Surface Area, Volume and Capacity

Q1: Multiple Choice Type

i. The lateral surface area of a cube is 100 cm²; then its volume is:

Step 1: Lateral Surface Area (LSA) of a cube = \(4a^2\), where \(a\) is the edge length.
Given \(4a^2 = 100\).
Step 2: Solve for \(a\): \[ a^2 = \frac{100}{4} = 25 \\ \Rightarrow a = 5 \, \text{cm} \]Step 3: Volume of cube \(V = a^3 = 5^3 = 125 \, \text{cm}^3\).
Answer: b. 125 cm³.

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:

Step 1: Original volume = \(l \times b \times h\).
New volume = \((2l) \times (2b) \times (2h) = 8 \times lbh\).
Answer: c. 8 : 1.

iii. The radius of a cylinder is doubled and its height is halved; then the new volume is:

Step 1: Original volume \(V = \pi r^2 h\).
New volume \(V’ = \pi (2r)^2 \times \frac{h}{2} = \pi \times 4r^2 \times \frac{h}{2} = 2 \pi r^2 h = 2V\).
Answer: b. 2 times.

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:

Step 1: Volume of cuboid = \(9 \times 8 \times 3 = 216 \, \text{cm}^3\).
Step 2: Volume of cube = \(a^3\), so \[ a^3 = 216 \\ \Rightarrow a = \sqrt[3]{216} = 6 \, \text{cm} \]Answer: a. 6 cm.

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:

Step 1: Let each side of base be \(a\) cm.
Volume = Base area × height = \(a^2 \times 7 = 448\).
Step 2: Solve for \(a\): \[ a^2 = \frac{448}{7} = 64 \\ \Rightarrow a = 8 \, \text{cm} \]Answer: d. 8 cm.

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?

Step 1: Volume of cylinder = \(\pi r^2 h\) (area of circle × height) → Statement 1 is TRUE.
Step 2: Lateral surface area of closed cylinder = \(2\pi r h\), and total surface area = \(2\pi r (h + r)\).
Statement 2 incorrectly states lateral surface area as \(2\pi r (h + r)\), which is the total surface area.
Answer: c. Statement 1 is true, and statement 2 is false.

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.

Step 1: For an open cuboid:
Internal length = \(10 – 2 \times 1 = 8\) cm ✓
Internal breadth = \(12 – 2 \times 1 = 10\) cm ✓
Internal height = \(6 – 1 = 5\) cm ✓
Both Assertion and Reason are correct.
Reason correctly explains the Assertion.
Answer: a. Both A and R are correct, and R is the correct explanation for A.

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 = \((l \times b \times h)\) cubic units.

Step 1: Volume of cubes: \[ V_1 = 6^3 = 216, \quad V_2 = 8^3 = 512, \quad V_3 = 10^3 = 1000 \] Total volume: \[ V = 216 + 512 + 1000 = 1728 \]Step 2: Volume of new cube = \(a^3 = 1728 \\ \Rightarrow a = \sqrt[3]{1728} = 12 \text{ cm}\).
Side of new cube = 12 cm.
Smallest cube side = 6 cm.
New cube side is 2 times the smallest cube side.
Reason is true but it describes volume of cuboid, not about cubes or melting.
Answer: b. Both A and R are correct, and R is not the correct explanation for A.

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 = \(2 \times h \times (l + b)\) square units.

Step 1: Calculate lateral surface area: \[ LSA = 2 \times h \times (l + b) = 2 \times 9 \times (15 + 12) = 18 \times 27 = 486 \, \text{cm}^2 \]Given lateral surface area = 846 cm², which does not match calculated 486 cm².
So, Assertion is false.
Reason’s formula for lateral surface area is correct.
Answer: d. A is false, but R is true.

x. Assertion (A): If radius of a right circular cylinder is doubled and the height is reduced to \(\frac{1}{2}\) 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 = \(\pi r^2 h\), where \(r\) is the radius of circular base and \(h\) is height.

Step 1: Original volume: \[ V = \pi r^2 h \] New volume: \[ V’ = \pi (2r)^2 \times \frac{h}{2} = \pi \times 4r^2 \times \frac{h}{2} = 2 \pi r^2 h = 2V \]So, new volume is twice the original volume, not equal.
Assertion is false.
Reason is true.
Answer: d. A is false, but R is true.

Q2: A cuboid is 8 m long, 12 m broad and 3.5 m high. Find its:

i. Total Surface Area (TSA)

Step 1: Recall the formula for TSA of cuboid: \[ TSA = 2(lb + bh + hl) \]Step 2: Substitute given values: \[ l = 8\, m, \quad b = 12\, m, \quad h = 3.5\, m \] Calculate each term: \[ lb = 8 \times 12 = 96 \\ bh = 12 \times 3.5 = 42 \\ hl = 3.5 \times 8 = 28 \]Step 3: Calculate TSA: \[ TSA = 2 (96 + 42 + 28) = 2 \times 166 = 332 \, m^2 \]Answer: Total Surface Area = 332 m².

ii. Lateral Surface Area (LSA)

Step 4: Recall formula for LSA of cuboid: \[ LSA = 2h(l + b) \]Step 5: Substitute values: \[ LSA = 2 \times 3.5 \times (8 + 12) = 7 \times 20 = 140 \, m^2 \]Answer: Lateral Surface Area = 140 m².

Q3: 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?

Step 1: Convert all dimensions to the same unit, here cm: \[ \text{Length of wall} = 16\, m = 1600\, cm \\ \text{Height of wall} = 3\, m = 300\, cm \\ \text{Thickness of wall} = 22.5\, cm \]Step 2: Calculate volume of the wall: \[ V_{\text{wall}} = \text{length} \times \text{height} \times \text{thickness} = 1600 \times 300 \times 22.5 = 10,800,000\, cm^3 \]Step 3: Calculate volume of one brick: \[ V_{\text{brick}} = 25 \times 11.25 \times 6 = 1687.5\, cm^3 \]Step 4: Number of bricks required: \[ N = \frac{V_{\text{wall}}}{V_{\text{brick}}} = \frac{10,800,000}{1687.5} = 6400 \]Answer: The number of bricks required is 6400.

Q4: 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.

Step 1: Let the common multiple be \(x\). Then, \[ l = 6x, \quad b = 5x, \quad h = 3x \]Step 2: Total Surface Area (TSA) of cuboid: \[ TSA = 2(lb + bh + hl) \] Substitute values: \[ 504 = 2[(6x \times 5x) + (5x \times 3x) + (3x \times 6x)] \\ 504 = 2[30x^2 + 15x^2 + 18x^2] = 2 \times 63x^2 = 126x^2 \]Step 3: Solve for \(x^2\): \[ 126x^2 = 504 \\ \Rightarrow x^2 = \frac{504}{126} = 4 \\ \Rightarrow x = 2 \]Step 4: Calculate volume \(V = l \times b \times h\): \[ V = 6x \times 5x \times 3x = 90x^3 = 90 \times 2^3 = 90 \times 8 = 720 \, \text{cm}^3 \]Answer: The volume of the cuboid is 720 cm³.

Q5: 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.

Step 1: Given:
External length \(l_e = 65\, cm\)
External breadth \(b_e = 34\, cm\)
External height \(h_e = 25\, cm\)
Thickness of wood \(t = 2\, cm\)
Since the box is open from the top, the internal height is: \[ h_i = h_e – t = 25 – 2 = 23\, cm \]Internal length and breadth: \[ l_i = l_e – 2t = 65 – 4 = 61\, cm \\ b_i = b_e – 2t = 34 – 4 = 30\, cm \]Step 2: Capacity of the box = Volume of the internal hollow space: \[ V_{\text{internal}} = l_i \times b_i \times h_i = 61 \times 30 \times 23 = 42090\, cm^3 \]Step 3: Volume of external box: \[ V_{\text{external}} = l_e \times b_e \times h_e = 65 \times 34 \times 25 = 55250\, cm^3 \]Step 4: Volume of wood used: \[ V_{\text{wood}} = V_{\text{external}} – V_{\text{internal}} = 55250 – 42090 = 13160\, cm^3 \]Answer: Capacity of the box = 42090 cm³.
Volume of wood used to make the box = 13160 cm³.

Q6: 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.

Step 1: Let the radius of the cylinder be \(r\) cm and the length (height) be \(h\) cm.
Given:
Curved Surface Area (CSA) = \(2 \pi r h = 132\) cm²
Volume \(V = \pi r^2 h = 462\) cm³
Using \(\pi = \frac{22}{7}\).
Step 2: From CSA formula: \[ 2 \pi r h = 132 \\ \Rightarrow 2 \times \frac{22}{7} \times r \times h = 132 \\ \Rightarrow \frac{44}{7} r h = 132 \\ \Rightarrow r h = \frac{132 \times 7}{44} = 21 \]Step 3: From volume formula: \[ \pi r^2 h = 462 \\ \Rightarrow \frac{22}{7} r^2 h = 462 \] Multiply both sides by 7: \[ 22 r^2 h = 462 \times 7 = 3234 \] Divide by 22: \[ r^2 h = \frac{3234}{22} = 147 \]Step 4: From Step 2, \(r h = 21\), so: \[ r^2 h = r \times (r h) = r \times 21 = 147 \\ \Rightarrow r = \frac{147}{21} = 7\, \text{cm} \]Step 5: Using \(r h = 21\), calculate \(h\): \[ 7 \times h = 21 \\ \Rightarrow h = 3\, \text{cm} \]Step 6: Diameter \(d = 2r = 2 \times 7 = 14\, \text{cm}\).
Answer: Diameter = 14 cm. and Length (height) = 3 cm.

Q7: 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.

Step 1: Given:
Perimeter of floor, \(P = 250\, m\)
Cost of painting walls = ₹15,000
Rate of painting = ₹10 per m²
Step 2: Calculate total area painted (area of four walls): \[ \text{Area} = \frac{\text{Cost}}{\text{Rate}} = \frac{15000}{10} = 1500\, m^2 \]Step 3: Let length = \(l\), breadth = \(b\), height = \(h\).
Perimeter of floor: \[ 2(l + b) = 250 \\ \Rightarrow l + b = 125 \]Area of four walls (lateral surface area) = perimeter × height: \[ \text{Area} = \text{Perimeter} \times h = 250 \times h \]Step 4: Equate area of walls to 1500: \[ 250 \times h = 1500 \\ \Rightarrow h = \frac{1500}{250} = 6\, m \]Answer: The height of the hall is 6 meters.

Q8: 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.

Step 1: Let the breadth of the hall be \(b\) meters.
Then length \(l = 2b\) meters.
Height \(h = 3\) meters.
Step 2: The area of four walls (lateral surface area) of a rectangular hall is: \[ \text{Area} = 2h(l + b) \]Substitute given values: \[ 108 = 2 \times 3 \times (2b + b) = 6 \times 3b = 18b \]Step 3: Solve for \(b\): \[ 18b = 108 \\ \Rightarrow b = \frac{108}{18} = 6\, m \]Calculate length: \[ l = 2b = 2 \times 6 = 12\, m \]Step 4: Calculate volume \(V = l \times b \times h\): \[ V = 12 \times 6 \times 3 = 216\, m^3 \]Answer: The volume of the hall is 216 cubic meters.

Q9: 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.

Step 1: Volume of the original cube: \[ V = a^3 \\ = 12^3 = 1728 \text{ cm}^3 \]Step 2: Let the side of each small cube be \(s\) cm.
Since the cube is cut into 8 identical cubes: \[ 12^3 = 8s^3 \\ 1728 = 8s^3 \\ s^3 = \frac{1728}{8} = 216 \\ s = 6 \text{ cm} \]Step 3: Surface area of the original cube: \[ SA = 6a^2 \\ = 6 \times 12^2 \\ = 6 \times 144 = 864 \text{ cm}^2 \]Step 4: Surface area of one small cube: \[ SA = 6s^2 \\ = 6 \times 6^2 \\ = 6 \times 36 = 216 \text{ cm}^2 \]Step 5: Total surface area of 8 small cubes: \[ = 8 \times 216 = 1728 \text{ cm}^2 \]Step 6: Ratio of surface areas: \[ 864 : 1728 \\ = 1 : 2 \]Answer: Side of each small cube = 6 cm
Ratio of surface areas = 1 : 2

Q10: 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.

Step 1: Given:
Diameter of roller, \(d = 1.4\, m\)
Length of roller, \(l = 2\, m\)
Number of revolutions = 50
Radius, \(r = \frac{d}{2} = \frac{1.4}{2} = 0.7\, m\)
Step 2: The roller rolls on the ground covering the curved surface area equivalent to the lateral surface area of the cylinder in each revolution.
Curved Surface Area (CSA) of the roller: \[ CSA = 2 \pi r h = 2 \pi r l \]Substitute values: \[ CSA = 2 \times \frac{22}{7} \times 0.7 \times 2 = 2 \times \frac{22}{7} \times 1.4 = 2 \times 4.4 = 8.8\, m^2 \]Step 3: Area covered in 50 revolutions: \[ \text{Area} = 50 \times 8.8 = 440\, m^2 \]Answer: The maximum area covered by the garden roller in 50 revolutions is 440 m².

Q11: 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.

Step 1: Given:
Number of pillars = 24
Radius of each pillar, \(r = 28\, cm = 0.28\, m\)
Height of each pillar, \(h = 4\, m\)
Rate of painting = ₹8 per m²
Step 2: Curved Surface Area (CSA) of one cylindrical pillar: \[ CSA = 2 \pi r h \]Using \(\pi = \frac{22}{7}\), substitute values: \[ CSA = 2 \times \frac{22}{7} \times 0.28 \times 4 = 2 \times \frac{22}{7} \times 1.12 \] Calculate: \[ 2 \times \frac{22}{7} \times 1.12 = 2 \times 3.142857 \times 1.12 = 7.04\, m^2 \]Step 3: Total curved surface area of 24 pillars: \[ \text{Total CSA} = 24 \times 7.04 = 168.96\, m^2 \]Step 4: Total cost of painting: \[ \text{Cost} = \text{Total CSA} \times \text{Rate} = 168.96 \times 8 = ₹1351.68 \]Answer: The total cost of painting the curved surface area of the 24 pillars is ₹1351.68.

Q12: 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.

Step 1: Let the radius of the cylinder be \(r\) cm and height be \(h\) cm.
Given: \[ \frac{\text{Curved Surface Area}}{\text{Total Surface Area}} = \frac{1}{2} \] Total Surface Area (TSA) = 616 cm².
Let: \[ CSA = x, \quad TSA = 2x \] Since TSA = 616 cm², then: \[ 2x = 616 \\ \Rightarrow x = \frac{616}{2} = 308\, \text{cm}^2 \]So, \[ CSA = 308\, \text{cm}^2, \quad TSA = 616\, \text{cm}^2 \]Step 2: Recall formulas: \[ CSA = 2 \pi r h \\ TSA = 2 \pi r (h + r) \]From the ratio: \[ \frac{CSA}{TSA} = \frac{2 \pi r h}{2 \pi r (h + r)} = \frac{h}{h + r} = \frac{1}{2} \]Step 3: Solve for \(h\): \[ \frac{h}{h + r} = \frac{1}{2} \\ \Rightarrow 2h = h + r \\ \Rightarrow h = r \]So height \(h = r\).
Step 4: Use TSA formula: \[ TSA = 2 \pi r (h + r) = 2 \pi r (r + r) = 2 \pi r \times 2r = 4 \pi r^2 \] Given TSA = 616 cm², substitute \(\pi = \frac{22}{7}\): \[ 4 \times \frac{22}{7} \times r^2 = 616 \\ \Rightarrow \frac{88}{7} r^2 = 616 \] Multiply both sides by 7: \[ 88 r^2 = 4312 \] Divide both sides by 88: \[ r^2 = \frac{4312}{88} = 49 \\ r = 7\, \text{cm} \]Since \(h = r\), height \(h = 7\, \text{cm}\).
Step 5: Find volume \(V\): \[ V = \pi r^2 h = \pi r^3 = \frac{22}{7} \times 7^3 = \frac{22}{7} \times 343 = 22 \times 49 = 1078\, \text{cm}^3 \]Answer: The volume of the cylinder is 1078 cm³.

Q13: The areas of three adjacent faces of a box are 120 cm², 72 cm² and 60 cm². Find the volume of the box.

Step 1: Let the length, breadth and height of the box be \(l\), \(b\) and \(h\) respectively.
Given areas of three adjacent faces: \[ lb = 120, \quad bh = 72, \quad hl = 60 \]Step 2: Multiply all three equations: \[ (lb) \times (bh) \times (hl) = 120 \times 72 \times 60 \\ (l^2 b^2 h^2) = 120 \times 72 \times 60 \\ (l b h)^2 = 120 \times 72 \times 60 \]Calculate the product: \[ 120 \times 72 = 8640 \\ 8640 \times 60 = 518400 \\ (l b h)^2 = 518400 \\ l b h = \sqrt{518400} = 720 \]Answer: The volume of the box is 720 cm³.

Q14: 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.

Step 1: Given:
Edge of each cube, \(a = 5\, cm\)
Number of cubes joined = 8
Step 2: Since cubes are joined end to end, the resulting cuboid will have:
Length, \(l = 8 \times 5 = 40\, cm\)
Breadth, \(b = 5\, cm\)
Height, \(h = 5\, cm\)
Step 3: Calculate total surface area (TSA) of the cuboid: \[ TSA = 2 (lb + bh + hl) \\ = 2 (40 \times 5 + 5 \times 5 + 40 \times 5) \\ = 2 (200 + 25 + 200) = 2 \times 425 = 850\, cm^2 \]Step 4: Calculate volume of the cuboid: \[ V = l \times b \times h = 40 \times 5 \times 5 = 1000\, cm^3 \]Answer: Total Surface Area = 850 cm². and Volume = 1000 cm³.

Q15: 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.

Step 1: Given:
Length of pool, \(l = 20\, m = 2000\, cm\)
Width of pool, \(b = 10\, m = 1000\, cm\)
Depth of pool, \(h = 2\, m = 200\, cm\)
Tile dimensions = 40 cm × 40 cm
Step 2: Find the surface area to be tiled.
The pool has 5 surfaces to be tiled: the bottom and the 4 walls (excluding the top).
Area of bottom: \[ A_{\text{bottom}} = l \times b = 2000 \times 1000 = 2,000,000\, cm^2 \]Area of 2 longer walls: \[ 2 \times (l \times h) = 2 \times (2000 \times 200) = 2 \times 400,000 = 800,000\, cm^2 \]Area of 2 shorter walls: \[ 2 \times (b \times h) = 2 \times (1000 \times 200) = 2 \times 200,000 = 400,000\, cm^2 \]Total area to be tiled: \[ A_{\text{total}} = 2,000,000 + 800,000 + 400,000 = 3,200,000\, cm^2 \]Step 3: Area of one tile: \[ A_{\text{tile}} = 40 \times 40 = 1600\, cm^2 \]Step 4: Number of tiles required: \[ \text{Number of tiles} = \frac{A_{\text{total}}}{A_{\text{tile}}} = \frac{3,200,000}{1600} = 2000 \]Answer: The number of tiles required is 2000.