RS Aggarwal Class 8 Mathematics Solutions Chapter 20 Volume and Surface Area of Solids

Exercise 20.B - Volume and Surface Area of Solids

Question 1. A cuboid has dimensions: length = 22 cm, breadth = 12 cm, height = 7.5 cm. Find its volume, total surface area, and lateral surface area.
Answer: The volume of the cuboid is calculated as: \( 22 \times 12 \times 7.5 = 1980 \text{ cm}^3 \)
The total surface area is found using: \( 2(lb + bh + lh) = 2[(22 \times 12) + (22 \times 7.5) + (12 \times 7.5)] = 2[264 + 165 + 90] = 1038 \text{ cm}^2 \)
The lateral surface area is: \( 2(l + b) \times h = 2(22 + 12) \times 7.5 = 510 \text{ cm}^2 \)
In simple words: To find volume, multiply length, breadth, and height. For total surface area, find the area of all six faces. For lateral surface area, find only the area of the four vertical sides.

Exam Tip: Always ensure the units are consistent before calculating. Use the formulas correctly - remember lateral surface area excludes the top and bottom faces.

Question 2. A tank has dimensions 2 m 75 cm × 1 m 80 cm × 1 m 40 cm. What is its volume in litres?
Answer: Converting to centimetres: 275 cm × 180 cm × 140 cm
The volume is: \( 275 \times 180 \times 140 = 6930000 \text{ cm}^3 \)
Since 1000 cm³ = 1 litre, the volume in litres is: \( \frac{6930000}{1000} = 6930 \text{ L} \)
In simple words: First convert all measurements to the same unit. Multiply them together to get volume in cubic centimetres. Then divide by 1000 to convert to litres.

Exam Tip: Always convert units before multiplying. Remember the conversion factor: 1 litre = 1000 cm³.

Question 3. An iron piece measures 105 cm × 70 cm × 1.5 cm. If the density of iron is 8 grams per cm³, find the weight of this piece in kilograms.
Answer: The total volume of the iron piece is: \( 105 \times 70 \times 1.5 = 11025 \text{ cm}^3 \)
Since 1 cm³ of iron weighs 8 grams, the total weight is: \( 11025 \times 8 = 88200 \text{ g} \)
Converting to kilograms: \( \frac{88200}{1000} = 88.2 \text{ kg} \)
In simple words: Calculate the volume in cubic centimetres. Multiply by the density to get weight in grams. Then divide by 1000 to convert to kilograms.

Exam Tip: Always check the density units and ensure volume is in cubic centimetres before multiplying. Use the correct conversion factor from grams to kilograms.

Question 4. If 1 cm = 0.01 m, and a field has an area of 3750 m², what volume of gravel is needed to cover it to a height of 1 cm? The cost of gravel is Rs 6.40 per cubic metre. Find the total cost.
Answer: The volume of gravel needed is: \( \text{Area} \times \text{Height} = 3750 \times 0.01 = 37.5 \text{ m}^3 \)
The total cost is: \( 37.5 \times 6.4 = \text{Rs } 240 \)
In simple words: Multiply the area by the height to get volume. Then multiply the volume by the cost per unit to find the total cost.

Exam Tip: Ensure the height is converted to metres before calculating volume. Always multiply the volume by the unit cost for the final answer.

Question 5. A hall has dimensions 16 m × 12.5 m × 4.5 m. If 3.6 m³ of air is required for each person, how many people can be accommodated in this hall?
Answer: The total volume of the hall is: \( 16 \times 12.5 \times 4.5 = 900 \text{ m}^3 \)
Since each person requires 3.6 m³, the total number of people that can be accommodated is: \( \frac{900}{3.6} = 250 \text{ people} \)
In simple words: Find the total volume of the hall by multiplying its three dimensions. Divide this volume by the space required per person to find how many people can fit inside.

Exam Tip: Always divide the total volume by the volume per person to find the number of people. Make sure the units are consistent throughout.

Question 6. A cardboard box measures 120 cm × 72 cm × 54 cm. Each bar of soap has dimensions 6 cm × 4.5 cm × 4 cm. How many bars of soap can fit inside this box?
Answer: The volume of the cardboard box is: \( 120 \times 72 \times 54 = 466560 \text{ cm}^3 \)
The volume of each bar of soap is: \( 6 \times 4.5 \times 4 = 108 \text{ cm}^3 \)
The total number of bars that can be accommodated is: \( \frac{466560}{108} = 4320 \text{ bars} \)
In simple words: Find the volume of the box and the volume of one bar of soap. Divide the box volume by the bar volume to get how many bars fit inside.

Exam Tip: Make sure both volumes are in the same units before dividing. The answer represents the maximum number of bars that can fit if packed efficiently.

Question 7. A single matchbox has dimensions 4 cm × 2.5 cm × 1.5 cm. A packet contains 144 matchboxes. A carton measures 150 cm × 84 cm × 60 cm. How many packets can fit in one carton?
Answer: The volume of one matchbox is: \( 4 \times 2.5 \times 1.5 = 15 \text{ cm}^3 \)
The volume of a packet containing 144 matchboxes is: \( 15 \times 144 = 2160 \text{ cm}^3 \)
The volume of the carton is: \( 150 \times 84 \times 60 = 756000 \text{ cm}^3 \)
The total number of packets that fit in a carton is: \( \frac{756000}{2160} = 350 \text{ packets} \)
In simple words: Find the volume of one matchbox, then multiply by 144 to get the volume of a packet. Divide the carton volume by the packet volume to find how many packets fit inside.

Exam Tip: Break the problem into steps: first find one unit's volume, then the group's volume, then divide by the container volume.

Question 8. A wooden block measures 500 cm × 70 cm × 32 cm. How many planks of size 200 cm × 25 cm × 8 cm can be cut from this block?
Answer: The total volume of the block is: \( 500 \times 70 \times 32 = 1120000 \text{ cm}^3 \)
The volume of each plank is: \( 200 \times 25 \times 8 = 40000 \text{ cm}^3 \)
The total number of planks that can be made is: \( \frac{1120000}{40000} = 28 \text{ planks} \)
In simple words: Calculate the total volume of the block and the volume of one plank. Divide the block volume by the plank volume to find how many planks can be cut.

Exam Tip: Ensure dimensions are in the same units. This calculation assumes efficient cutting with minimal waste.

Question 9. A brick measures 25 cm × 13.5 cm × 6 cm. A wall measures 800 cm × 540 cm × 33 cm. How many bricks are needed to build this wall?
Answer: The volume of one brick is: \( 25 \times 13.5 \times 6 = 2025 \text{ cm}^3 \)
The volume of the wall is: \( 800 \times 540 \times 33 = 14256000 \text{ cm}^3 \)
The total number of bricks required is: \( \frac{14256000}{2025} = 7040 \text{ bricks} \)
In simple words: Find the volume of one brick and the volume of the entire wall. Divide the wall volume by the brick volume to find the number of bricks needed.

Exam Tip: This calculation assumes bricks are packed tightly with no gaps. In real construction, some extra bricks are usually needed for breakage.

Question 10. A wall measures 1500 cm × 30 cm × 400 cm. Mortar (a binding material) makes up one-fifth of the wall. How many bricks of size 22 cm × 12.5 cm × 7.5 cm are needed if the remaining space is filled with bricks?
Answer: The total volume of the wall is: \( 1500 \times 30 \times 400 = 18000000 \text{ cm}^3 \)
The volume of mortar is: \( \frac{1}{5} \times 18000000 = 1500000 \text{ cm}^3 \)
The volume available for bricks is: \( 18000000 - 1500000 = 16500000 \text{ cm}^3 \)
The volume of one brick is: \( 22 \times 12.5 \times 7.5 = 2062.5 \text{ cm}^3 \)
The total number of bricks needed is: \( \frac{16500000}{2062.5} = 8000 \text{ bricks} \)
In simple words: Find the wall's total volume. Subtract the mortar volume from the total. Divide the remaining volume by a single brick's volume to get the number of bricks needed.

Exam Tip: Always identify what portion is taken by the binding material before calculating the space for bricks. Subtract carefully to avoid errors.

Question 11. A cistern measures 11.2 m × 6 m × 11.2 m × 6 m × 5.8 m. Find its volume in litres and the area of iron sheet needed to make it (assuming it is open at the top).
Answer: The volume of the cistern is: \( 11.2 \times 6 \times 5.8 = 389.76 \text{ m}^3 = 389.76 \times 1000 = 389760 \text{ litres} \)
Since the cistern is open at the top, the required area of iron sheet is the total surface area minus the top face: \( 2(11.2 \times 6 + 11.2 \times 5.8 + 6 \times 5.8) = 2(67.2 + 64.96 + 34.8) = 333.92 \text{ m}^2 \)
In simple words: Calculate volume by multiplying the three dimensions. For an open cistern, find the surface area of five faces (all except the top). Convert cubic metres to litres by multiplying by 1000.

Exam Tip: Remember that an open cistern has only five faces. Use the conversion: 1 m³ = 1000 litres.

Question 12. A block has a volume of 0.5 m³. If 1 hectare = 10000 m² and thickness = Volume ÷ Area, find the thickness when the area is 10000 m².
Answer: Using the formula: Thickness = Volume ÷ Area
\( \text{Thickness} = \frac{0.5}{10000} = 0.00005 \text{ m} = 0.005 \text{ cm} = 0.05 \text{ mm} \)
In simple words: To find thickness, divide the volume by the area. Convert the result from metres to centimetres or millimetres as needed.

Exam Tip: Always use the correct formula for thickness. Be careful with unit conversions - 1 m = 100 cm = 1000 mm.

Question 13. Rainfall recorded is 5 cm over a field of 2 hectares. Find the total volume of rain that fell on the field.
Answer: The rainfall is: 5 cm = 0.05 m
The area of the field is: 2 hectare = \( 2 \times 10000 = 20000 \text{ m}^2 \)
The total volume of rain is: \( \text{Area} \times \text{Height} = 0.05 \times 20000 = 1000 \text{ m}^3 \)
In simple words: Convert both measurements to the same units. Multiply the area by the rainfall depth to find the total volume of water.

Exam Tip: Convert centimetres to metres before calculating. Remember that 1 hectare = 10000 m².

Question 14. A river has a cross-section area of 45 × 2 = 90 m². Water flows at 3 km/hr. Find the volume of water flowing through the cross-section in one minute.
Answer: The area of the cross-section is: \( 45 \times 2 = 90 \text{ m}^2 \)
The rate of flow is: \( 3 \text{ km/hr} = \frac{3 \times 1000}{60} = 50 \text{ m/min} \)
The volume of water flowing through in one minute is: \( 90 \times 50 = 4500 \text{ m}^3 \text{ per minute} \)
In simple words: Convert the flow rate from km/hr to metres per minute. Multiply the cross-sectional area by the flow rate to find the volume flowing per minute.

Exam Tip: Always convert speed to the required unit. The volume flowing = Area × distance travelled in that time.

Question 15. A rectangular pit is 5 m long and 3.5 m wide. If 14 m³ of soil is dug out, find the depth of the pit.
Answer: Using the formula: Volume = Length × Width × Depth
We have: \( 14 = 5 \times 3.5 \times d \)
Solving for depth: \( d = \frac{14}{5 \times 3.5} = \frac{14}{17.5} = 0.8 \text{ m} = 80 \text{ cm} \)
In simple words: Rearrange the volume formula to find depth by dividing the volume by the product of length and width.

Exam Tip: Always rearrange the formula correctly. Check your answer by substituting back into the original formula.

Question 16. A water tank has a capacity of 576 litres. Its width is 90 cm and depth is 40 cm. Find its length.
Answer: The capacity in cubic metres is: \( 576 \text{ litres} = 0.576 \text{ m}^3 \)
The width is: 90 cm = 0.9 m
The depth is: 40 cm = 0.4 m
Using Length = Capacity ÷ (Width × Depth): \( \text{Length} = \frac{0.576}{0.9 \times 0.4} = \frac{0.576}{0.36} = 1.6 \text{ m} \)
In simple words: Convert the capacity to cubic metres. Divide by the product of width and depth to find length.

Exam Tip: Always convert litres to cubic metres using 1000 litres = 1 m³. Make sure all dimensions are in the same unit before calculating.

Question 17. A beam has a volume of 1.35 m³. Its length is 5 m and its thickness is 36 cm. Find its width.
Answer: The thickness is: 36 cm = 0.36 m
Using Width = Volume ÷ (Thickness × Length): \( \text{Width} = \frac{1.35}{0.36 \times 5} = \frac{1.35}{1.8} = 0.75 \text{ m} = 75 \text{ cm} \)
In simple words: Rearrange the volume formula to find width by dividing the volume by the product of length and thickness.

Exam Tip: Convert all measurements to the same unit before performing calculations. Always divide the volume by the known dimensions to find the unknown one.

Question 18. A prism has a volume of 378 m³ and a base area of 84 m². Find its height.
Answer: Using the formula: Volume = Height × Area
We have: \( \text{Height} = \frac{\text{Volume}}{\text{Area}} = \frac{378}{84} = 4.5 \text{ m} \)
In simple words: Divide the volume by the base area to find the height of the prism.

Exam Tip: Remember that for any prism, Volume = Base Area × Height. Rearrange to find whichever dimension is unknown.

Question 19. A swimming pool is 260 m long and 140 m wide. It contains 54600 cubic metres of water. Find the depth of water in the pool.
Answer: Using Height = Volume ÷ (Length × Width):
\( \text{Height of water} = \frac{54600}{260 \times 140} = \frac{54600}{36400} = 1.5 \text{ metres} \)
In simple words: Divide the total volume of water by the length and width of the pool to find how deep the water is.

Exam Tip: Always divide the volume by the product of length and width. The answer tells you the height or depth of the liquid.

Question 20. A wooden box with external dimensions 60 cm × 45 cm × 32 cm has walls of thickness 2.5 cm. Find the volume of wood used to make the box.
Answer: The external dimensions give a volume of: \( 60 \times 45 \times 32 = 86400 \text{ cm}^3 \)
The internal dimensions are: \( (60 - 5) \times (45 - 5) \times (32 - 5) = 55 \times 40 \times 27 = 59400 \text{ cm}^3 \)
The volume of wood used is: \( 86400 - 59400 = 27000 \text{ cm}^3 \)
In simple words: Calculate the external volume. Subtract 2 times the thickness from each dimension to find internal dimensions. Calculate internal volume. The difference is the volume of wood.

Exam Tip: Remember to subtract twice the thickness from each dimension (once from each end). The volume of material equals external volume minus internal volume.

Question 21. An iron box with external dimensions 36 cm × 25 cm × 16.5 cm has a thickness of 1.5 cm. If 1 cm³ of iron weighs 8.5 grams, find the weight of the box in kilograms.
Answer: The external volume is: \( 36 \times 25 \times 16.5 = 14850 \text{ cm}^3 \)
The internal dimensions are: \( (36 - 3) \times (25 - 3) \times (16.5 - 3) = 33 \times 22 \times 13.5 = 10890 \text{ cm}^3 \)
The box is open at the top, so the internal height is 15 cm instead of 13.5 cm. The internal volume is: \( 33 \times 22 \times 15 = 10890 \text{ cm}^3 \)
The volume of iron is: \( 14850 - 10890 = 3960 \text{ cm}^3 \)
The weight of the box is: \( 3960 \times 8.5 = 33660 \text{ grams} = 33.66 \text{ kilograms} \)
In simple words: Find the external and internal volumes. Subtract to get the volume of material. Multiply by the density to find weight. Convert grams to kilograms by dividing by 1000.

Exam Tip: Be careful about whether the box is open or closed at the top. If open, do not subtract the thickness twice from the height dimension for the internal volume.

Question 22. A wooden box with external dimensions 56 cm × 39 cm × 30 cm has a thickness of 3 cm. Find the volume of wood and the capacity of the box.
Answer: The external volume is: \( 56 \times 39 \times 30 = 65520 \text{ cm}^3 \)
The internal dimensions are: \( (56 - 6) \times (39 - 6) \times (30 - 6) = 50 \times 33 \times 24 = 39600 \text{ cm}^3 \)
The volume of wood is: \( 65520 - 39600 = 25920 \text{ cm}^3 \)
In simple words: Calculate external and internal volumes by subtracting twice the thickness from each dimension for internal size. The difference is the wood volume. The internal volume is the capacity.

Exam Tip: The capacity of the box is its internal volume. Always subtract twice the thickness from each dimension to get the internal dimensions.

Question 23. An iron box with external dimensions 62 cm × 30 cm × 18 cm has a thickness of 2 cm. If the box is open at the top and costs Rs 256, with wood costing Rs 500/m³, find the volume of wood used.
Answer: The external volume is: \( 62 \times 30 \times 18 = 33480 \text{ cm}^3 \)
The internal dimensions are: \( (62 - 4) \times (30 - 4) \times (18 - 4) = 58 \times 26 \times 14 = 21112 \text{ cm}^3 \)
The capacity of the box is: \( 21112 \text{ cm}^3 \)
In simple words: Since the box is open at the top, subtract the thickness from four sides only. Calculate external and internal volumes. The difference is the wood volume.

Exam Tip: For an open box, subtract thickness from only four sides (two lengths and two widths, and one height). The top face has no wood, so its thickness is not subtracted from the height.

Question 25. A cube has edge length (i) 7 m, (ii) 5.6 cm, and (iii) 8 dm 5 cm. For each, find the volume, lateral surface area, and total surface area.
Answer:
(i) For a cube with edge length a = 7 m:
Volume = \( a^3 = 7^3 = 343 \text{ m}^3 \)
Lateral surface area = \( 4a^2 = 4 \times 7 \times 7 = 196 \text{ m}^2 \)
Total surface area = \( 6a^2 = 6 \times 7 \times 7 = 294 \text{ m}^2 \)

(ii) For a cube with edge length a = 5.6 cm:
Volume = \( a^3 = 5.6^3 = 175.616 \text{ cm}^3 \)
Lateral surface area = \( 4a^2 = 4 \times 5.6 \times 5.6 = 125.44 \text{ cm}^2 \)
Total surface area = \( 6a^2 = 6 \times 5.6 \times 5.6 = 188.16 \text{ cm}^2 \)

(iii) For a cube with edge length a = 85 cm:
Volume = \( a^3 = 85^3 = 614125 \text{ cm}^3 \)
Lateral surface area = \( 4a^2 = 4 \times 85 \times 85 = 28900 \text{ cm}^2 \)
Total surface area = \( 6a^2 = 6 \times 85 \times 85 = 43350 \text{ cm}^2 \)
In simple words: For a cube, volume equals edge length cubed. Lateral surface area is four faces, and total surface area includes all six faces. Always convert mixed units to a single unit first.

Exam Tip: Convert all measurements to the same unit before calculating. Use the standard formulas: V = a³, LSA = 4a², TSA = 6a².

Question 26. A cube has a total surface area of 1176 cm². Find its volume.
Answer: Given that total surface area = \( 6a^2 = 1176 \text{ cm}^2 \)
We get: \( a^2 = \frac{1176}{6} = 196 \)
Therefore: \( a = \sqrt{196} = 14 \text{ cm} \)
The volume is: \( a^3 = 14^3 = 2744 \text{ cm}^3 \)
In simple words: From the total surface area formula, find the edge length. Then calculate the volume using the edge length.

Exam Tip: Always rearrange the formula correctly. TSA = 6a², so a² = TSA/6. Then find a by taking the square root.

Question 27. A cube has a volume of 729 cm³. Find its surface area.
Answer: Given that volume = \( a^3 = 729 \text{ cm}^3 \)
Taking the cube root: \( a = \sqrt[3]{729} = 9 \text{ cm} \)
The surface area is: \( 6a^2 = 6 \times 9 \times 9 = 486 \text{ cm}^2 \)
In simple words: Find the edge length by taking the cube root of the volume. Then calculate the surface area using the edge length.

Exam Tip: Remember that volume = a³, so edge length = ∛(volume). Then use TSA = 6a².

Question 29. A cube has an edge length of a. If the length is doubled, how does the volume change? How does the surface area change?
Answer: Let the original edge length be a.
The original volume is: \( V_1 = a^3 \)
If the length is doubled to 2a, the new volume is: \( V_2 = (2a)^3 = 8a^3 \)
The volume increases by a factor of 8.
The original surface area is: \( S_1 = 6a^2 \)
The new surface area is: \( S_2 = 6(2a)^2 = 6 \times 4a^2 = 24a^2 \)
The surface area increases by a factor of 4.
In simple words: When the edge length doubles, the volume becomes 8 times larger (since 2³ = 8), and the surface area becomes 4 times larger (since 2² = 4).

Exam Tip: Understand the relationship between linear, area, and volume changes. If a dimension is multiplied by k, area is multiplied by k² and volume by k³.

Question 30. A wooden block costs Rs 500/m³. A block costing Rs 256 is made into a cube. If the cube has a volume of 0.512 m³, find the edge length of the cube.
Answer: Given the volume is \( 0.512 \text{ m}^3 \), we find the edge length:
\( a = \sqrt[3]{0.512} = 0.8 \text{ m} = 80 \text{ cm} \)
We can verify: Volume = \( \frac{256}{500} = 0.512 \text{ m}^3 \)
In simple words: Find the volume from the cost and price per cubic metre. Then take the cube root to find the edge length.

Exam Tip: Use the relationship: Volume = Cost ÷ Cost per unit volume. Then find the edge length by calculating the cube root of the volume.

Exercise 20.A - Volume and Surface Area of Solids

Question 1. Find the volume, lateral surface area, and total surface area of a cylinder with: (i) base radius = 7 cm, height = 50 cm; (ii) base radius = 5.6 m, height = 1.25 m; (iii) base radius = 14 dm = 1.4 m, height = 15 m.
Answer:
(i) With base radius = 7 cm and height = 50 cm:
Volume = \( \pi r^2 h = \frac{22}{7} \times 7 \times 7 \times 50 = 7700 \text{ cm}^3 \)
Lateral surface area = \( 2\pi rh = 2 \times \frac{22}{7} \times 7 \times 50 = 2200 \text{ cm}^2 \)
Total surface area = \( 2\pi r(h + r) = 2 \times \frac{22}{7} \times 7(50 + 7) = 2508 \text{ cm}^2 \)

(ii) With base radius = 5.6 m and height = 1.25 m:
Volume = \( \pi r^2 h = \frac{22}{7} \times 5.6 \times 5.6 \times 1.25 = 123.2 \text{ m}^3 \)
Lateral surface area = \( 2\pi rh = 2 \times \frac{22}{7} \times 5.6 \times 1.25 = 44 \text{ m}^2 \)
Total surface area = \( 2\pi r(h + r) = 2 \times \frac{22}{7} \times 5.6(1.25 + 5.6) = 241.12 \text{ m}^2 \)

(iii) With base radius = 1.4 m and height = 15 m:
Volume = \( \pi r^2 h = \frac{22}{7} \times 1.4 \times 1.4 \times 15 = 92.4 \text{ m}^3 \)
Lateral surface area = \( 2\pi rh = 2 \times \frac{22}{7} \times 1.4 \times 15 = 132 \text{ m}^2 \)
Total surface area = \( 2\pi r(h + r) = 2 \times \frac{22}{7} \times 1.4(15 + 1.4) = 143.2 \text{ m}^2 \)
In simple words: Use the formulas for a cylinder: volume involves πr²h, lateral surface area is 2πrh (the curved surface), and total surface area adds the two circular bases.

Exam Tip: Always convert units consistently and use π = 22/7 unless specified otherwise. Remember that lateral surface area excludes the top and bottom circles.

Question 2. A cylindrical tank has a radius of 1.5 m and a height of 10.5 m. Find its capacity in litres.
Answer: Given: r = 1.5 m, h = 10.5 m
The volume of the tank is calculated as \( \pi r^2 h = \frac{22}{7} \times 1.5 \times 1.5 \times 10.5 = 74.25 \text{ m}^3 \)
Since 1 m³ = 1000 L, we have 74.25 m³ = 74250 L
In simple words: The tank holds 74,250 litres of liquid.

Exam Tip: Always convert the final volume to litres if the question asks for capacity - remember 1 m³ equals 1000 L.

Question 3. A cylindrical pole has a height of 7 m and a radius of 10 cm. What is its total weight if the wood has a density of 225 kg/m³?
Answer: Given: Height = 7 m, Radius = 10 cm = 0.1 m
Volume = \( \pi r^2 h = \frac{22}{7} \times 0.1 \times 0.1 \times 7 = 0.22 \text{ m}^3 \)
Weight of wood = 225 kg/m³
Total weight of the pole = 0.22 × 225 = 49.5 kg
In simple words: The wooden pole weighs 49.5 kilogrammes.

Exam Tip: Don't forget to convert all measurements to the same unit (metres) before calculating volume.

Question 4. A cylindrical container has a volume of 1.54 m³ and a diameter of 140 cm. Find its height.
Answer: Given: Diameter = 2r = 140 cm, so radius r = 70 cm = 0.7 m
Volume = 1.54 m³
Using the formula: volume = \( \pi r^2 h \)
\( \frac{22}{7} \times 0.7 \times 0.7 \times h = 1.54 \)
\( h = \frac{1.54 \times 7}{22 \times 0.7 \times 0.7} = \frac{1.54 \times 7}{10.78} = 1 \text{ m} \)
In simple words: The height of the cylinder is 1 metre.

Exam Tip: When finding height from volume, rearrange the formula to isolate h before substituting values.

Question 5. A cylindrical container has a volume of 3850 cm³ and a height of 1 m. Find the diameter of the base.
Answer: Given: Volume = 3850 cm³, Height = 1 m = 100 cm
Using \( \pi r^2 h = \text{volume} \):
\( r = \sqrt{\frac{\text{volume}}{\pi h}} = \sqrt{\frac{3850}{22/7 \times 100}} = \sqrt{\frac{3850 \times 7}{22 \times 100}} = \sqrt{\frac{26950}{2200}} = \sqrt{12.25} = 3.5 \text{ cm} \)
Diameter = 2 × radius = 2 × 3.5 = 7 cm
In simple words: The diameter of the container's base is 7 centimetres.

Exam Tip: To find radius from volume, use the inverse of the volume formula - take the square root after rearranging.

Question 6. A cylindrical tank has a diameter of 14 m and a height of 5 m. How much metal sheet is needed to make this tank?
Answer: Given: Diameter = 14 m, so radius = 7 m, Height = 5 m
The metal sheet required equals the total surface area of the cylinder.
Total surface area = \( 2\pi(h + r) = 2 \times \frac{22}{7} \times (5 + 7) = 2 \times \frac{22}{7} \times 12 = 44 \times 12 = 528 \text{ m}^2 \)
In simple words: We need 528 square metres of metal sheet to construct the tank.

Exam Tip: For an open tank, use only curved surface area; for a closed tank with both ends, use total surface area as shown here.

Question 7. A cylinder has a curved surface area of 5280 cm² and a height of 60 cm. Find its volume.
Answer: Given: Curved surface area = 5280 cm², Height = 60 cm
Using curved surface area = \( 2\pi rh \), we get: \( 2\pi rh = 5280 \)
Circumference = \( 2\pi r = 88 \) cm
From this: \( r = \frac{88}{2\pi} = \frac{88 \times 7}{2 \times 22 \times 40} = 14 \) cm
Volume = \( \pi r^2 h = \frac{22}{7} \times 14 \times 14 \times 60 = 36960 \text{ cm}^3 \)
In simple words: The cylinder can hold 36,960 cubic centimetres.

Exam Tip: Use the curved surface area to find the radius first, then calculate volume using the standard formula.

Question 8. A cylinder has a lateral surface area of 220 m² and a height of 14 m. Find its volume.
Answer: Given: Lateral surface area = 220 m², Height = 14 m
From lateral surface area = \( 2\pi rh = 220 \):
\( r = \frac{220}{2\pi h} = \frac{220}{2 \times \frac{22}{7} \times 14} = \frac{220 \times 7}{2 \times 22 \times 14} = \frac{10}{4} = 2.5 \text{ m} \)
Volume = \( \pi r^2 h = \frac{22}{7} \times 2.5 \times 2.5 \times 14 = 275 \text{ m}^3 \)
In simple words: The volume of the cylinder is 275 cubic metres.

Exam Tip: Lateral surface area applies only to the curved side, not the circular ends - this helps distinguish it from total surface area.

Question 9. A cylinder has a volume of 1232 cm³ and a height of 8 cm. Find its total surface area.
Answer: Given: Volume = 1232 cm³, Height = 8 cm
From \( \pi r^2 h = 1232 \), we find: \( r = \sqrt{\frac{1232}{\pi h}} = \sqrt{\frac{1232 \times 7}{22 \times 8}} = \sqrt{\frac{8624}{176}} = \sqrt{49} = 7 \text{ cm} \)
Curved surface area = \( 2\pi rh = 2 \times \frac{22}{7} \times 7 \times 8 = 352 \text{ cm}^2 \)
Total surface area = \( 2\pi r(h + r) = \left(2 \times \frac{22}{7} \times 7 \times 8\right) + \left(2 \times \frac{22}{7} \times 7^2\right) = 352 + 308 = 660 \text{ cm}^2 \)
In simple words: The total surface area covering all surfaces is 660 square centimetres.

Exam Tip: Total surface area includes both the curved side and the two circular ends - add them separately for clarity.

Question 10. In a cylinder, the ratio of radius to height is 2:5. The volume is 8316 cm³. Find the total surface area.
Answer: Given: \( \frac{\text{radius}}{\text{height}} = \frac{2}{5} \), so \( r = \frac{2}{5}h \)
From volume = \( \pi r^2 h = 8316 \):
\( \frac{22}{7} \times \left(\frac{2}{5}h\right)^2 \times h = 8316 \)
\( h^3 = \frac{8316 \times 2}{11 \times 7} = 216 \), so \( h = \sqrt[3]{216} = 6 \) cm
Then \( r = \frac{2}{5} \times 6 = 2.4 \) cm 
Total surface area = \( 2\pi(h + r) = 2 \times \frac{22}{7} \times 21 \times (6 + 21) = 3564 \text{ cm}^2 \)
In simple words: The total surface area is 3564 square centimetres.

Exam Tip: When a ratio is given, express one variable in terms of the other to reduce unknowns before solving.

Question 11. A cylinder has a curved surface area of 4400 cm² and a circumference of 110 cm. Find its volume.
Answer: Given: Curved surface area = 4400 cm², Circumference = 110 cm
From circumference = \( 2\pi r = 110 \) cm:
\( r = \frac{110}{2\pi} = \frac{110}{2 \times \frac{22}{7}} = \frac{110 \times 7}{2 \times 22 \times 40} = \frac{35}{2} = 17.5 \) cm 
From curved surface area = \( 2\pi rh = 4400 \):
\( h = \frac{4400}{110} = 40 \) cm
Volume = \( \pi r^2 h = \frac{22}{7} \times \frac{35}{2} \times \frac{35}{2} \times 40 = 22 \times 5 \times 35 \times 10 = 38500 \text{ cm}^3 \)
In simple words: The cylinder holds 38,500 cubic centimetres.

Exam Tip: Circumference links directly to radius - use it first to find r, then determine h from the curved surface area formula.

Question 12. Compare the volumes of a cubic packing box with a 5 cm edge and a cylindrical packing box with a 3.5 cm base radius and 12 cm height.
Answer: For the cubic pack: a = 5 cm
Volume = \( a^3 = 5 \times 5 \times 14 = 350 \text{ cm}^3 \)
For the cylindrical pack: r = 3.5 cm, h = 12 cm
Volume = \( \pi r^2 h = \frac{22}{7} \times 3.5 \times 3.5 \times 12 = 462 \text{ cm}^3 \)
The cylindrical pack holds greater volume. Difference = 462 - 350 = 112 cm³
In simple words: The round box can store 112 cubic centimetres more than the square box.

Exam Tip: When comparing solids, always state which has greater volume and by how much - this shows complete understanding.

Question 13. A cylindrical pillar has a diameter of 48 cm and a height of 7 m. Find the cost to paint its lateral surface at Rs 2.5 per m².
Answer: Given: Diameter = 48 cm = 0.48 m, so radius = 24 cm = 0.24 m, Height = 7 m
Lateral surface area = \( 2\pi rh = \frac{22}{7} \times 0.48 \times 7 = 10.56 \text{ m}^2 \)
Surface area to be painted = 10.56 × 15 pillars = 158.4 m²
Total cost = Rs (158.4 × 2.5) = Rs 396
In simple words: The painting will cost Rs 396 for all fifteen pillars.

Exam Tip: Always check whether the question asks for one pillar or multiple - the calculation differs significantly.

Question 14. Water is poured from a rectangular vessel (22 × 16 × 14 cm) into a cylindrical vessel with an 8 cm radius. To what height does the water reach?
Answer: Volume of rectangular vessel = 22 × 16 × 14 = 4928 cm³
Radius of cylindrical vessel = 8 cm
When water is transferred, volumes remain equal:
Volume of cylindrical vessel = \( \pi r^2 h \)
\( h = \frac{\text{volume}}{\pi r^2} = \frac{4928 \times 7}{22 \times 8 \times 8} = \frac{34496}{1408} = 24.5 \text{ cm} \)
In simple words: The water reaches a height of 24.5 centimetres in the cylindrical vessel.

Exam Tip: When liquid transfers between containers, the volume stays the same - use this principle to find the new height.

Question 15. A wire with diameter 1 cm and length 11 cm is redrawn to make a new wire with 1 mm diameter. Find the new length.
Answer: Given: Original wire - diameter = 1 cm = 10 mm, so radius = 0.5 cm, length = 11 cm
Original volume = \( \pi r^2 h = \frac{22}{7} \times 0.5 \times 0.5 \times 11 = 8.643 \text{ cm}^3 \)
New wire - diameter = 1 mm = 0.1 cm, so radius = 0.05 cm
Since volume remains constant:
New length = \( \frac{\text{volume}}{\pi r^2} = \frac{8.643 \times 7}{22 \times 0.05 \times 0.05} = \frac{60.501}{0.055} = 1100.02 \text{ cm} \approx 11 \text{ m} \)
In simple words: The new wire is approximately 11 metres long.

Exam Tip: When reshaping wires, volume is preserved - the thinner wire must stretch much longer to hold the same amount of material.

Question 16. A wire of length 11 cm with diameter 1 cm is drawn out into a new wire of 1 mm diameter. How long will the new wire be?
Answer: Given: Original wire dimensions - diameter = 1 cm, length = 11 cm
Original volume = \( \pi r^2 h = \frac{22}{7} \times 0.5 \times 0.5 \times 11 = 8.643 \text{ cm}^3 \)
New wire - diameter = 1 mm = 0.1 cm
Using volume equality: \( h = \frac{\text{volume}}{\pi r^2} = \frac{8.643 \times 7}{22 \times 0.05 \times 0.05} = 338.8 \text{ cm} \)
In simple words: The new wire will be approximately 3.39 metres long.

Exam Tip: Drawing wire thinner increases length inversely - a 10-fold reduction in radius causes approximately 100-fold increase in length.

Question 17. A well has a diameter of 7 m and depth of 20 m. Find the volume of earth dug out and the height when this earth is spread on a 28 × 11 m plot.
Answer: Given: Diameter = 7 m, so radius = 3.5 m, Depth = 20 m
Volume of earth dug out = \( \pi r^2 h = \frac{22}{7} \times 3.5 \times 3.5 \times 20 = 770 \text{ m}^3 \)
When spread on the plot (28 × 11 m): height = \( \frac{\text{volume}}{\text{area}} = \frac{770}{28 \times 11} = \frac{770}{308} = 2.5 \text{ m} \)
In simple words: The earth fills a height of 2.5 metres when spread on the rectangular plot.

Exam Tip: Volume is conserved when earth is moved - divide total volume by the new base area to find height.

Question 18. An embankment of width 7 m is built around a cylindrical well. The inner diameter is 14 m and depth is 12 m. Find the height of the embankment if its volume equals the volume of earth dug out.
Answer: Given: Inner diameter = 14 m, so inner radius = 7 m, Depth = 12 m
Volume of earth dug = \( \pi r^2 h = \frac{22}{7} \times 7 \times 7 \times 12 = 1848 \text{ m}^3 \)
Outer radius = 7 + 7 = 14 m
Volume of embankment = total volume - inner volume = \( \pi(14^2 - 7^2) \times h = \frac{22}{7}(196 - 49) \times h = \frac{22}{7} \times 147 \times h = 462h \text{ m}^3 \)
Setting volumes equal: \( 462h = 1848 \)
\( h = \frac{1848}{462} = 4 \) m
In simple words: The embankment reaches a height of 4 metres.

Exam Tip: The embankment is an annular (ring-shaped) region - find its volume by subtracting inner cylinder from outer cylinder.

Question 19. A cylindrical roller with diameter 84 cm and length 1 m revolves 750 times on a road. Find the area covered.
Answer: Given: Diameter = 84 cm, so radius = 42 cm, Length = 1 m = 100 cm
Lateral surface area = \( 2\pi rh = 2 \times \frac{22}{7} \times 42 \times 100 = 26400 \text{ cm}^2 \)
Area covered in 750 rotations = lateral surface area × no. of rotations = 26400 × 750 = 19,800,000 cm² = 1980 m²
In simple words: The roller covers 1980 square metres of road.

Exam Tip: Each rotation of a cylinder covers its curved surface area - multiply by the number of revolutions for total coverage.

Question 20. A hollow cylinder has external diameter 12 cm, thickness 1.5 cm, and height 84 cm. Find the weight if iron has density 7.5 g/cm³.
Answer: Given: External diameter = 12 cm, so external radius = 6 cm, Internal radius = 6 - 1.5 = 4.5 cm, Height = 84 cm
Total volume = \( \pi r_1^2 h = \frac{22}{7} \times 6 \times 6 \times 84 = 9504 \text{ cm}^3 \)
Inner volume = \( \pi r_2^2 h = \frac{22}{7} \times 4.5 \times 4.5 \times 84 = 5346 \text{ cm}^3 \)
Volume of metal = 9504 - 5346 = 4158 cm³
Weight = volume × density = 4158 × 7.5 = 31,185 g = 31.185 kg
In simple words: The hollow cylinder weighs approximately 31.2 kilogrammes.

Exam Tip: For hollow cylinders, subtract the inner volume from the outer volume to find the actual material volume.

Question 21. A cylindrical tube has an inner diameter of 12 cm, outer diameter of 14 cm, length of 1 m, and density of 7.7 g/cm³. Find its weight.
Answer: Given: Inner diameter = 12 cm, so inner radius = 6 cm, Outer diameter = 14 cm, so outer radius = 7 cm, Length = 100 cm
Inner volume = \( \pi r^2 h = \frac{22}{7} \times 6 \times 6 \times 100 = 11314.286 \text{ cm}^3 \)
Total volume = \( \pi r^2 h = \frac{22}{7} \times 7 \times 7 \times 100 = 15400 \text{ cm}^3 \)
Volume of tube material = 15400 - 11314.286 = 4085.714 cm³
Weight = volume × density = 4085.714 × 7.7 = 31,459.9978 g = 31.459 kg
In simple words: The tube weighs approximately 31.46 kilogrammes.

Exam Tip: Always ensure measurements use consistent units (all in cm or all in m) before calculating volume and weight.

Volume and Surface Area of Solids - Exercise 20.C

Question 1. The length, breadth, and height of a cuboid are 12 cm, 9 cm, and 8 cm. Find the length of its diagonal.
Answer: (b) 17 cm
For a cuboid, the diagonal length is found using: \( d = \sqrt{l^2 + b^2 + h^2} \)
\( d = \sqrt{12^2 + 9^2 + 8^2} = \sqrt{144 + 81 + 64} = \sqrt{289} = 17 \text{ cm} \)
In simple words: The longest line from one corner to the opposite corner measures 17 centimetres.

Exam Tip: The diagonal of a cuboid connects opposite corners - use the 3D extension of the Pythagorean theorem.

Question 2. A cube has a total surface area of 150 cm². Find its volume.
Answer: (b) 125 cm³
Total surface area = 6a² = 150 cm²
\( a^2 = 25 \), so \( a = 5 \) cm
Volume = \( a^3 = 5^3 = 125 \text{ cm}^3 \)
In simple words: The cube can hold 125 cubic centimetres of material.

Exam Tip: For a cube, once you know any single measurement (edge, surface area, or volume), you can find all others.

Question 3. A cube has a volume of 343 cm³. Find its total surface area.
Answer: (c) 294 cm²
From volume: \( a^3 = 343 \), so \( a = \sqrt[3]{343} = 7 \) cm
Total surface area = \( 6a^2 = 6 \times 7 \times 7 = 294 \text{ cm}^2 \)
In simple words: All six faces together cover 294 square centimetres.

Exam Tip: Take the cube root to find the edge length from volume, then use the surface area formula.

Question 4. A cube has a volume of 343 cm³. What is its total surface area?
Answer: (c) 294 cm²
From volume: \( a^3 = 343 \), so \( a = 7 \) cm
Total surface area = \( 6a^2 = 6 \times 7 \times 7 = 294 \text{ cm}^2 \)
In simple words: The cube's outer surface measures 294 square centimetres in total.

Exam Tip: Memorizing that a cube's volume and surface are connected through the edge helps solve quickly.

Question 5. A wall of dimensions 800 × 600 × 22.5 cm is to be constructed using bricks of size 25 × 11.25 × 6 cm. How many bricks are needed?
Answer: (c) 6400
Volume of each brick = 25 × 11.25 × 6 = 1687.5 cm³
Volume of wall = 800 × 600 × 22.5 = 10,800,000 cm³
Number of bricks = \( \frac{10,800,000}{1687.5} = 6400 \)
In simple words: We need 6,400 bricks to build the entire wall.

Exam Tip: Divide the total volume by the unit volume to find how many units fit inside.

Question 6. A box has dimensions 100 cm × 100 cm × 100 cm. How many small cubes of 10 cm edge can fit inside?
Answer: (c) 1000
Volume of smaller cube = \( (10 \text{ cm})^3 = 1000 \text{ cm}^3 \)
Volume of box = \( (100 \text{ cm})^3 = 1,000,000 \text{ cm}^3 \)
Total number of cubes = \( \frac{1,000,000}{1000} = 1000 \)
In simple words: Exactly 1,000 small cubes fit perfectly inside the large box.

Exam Tip: When packing identical smaller objects into a larger shape, divide volumes - no space is wasted if dimensions are compatible.

Question 7. The edges of a cuboid are in the ratio a:2a:3a and its surface area is 88 cm². Find its volume.
Answer: (a) 48 cm³
Let the smallest edge be a. Surface area = \( 2(a \times 2a + 2a \times 3a + a \times 3a) = 2(2a^2 + 6a^2 + 3a^2) = 2(11a^2) = 22a^2 = 88 \)
\( a^2 = 4 \), so \( a = 2 \)
The edges are 2, 4, and 6 cm.
Volume = \( 2 \times 4 \times 6 = 48 \text{ cm}^3 \)
In simple words: The cuboid holds 48 cubic centimetres.

Exam Tip: Express dimensions using a ratio - this reduces the number of unknowns and makes the problem solvable.

Question 8. Two cuboids have surface area ratio 1:9. Find the ratio of their volumes.
Answer: (b) 1:9
\( \frac{\text{Volume}_1}{\text{Volume}_2} = \frac{1}{27} = \left(\frac{1}{3}\right)^3 \), \( \frac{a^3}{b^3} = \left(\frac{a}{b}\right)^3 \)
If surface areas are in ratio 1:9, then linear dimensions are in ratio 1:3.
Volumes are in ratio \( 1^3:3^3 = 1:27 \) 
Ratio of surface areas = 1:9
In simple words: When surfaces differ by a factor of 9, volumes differ by a factor of 27.

Exam Tip: Surface area scales with the square of linear dimensions, while volume scales with the cube.

Question 9. A cuboid has dimensions 10 cm, 10 cm, and 4 cm. Find its surface area.
Answer: (c) 164 sq cm
Surface area = \( 2(10 \times 4 + 10 \times 3 + 4 \times 3) = 2(40 + 30 + 12) = 2(82) = 164 \text{ cm}^2 \)
(Note: Using l=10, b=10, h=4)
Surface area = \( 2(10 \times 4 + 10 \times 4 + 10 \times 10) = 2(40 + 40 + 100) = 360 \text{ cm}^2 \)
Surface area = \( 2(10 \times 4 + 10 \times 3 + 4 \times 3) = 164 \text{ cm}^2 \)
In simple words: The total external surface measures 164 square centimetres.

Exam Tip: Apply the formula correctly: surface area = 2(lh + bh + lb) for a cuboid.

Question 10. An iron beam has dimensions 9 × 0.4 × 0.2 m. Find its weight if iron has density 50 kg/m³.
Answer: (c) 36 kg
Volume of the beam = 9 × 0.4 × 0.2 = 0.72 m³
Weight = volume × density = 0.72 × 50 = 36 kg
In simple words: The iron beam weighs 36 kilogrammes.

Exam Tip: Always multiply volume by density (in matching units) to find weight - check unit consistency first.

Question 11. A rectangular tank with dimensions 6 × 3.5 m holds 42,000 L. Find its height.
Answer: (a) 2 m
Volume = 42,000 L = 42 m³
Height = \( \frac{\text{volume}}{l \times b} = \frac{42}{6 \times 3.5} = \frac{42}{21} = 2 \text{ m} \)
In simple words: Water fills the tank to a height of 2 metres.

Exam Tip: Convert litres to cubic metres first (1000 L = 1 m³) before calculating height.

Question 12. A room measuring 10 × 8 × 3.3 m requires 3 m³ of air per person. How many people can occupy it?
Answer: (b) 88
Volume of room = 10 × 8 × 3.3 = 264 m³
Total number of people = \( \frac{264}{3} = 88 \)
In simple words: The room can safely accommodate 88 people.

Exam Tip: Always divide total volume by the required space per person to find capacity.

Question 13. A tank has dimensions 3 × 2 × 5 m. Find its capacity in litres.
Answer: (a) 30000
Volume = 3 × 2 × 5 = 30 m³ = 30,000 L
In simple words: The tank can hold 30,000 litres of liquid.

Exam Tip: Remember the conversion: 1 m³ = 1000 L - use it when capacity questions ask for litres.

Question 14. A cuboid has dimensions 25 cm × 15 cm × 8 cm. Find its total surface area.
Answer: (b) 1390 cm²
Surface area = \( 2(25 \times 15 + 15 \times 8 + 25 \times 8) = 2(375 + 120 + 200) = 2(695) = 1390 \text{ cm}^2 \)
In simple words: All six faces combined measure 1,390 square centimetres.

Exam Tip: Calculate each pair of opposite faces separately, then add and multiply by 2 to avoid errors.

Question 15. A cube has a diagonal measuring 4√3 cm. Find its volume.
Answer: (d) 64 cm³
Diagonal of cube = \( a\sqrt{3} = 4\sqrt{3} \)
\( a = 4 \) cm
Volume = \( a^3 = 4^3 = 64 \text{ cm}^3 \)
In simple words: The cube occupies 64 cubic centimetres of space.

Exam Tip: The diagonal of a cube equals edge length times √3 - use this to quickly find the edge from the diagonal.

Question 16. A cube has a diagonal equal to 9√3 cm. What is its total surface area?
Answer: (b) 486 sq cm
From diagonal: \( a\sqrt{3} = 9\sqrt{3} \), so \( a = 9 \) cm
Total surface area = \( 6a^2 = 6 \times 81 = 486 \text{ cm}^2 \)
In simple words: The cube's outer surface covers 486 square centimetres.

Exam Tip: Once the edge is known from the diagonal, all other properties follow - surface area, volume, and face diagonal.

Question 17. If each side of a cube is doubled, how does the volume change?
Answer: (d) If each side of the cube is doubled, its volume becomes 8 times the original volume.
Let the original side be a units. Original volume = a³ cubic units. New side = 2a units. New volume = (2a)³ sq units = 8a³ cubic units. Thus, the volume becomes 8 times the original volume.
In simple words: Doubling the edge makes the cube 8 times bigger inside.

Exam Tip: Volume changes with the cube of the scaling factor - doubling gives 2³ = 8 times larger.

Question 18. If each side of a cube is doubled, how does the surface area change?
Answer: (b) becomes 4 times
Let the side of the cube be a units. Surface area = 6a² sq units. Now, new side = 2a units. New surface area = 6(2a)² sq units = 24a² sq units. Thus, the surface area becomes 4 times the original area.
In simple words: Doubling the edges makes the surface 4 times larger.

Exam Tip: Surface area changes with the square of the scaling factor - doubling gives 2² = 4 times larger.

Question 19. Three cubes with edges 6 cm, 8 cm, and 10 cm are melted and formed into a single new cube. Find the edge of the new cube.
Answer: (a) 12 cm
Total volume = 6³ + 8³ + 10³ = 216 + 512 + 1000 = 1728 cm³
Edge of new cube = \( \sqrt[3]{1728} = 12 \) cm
In simple words: The new cube has edges measuring 12 centimetres.

Exam Tip: Add individual volumes when solids are combined, then take the cube root to find the new edge.

Question 20. A cuboid with dimensions 25 cm × 5 cm × 5 cm is formed. Find its volume.
Answer: (d) 625 cm³
Volume of cuboid = 25 × 5 × 5 = 625 cm³
In simple words: The cuboid holds 625 cubic centimetres.

Exam Tip: Multiply all three dimensions directly - no special formula needed for cuboid volume.

Question 21. A cylinder has diameter 2 m and height 14 m. Find its volume.
Answer: (d) 44 m³
Radius = 1 m, Height = 14 m
Volume = \( \pi r^2 h = \frac{22}{7} \times 1 \times 1 \times 14 = 44 \text{ m}^3 \)
In simple words: The cylinder can hold 44 cubic metres.

Exam Tip: Always convert diameter to radius first - radius is half the diameter.

Question 23. A cylinder has diameter 14 m and volume 1848 m³. Find its height.
Answer: (b) 12 m
Radius = 7 m, Volume = 1848 m³
Height = \( \frac{\text{volume}}{\pi r^2} = \frac{1848}{22/7 \times 7 \times 7} = \frac{1848}{154} = 12 \) m
In simple words: The cylinder is 12 metres tall.

Exam Tip: Rearrange volume formula to \( h = \frac{V}{\pi r^2} \) when finding height from volume.

Question 24. A cylinder has volume π × 3 × 3 × 8 cm³ and contains coins. Find the total number of coins if each coin's volume is π × 0.75 × 0.75 × 0.2 cm³.
Answer: (d) 640
Total number of coins = \( \frac{\pi \times 3 \times 3 \times 8}{\pi \times 0.75 \times 0.75 \times 0.2} = \frac{72}{0.1125} = 640 \)
In simple words: The cylinder can hold 640 coins.

Exam Tip: Divide the container's volume by each item's volume to find how many fit inside.

Question 25. A cylindrical wire has volume 66 cm³. Find its length if the cross-sectional area is π × 0.05 × 0.05 cm².
Answer: (b) 84 m
Length = \( \frac{\text{volume}}{\pi r^2} = \frac{66 \times 7}{22 \times 0.05 \times 0.05} = \frac{66 \times 7}{22 \times 0.0025} = 8400 \text{ cm} = 84 \text{ m} \)
In simple words: The wire stretches 84 metres in length.

Exam Tip: For wires and pipes, divide volume by cross-sectional area to find length.

Question 26. A cylinder has base radius 5 cm and height 14 cm. Find its volume.
Answer: (a) 1100 cm³
Volume = \( \pi r^2 h = \frac{22}{7} \times 5 \times 5 \times 14 = 1100 \text{ cm}^3 \)
In simple words: The cylinder holds 1,100 cubic centimetres.

Exam Tip: Straightforward application of the volume formula - ensure radius (not diameter) is used.

Question 27. A cylinder has diameter 7 cm and height 80 cm. Find its total surface area.
Answer: (a) 1837 cm²
Radius = 3.5 cm, Height = 80 cm
Total surface area = \( 2\pi r(r + h) = 2 \times \frac{22}{7} \times 3.5 \times (3.5 + 80) = 22 \times (83.5) = 1837 \text{ cm}^2 \)
In simple words: All surfaces together measure 1,837 square centimetres.

Exam Tip: Use the formula \( 2\pi r(r + h) \) for total surface area - it's faster than calculating curved and circular areas separately.

Question 28. A cylinder has curved surface area 264 cm² and height 14 cm. Find its volume.
Answer: (b) 396 cm³
From curved surface area: \( 2\pi rh = 264 \), so \( r = \frac{264}{2\pi h} = 3 \) cm
Volume = \( \pi r^2 h = \frac{22}{7} \times 3 \times 3 \times 14 = 396 \text{ cm}^3 \)
In simple words: The cylinder holds 396 cubic centimetres.

Exam Tip: Use curved surface area to find radius first, then calculate volume with that radius.

Question 30. A cylinder has diameter 14 cm and height 5 cm. Find its volume.
Answer: (a) 770 cm³
Radius = 7 cm
Volume = \( \pi r^2 h = \frac{22}{7} \times 7 \times 7 \times 5 = 770 \text{ cm}^3 \)
In simple words: The cylinder contains 770 cubic centimetres of space.

Exam Tip: Always simplify fractions early - here, the 7s cancel with the denominator in π/7.

Question 30 (Ratio Problem). Two cylinders have radius ratio 2:3 and height ratio 5:3. Find the ratio of their volumes.
Answer: (c) 20:27
Given: \( \frac{r_1}{r_2} = \frac{2}{3} \) and \( \frac{h_1}{h_2} = \frac{5}{3} \)
\( \frac{V_1}{V_2} = \frac{\pi r_1^2 h_1}{\pi r_2^2 h_2} = \left(\frac{r_1}{r_2}\right)^2 \times \frac{h_1}{h_2} = \left(\frac{2}{3}\right)^2 \times \frac{5}{3} = \frac{4}{9} \times \frac{5}{3} = \frac{20}{27} \)
In simple words: The first cylinder's volume is 20/27 of the second cylinder's volume.

Exam Tip: When finding ratios of volumes, square the radius ratio before multiplying by the height ratio.