 7) / 88 = 49 ⇒ r = 7 cm
h = 7 cm
Volume = πr²h = (22/7) × 49 × 7 = 1078 cm³

13. The areas of three adjacent faces of a box are 120 cm², 72 cm² and 60 cm². Find the volume of the box.
Solution:
lb = 120, bh = 72, hl = 60
Multiply all: (lb)(bh)(hl) = l²b²h² = (lbh)²
(Volume)² = 120 × 72 × 60 = 518,400
Volume = √518400 = 720 cm³

14. Eight identical cubes, each of edge 5 cm, are joined end to end. Find the total surface area and the volume of the resulting cuboid.
Solution:
Length = 8 × 5 = 40 cm, Breadth = 5 cm, Height = 5 cm
TSA = 2(40×5 + 5×5 + 5×40) = 2(200 + 25 + 200) = 2(425) = 850 cm²
Volume = 40 × 5 × 5 = 1000 cm³

15. A rectangular swimming pool 20 m long, 10 m wide and 2 m deep is to be tiled. If each tile is 40 cm × 40 cm, find the number of tiles required.
Solution:
Area to tile = Floor + 4 Walls
= (20 × 10) + 2(2)(20 + 10)
= 200 + 4(30) = 320 m²
Area of 1 tile = 0.4 m × 0.4 m = 0.16 m²
Number of tiles = 320 / 0.16 = 2000 tiles