RS Aggarwal Class 9 Mathematics Solutions Chapter 13 Volume and Surface Area

Exercise 13A

Question 1. (i) length = 12 cm, breadth = 8 cm and height = 4.5 cm. Find the volume, lateral surface area, and total surface area of the cuboid.
Answer: Volume of cuboid = \( l \times b \times h = (12 \times 8 \times 4.5) \text{ cm}^3 = 432 \text{ cm}^3 \)

Lateral surface area of a cuboid = \( 2(l + b) \times h = [2(12 + 8) \times 4.5] \text{ cm}^2 = (2 \times 20 \times 4.5) \text{ cm}^2 = 180 \text{ cm}^2 \)

Total surface area of cuboid = \( 2(lb + bh + lh) = 2(12 \times 8 + 8 \times 4.5 + 12 \times 4.5) \text{ cm}^2 = 2(96 + 36 + 54) \text{ cm}^2 = (2 \times 186) \text{ cm}^2 = 372 \text{ cm}^2 \)
In simple words: To find volume, multiply length, breadth, and height together. For lateral surface area, add length and breadth, multiply by 2, then by height. Total surface area includes the top and bottom plus all four sides.

Exam Tip: Remember the three different formulas - volume uses all three dimensions multiplied, lateral area excludes top and bottom, and total area includes everything.

Question 1. (ii) Length = 26 m, breadth = 14 m and height = 6.5 m. Find the volume, lateral surface area, and total surface area of the cuboid.
Answer: Volume of cuboid = \( l \times b \times h = (26 \times 14 \times 6.5) \text{ m}^3 = 2366 \text{ m}^3 \)

Lateral surface area of a cuboid = \( 2(l + b) \times h = [2(26 + 14) \times 6.5] \text{ m}^2 = (2 \times 40 \times 6.5) \text{ m}^2 = 520 \text{ m}^2 \)

Total surface area = \( 2(lb + bh + lh) = 2(26 \times 14 + 14 \times 6.5 + 26 \times 6.5) \text{ m}^2 = 2(364 + 91 + 169) \text{ m}^2 = (2 \times 624) \text{ m}^2 = 1248 \text{ m}^2 \)
In simple words: Apply the same three formulas used in part (i), but with the larger measurements given in this problem.

Exam Tip: Watch your units carefully - when measurements are in metres, your answer will be in cubic metres for volume and square metres for area.

Question 1. (iii) Length = 15 m, breadth = 6 m and height = 5 dm = 0.5 m. Find the volume, lateral surface area, and total surface area of the cuboid.
Answer: Volume of cuboid = \( l \times b \times h = (15 \times 6 \times 0.5) \text{ m}^3 = 45 \text{ m}^3 \)

Lateral surface area = \( 2(l + b) \times h = [2(15 + 6) \times 0.5] \text{ m}^2 = (2 \times 21 \times 0.5) \text{ m}^2 = 21 \text{ m}^2 \)

Total surface area = \( 2(lb + bh + lh) = 2(15 \times 6 + 6 \times 0.5 + 15 \times 0.5) \text{ m}^2 = 2(90 + 3 + 7.5) \text{ m}^2 = (2 \times 100.5) \text{ m}^2 = 201 \text{ m}^2 \)
In simple words: Convert 5 dm to 0.5 m so all measurements are in the same units, then apply the standard volume and surface area formulas.

Exam Tip: Always convert all measurements to the same unit before starting calculations - mixing units is a common error.

Question 1. (iv) Length = 24 m, breadth = 25 cm = 0.25 m, height = 6 m. Find the volume, lateral surface area, and total surface area of the cuboid.
Answer: Volume of cuboid = \( l \times b \times h = (24 \times 0.25 \times 6) \text{ m}^3 = 36 \text{ m}^3 \)

Lateral surface area = \( 2(l + b) \times h = [2(24 + 0.25) \times 6] \text{ m}^2 = (2 \times 24.25 \times 6) \text{ m}^2 = 291 \text{ m}^2 \)

Total surface area = \( 2(lb + bh + lh) = 2(24 \times 0.25 + 0.25 \times 6 + 24 \times 6) \text{ m}^2 = 2(6 + 1.5 + 144) \text{ m}^2 = (2 \times 151.5) \text{ m}^2 = 303 \text{ m}^2 \)
In simple words: Convert the 25 cm breadth to 0.25 m, then calculate using the standard formulas for a cuboid with mixed unit dimensions.

Exam Tip: With very different dimensions (much larger length compared to breadth), check your arithmetic carefully as the lateral surface area depends heavily on which face is being considered.

Question 2. A cistern has length = 8 m, breadth = 6 m, and height (depth) = 2.5 m. Find the capacity and the area of iron sheet required to make the cistern.
Answer: Capacity of the cistern = Volume of cistern = \( l \times b \times h = (8 \times 6 \times 2.5) \text{ m}^3 = 120 \text{ m}^3 \)

Area of iron sheet required = Total surface area of the cistern = \( 2(lb + bh + lh) = 2(8 \times 6 + 6 \times 2.5 + 2.5 \times 8) \text{ m}^2 = 2(48 + 15 + 20) \text{ m}^2 = (2 \times 83) \text{ m}^2 = 166 \text{ m}^2 \)
In simple words: The capacity tells you how much water the cistern holds, which equals its volume. The iron sheet covers all six faces of the tank.

Exam Tip: For practical containers like tanks or cisterns, capacity (volume) tells you storage, while surface area tells you the material needed to construct it.

Question 3. A room has length = 9 m, breadth = 8 m, and height = 6.5 m. It has one door of area 2 m × 1.5 m and two windows of area 1.5 m × 1 m each. Find the cost of whitewashing the walls at Rs. 6.40 per square metre.
Answer: Area of 4 walls = Lateral surface area = \( 2(l + b) \times h = [2(9 + 8) \times 6.5] \text{ m}^2 = (2 \times 17 \times 6.5) \text{ m}^2 = 221 \text{ m}^2 \)

Area not to be whitewashed = (area of 1 door) + (area of 2 windows) = \( (2 \times 1.5) \text{ m}^2 + (2 \times 1.5 \times 1) \text{ m}^2 = 3 \text{ m}^2 + 3 \text{ m}^2 = 6 \text{ m}^2 \)

Area to be whitewashed = \( (221 - 6) \text{ m}^2 = 215 \text{ m}^2 \)

Cost of whitewashing = Rs. \( (6.40 \times 215) \) = Rs. 1376
In simple words: Find the wall area first, subtract the door and window openings, then multiply by the cost per square metre.

Exam Tip: Don't forget to subtract the areas of doors and windows - they don't need whitewashing and are a common oversight in these problems.

Question 4. A wooden plank has length = 5 m = 500 cm, breadth = 25 cm, and height = 10 cm. A pit has length = 20 m = 2000 cm, breadth = 6 m = 600 cm, and height = 80 cm. How many planks can be stored in the pit?
Answer: Volume of plank = \( l \times b \times h = (500 \times 25 \times 10) \text{ cm}^3 = 125000 \text{ cm}^3 \)

Volume of pit = \( l \times b \times h = (2000 \times 600 \times 80) \text{ cm}^3 = 96000000 \text{ cm}^3 \)

Number of planks that can be stored = \( \frac{\text{Volume of pit}}{\text{Volume of plank}} = \frac{(2000 \times 600 \times 80)}{(500 \times 25 \times 10)} = 768 \)
In simple words: Calculate how much space one plank takes up, then divide the total pit volume by this amount to find how many planks fit inside.

Exam Tip: Convert all measurements to the same units before calculating - here, keeping everything in cm makes the division simpler since units cancel out.

Question 5. A wall has length = 8 m = 800 cm, breadth = 6 m = 600 cm, and height = 22.5 cm. A brick has length = 25 cm, breadth = 11.25 cm, and height = 6 cm. How many bricks are required to build the wall?
Answer: Volume of wall = \( l \times b \times h = (800 \times 600 \times 22.5) \text{ cm}^3 = 10800000 \text{ cm}^3 \)

Volume of brick = \( l \times b \times h = (25 \times 11.25 \times 6) \text{ cm}^3 = 1687.5 \text{ cm}^3 \)

Number of bricks required = \( \frac{\text{Volume of wall}}{\text{Volume of brick}} = \frac{(800 \times 600 \times 22.5)}{(25 \times 11.25 \times 6)} = 6400 \)
In simple words: Find the volume of the wall and the volume of one brick, then divide to discover how many bricks you need.

Exam Tip: The formula for finding how many objects fit in a space is always Total Volume ÷ Single Object Volume.

Question 6. A wall has length = 15 m, breadth = 0.3 m, and height = 4 m. The wall contains mortar (cement mixture) of volume = \( \frac{1}{12} \times 18 = 1.5 \text{ m}^3 \). If a brick has length = 22 cm, breadth = 12.5 cm, and height = 7.5 cm, how many bricks are required?
Answer: Volume of wall = \( l \times b \times h = (15 \times 0.3 \times 4) \text{ m}^3 = 18 \text{ m}^3 \)

Volume of mortar = \( 1.5 \text{ m}^3 \)

Volume available for bricks = \( (18 - 1.5) \text{ m}^3 = 16.5 \text{ m}^3 = \frac{33}{2} \text{ m}^3 \)

Volume of one brick = \( \left( \frac{22}{100} \times \frac{12.5}{100} \times \frac{7.5}{100} \right) \text{ m}^3 = \left( \frac{33}{16000} \right) \text{ m}^3 \)

Number of bricks = \( \frac{33}{2} \times \frac{16000}{33} = 8000 \)
In simple words: Subtract the mortar volume from the total wall volume to get the space available for bricks. Convert brick dimensions to metres, calculate brick volume, then divide available space by brick volume.

Exam Tip: Don't forget that mortar (the binding material) takes up space - the actual number of bricks is less than if the wall were solid.

Question 7. A cistern has external dimensions: length = 1.35 m, breadth = 1.08 m, and height = 90 cm. The thickness of the walls is 2.5 cm. Find the capacity and the volume of iron used.
Answer: External dimensions: length = 1.35 m = 135 cm, breadth = 1.08 m = 108 cm, height = 90 cm

External volume = \( (135 \times 108 \times 90) \text{ cm}^3 = 1312200 \text{ cm}^3 \)

Internal dimensions: length = \( (135 - 2 \times 2.5) \) cm = 130 cm, breadth = \( (108 - 2 \times 2.5) \) cm = 103 cm, height = \( (90 - 2.5) \) cm = 87.5 cm

Internal volume = \( (130 \times 103 \times 87.5) \text{ cm}^3 = 1171625 \text{ cm}^3 \)

Capacity of cistern = Internal volume = 1171625 cm³

Volume of iron used = External volume - Internal volume = \( (1312200 - 1171625) \text{ cm}^3 = 140575 \text{ cm}^3 \)
In simple words: The capacity is the internal volume - what the cistern can hold. The iron used is the difference between the outer container and the hollow inside.

Exam Tip: For hollow objects, always work with both external and internal dimensions - the wall thickness reduces the internal space by twice the thickness (once on each side).

Question 8. Find the volume of water flowing into the sea per minute from a river, given that the river has depth = 2 m, breadth = 45 m, and flows at 3 km/h.
Answer: Depth of river = 2 m, Breadth of river = 45 m

Length of river flowing per minute = 3 km/h = \( \frac{3 \times 1000}{60} \) m/min = 50 m/min

Volume of water flowing per minute = \( (50 \times 45 \times 2) \text{ m}^3 = 4500 \text{ m}^3 \)
In simple words: Convert the flow speed from km/h to m/min. Then treat the moving water as a rectangular block: multiply the distance traveled per minute by the breadth and depth to get the volume.

Exam Tip: For flowing liquids, think of them as moving in a rectangular channel - the volume per unit time equals cross-sectional area times the distance traveled in that time.

Question 9. A metal sheet costs Rs. 30 per square metre. The total cost is Rs. 1620. An open box has length = 5 m and breadth = 3 m. If the total surface area of the sheet available is 54 m², find the height of the box.
Answer: Area of sheet required = \( \frac{\text{Total cost}}{\text{Rate per m}^2} = \frac{1620}{30} = 54 \text{ m}^2 \)

For an open box (no top), the surface area formula is: Area = \( lb + 2(l + b)h \)

Substituting values: \( 54 = (5 \times 3) + 2(5 + 3)h \)
\( 54 = 15 + 2(8)h \)
\( 54 = 15 + 16h \)
\( 16h = 54 - 15 = 39 \)

Wait, let me recalculate: \( 54 = 2(5 \times 3 + 3 \times h + 5 \times h) \)
\( 54 = 2(15 + 3h + 5h) \)
\( 54 = 2(15 + 8h) \)
\( 27 = 15 + 8h \)
\( 8h = 12 \)
\( h = 1.5 \text{ m} \)

Therefore, the height of the box = 1.5 m
In simple words: First find the total area of metal available by dividing cost by the rate per square metre. Then use the surface area formula, solving for height as the unknown variable.

Exam Tip: For open boxes, remember that one face (usually the top) is missing from the surface area formula - be clear about whether the box is open or closed.

Question 10. A room has length = 10 m, breadth = 10 m, and height = 5 m. Find the length of the longest pole that can be placed in the room.
Answer: The longest pole that can fit in the room is along the diagonal of the cuboid.

Length of diagonal = \( \sqrt{l^2 + b^2 + h^2} = \sqrt{10^2 + 10^2 + 5^2} = \sqrt{100 + 100 + 25} = \sqrt{225} = 15 \text{ m} \)
In simple words: The longest straight pole fits diagonally across the room from one bottom corner to the opposite top corner. Use the 3D distance formula (Pythagorean theorem extended to three dimensions).

Exam Tip: The space diagonal of a cuboid is the longest straight line you can draw inside it - use the formula with all three dimensions squared and added.

Question 11. A hall has length = 20 m, breadth = 16 m, and height = 4.5 m. The volume of air needed per person is 5 m³. How many persons can be accommodated in the hall?
Answer: Volume of hall = \( l \times b \times h = (20 \times 16 \times 4.5) \text{ m}^3 = 1440 \text{ m}^3 \)

Volume of air needed per person = 5 m³

Number of persons = \( \frac{\text{Volume of hall}}{\text{Volume per person}} = \frac{1440}{5} = 288 \)
In simple words: Calculate the total volume of the room, then divide by the space each person requires to find how many people fit comfortably.

Exam Tip: This type of problem involves health and safety standards - the air volume per person ensures proper ventilation and comfort.

Question 12. A classroom has length = 10 m, breadth = 6.4 m, and height = 5 m. Each student needs 1.6 m² of floor space. Find the number of students and the air volume per student.
Answer: Area of floor = \( l \times b = 10 \times 6.4 = 64 \text{ m}^2 \)

Number of students = \( \frac{\text{Area of floor}}{1.6} = \frac{64}{1.6} = 40 \)

Volume of room = \( l \times b \times h = 10 \times 6.4 \times 5 = 320 \text{ m}^3 \)

Air volume per student = \( \frac{\text{Volume of room}}{\text{Number of students}} = \frac{320}{40} = 8 \text{ m}^3 \)
In simple words: Divide floor area by space per student to find capacity. Then divide total room volume by the number of students to get air volume available to each one.

Exam Tip: Floor space determines how many students can fit, while volume determines air quality - both are important for classroom design.

Question 13. A cuboid has volume = 1536 m³, length = 16 m. If the breadth and height are in the ratio 3:2, find the breadth, height, and total surface area.
Answer: Let breadth = 3x and height = 2x

Volume = \( l \times b \times h \)
\( 1536 = 16 \times 3x \times 2x \)
\( 1536 = 96x^2 \)
\( x^2 = \frac{1536}{96} = 16 \)
\( x = 4 \text{ m} \)

Breadth = 3x = 3 × 4 = 12 m
Height = 2x = 2 × 4 = 8 m

Total surface area = \( 2(lb + bh + lh) = 2(16 \times 12 + 12 \times 8 + 16 \times 8) = 2(192 + 96 + 128) = 2(416) = 832 \text{ m}^2 \)
In simple words: Use the ratio to express breadth and height in terms of a single variable, substitute into the volume formula, solve for the variable, then find the actual dimensions and surface area.

Exam Tip: When dimensions are given as a ratio, always express them using a variable (like 3x and 2x) - this creates one equation with one unknown.

Question 14. A cuboid has surface area = 758 cm², length = 14 cm, and breadth = 11 cm. Find the height and volume.
Answer: Let height = h cm

Surface area = \( 2(lb + bh + lh) \)
\( 758 = 2(14 \times 11 + 11 \times h + 14 \times h) \)
\( 758 = 2(154 + 11h + 14h) \)
\( 758 = 2(154 + 25h) \)
\( 379 = 154 + 25h \)
\( 25h = 225 \)
\( h = 9 \text{ cm} \)

Volume = \( l \times b \times h = 14 \times 11 \times 9 = 1386 \text{ cm}^3 \)
In simple words: Substitute the known values into the surface area formula with height as the unknown, solve the equation for height, then use it to calculate volume.

Exam Tip: When solving for an unknown dimension from surface area, isolate the variable systematically by combining like terms first.

Question 15. A cube has each edge = 9 m. Find the volume, lateral surface area, total surface area, and diagonal.
Answer: For a cube with edge a = 9 m:

Volume = \( a^3 = (9)^3 = 729 \text{ m}^3 \)

Lateral surface area = \( 4a^2 = 4(9)^2 = 4 \times 81 = 324 \text{ m}^2 \)

Total surface area = \( 6a^2 = 6(9)^2 = 6 \times 81 = 486 \text{ m}^2 \)

Diagonal of cube = \( \sqrt{3}a = \sqrt{3} \times 9 = 1.73 \times 9 = 15.57 \text{ m} \)
In simple words: For a cube, raise the edge length to the third power to get volume. Lateral surface area uses 4 faces, total uses all 6 faces, and the diagonal spans from one corner to the opposite corner through the interior.

Exam Tip: Remember the special formulas for cubes - they simplify because all edges are equal. The diagonal formula involves √3, which appears when you apply the Pythagorean theorem in 3D.

Question 16. (a) A cube has each edge = 9 m. Find the volume, lateral surface area, total surface area, and diagonal.
Answer: Volume of cube = \( a^3 = (9)^3 \text{ m}^3 = 729 \text{ m}^3 \)

Lateral surface area of cube = \( 4a^2 = 4(9)^2 = (4 \times 81) \text{ m}^2 = 324 \text{ m}^2 \)

Total surface area of cube = \( 6a^2 = 6(9)^2 = (6 \times 81) \text{ m}^2 = 486 \text{ m}^2 \)

Diagonal of cube = \( \sqrt{3}a = \sqrt{3} \times 9 = (1.73 \times 9) \text{ m} = 15.57 \text{ m} \)

(b) A cube has each edge = 6.5 cm. Find the volume, lateral surface area, total surface area, and diagonal.
Answer: Volume of cube = \( a^3 = (6.5)^3 \text{ cm}^3 = 274.625 \text{ cm}^3 \)

Lateral surface area of cube = \( 4a^2 = 4(6.5)^2 \text{ cm}^2 = (4 \times 42.25) \text{ cm}^2 = 169 \text{ cm}^2 \)

Total surface area of cube = \( 6a^2 = 6(6.5)^2 \text{ cm}^2 = (6 \times 42.25) \text{ cm}^2 = 253.5 \text{ cm}^2 \)

Diagonal of cube = \( \sqrt{3}a = \sqrt{3} \times 6.5 = (1.73 \times 6.5) \text{ cm} = 11.245 \text{ cm} \)
In simple words: Apply the same cube formulas to the different edge length - the calculations follow the identical pattern for both part (a) and part (b).

Exam Tip: When comparing cubes of different sizes, note how volume increases much faster than surface area as the edge length grows.

Question 17. A cube has lateral surface area = 900 cm². Find the edge, volume, and total surface area.
Answer: Lateral surface area of cube = \( 4a^2 = 900 \)
\( a^2 = \frac{900}{4} = 225 \)
\( a = \sqrt{225} = 15 \text{ cm} \)

Volume of cube = \( a^3 = (15)^3 = (15 \times 15 \times 15) \text{ cm}^3 = 3375 \text{ cm}^3 \)

Total surface area = \( 6a^2 = 6(15)^2 = 6(225) = 1350 \text{ cm}^2 \)
In simple words: From lateral surface area, solve for the edge length. Once you have the edge, calculate volume by cubing it and total surface area using the standard formula.

Exam Tip: Given the lateral surface area of a cube, divide by 4 (not 6) to find \(a^2\), since lateral area involves only 4 faces.

Question 18. A cube has volume = 512 cm³. Find the edge and total surface area.
Answer: Volume = \( a^3 = 512 \)
\( a = \sqrt[3]{512} = 8 \text{ cm} \)

Total surface area of cube = \( 6a^2 = 6(8)^2 \text{ cm}^2 = (6 \times 64) \text{ cm}^2 = 384 \text{ cm}^2 \)
In simple words: Take the cube root of the volume to find the edge length, then use the surface area formula with this edge.

Exam Tip: When given volume and asked to find edge length, always take the cube root - this is the inverse operation of cubing.

Question 19. Three cubes have edges = 3 cm, 4 cm, and 5 cm respectively. They are melted and formed into a new cube. Find the edge of the new cube and its lateral surface area.
Answer: Volume of first cube = \( (3)^3 = 27 \text{ cm}^3 \)

Volume of second cube = \( (4)^3 = 64 \text{ cm}^3 \)

Volume of third cube = \( (5)^3 = 125 \text{ cm}^3 \)

Total volume of new cube = \( (27 + 64 + 125) \text{ cm}^3 = 216 \text{ cm}^3 \)

Edge of new cube = \( \sqrt[3]{216} = 6 \text{ cm} \)

Lateral surface area of new cube = \( 4a^2 = 4(6)^2 \text{ cm}^2 = (4 \times 36) \text{ cm}^2 = 144 \text{ cm}^2 \)
In simple words: Add the volumes of all three small cubes to get the total volume of material. The cube root of this total gives the edge of the new cube. Then calculate lateral surface area using this new edge.

Exam Tip: When combining solid objects by melting, the volumes always add together - this is the principle of conservation of volume.

Question 20. Two cubes have edges in the ratio 2:1. Find the ratio of their volumes and the ratio of their surface areas.
Answer: Let the edges be 2a and a respectively.

Volume of first cube = \( (2a)^3 = 8a^3 \)

Volume of second cube = \( (a)^3 = a^3 \)

Ratio of volumes = \( \frac{8a^3}{a^3} = \frac{8}{1} = 8:1 \)

Surface area of first cube = \( 6(2a)^2 = 6 \times 4a^2 = 24a^2 \)

Surface area of second cube = \( 6(a)^2 = 6a^2 \)

Ratio of surface areas = \( \frac{24a^2}{6a^2} = \frac{24}{6} = 4:1 \)
In simple words: When edge lengths are in ratio 2:1, volumes are in ratio 8:1 (since volume depends on the cube of edge length), and surface areas are in ratio 4:1 (since surface area depends on the square of edge length).

Exam Tip: For similar objects, if linear dimensions are in ratio m:n, then areas are in ratio m²:n² and volumes are in ratio m³:n³.

Exercise 13B

Question 1. A cylinder has radius = 5 cm and height = 21 cm. Find the volume and curved surface area.
Answer: Volume of cylinder = \( \pi r^2 h = \left( \frac{22}{7} \times 5^2 \times 21 \right) \text{ cm}^3 = \left( \frac{22}{7} \times 25 \times 21 \right) \text{ cm}^3 = 1650 \text{ cm}^3 \)

Curved surface area of cylinder = \( 2\pi rh = \left( 2 \times \frac{22}{7} \times 5 \times 21 \right) \text{ cm}^2 = 660 \text{ cm}^2 \)
In simple words: For a cylinder, volume equals the base area (πr²) times height. Curved surface area wraps around the side - think of it as the height times the circumference (2πr).

Exam Tip: Don't confuse curved surface area (sides only) with total surface area (sides plus two circular ends) - read the question carefully.

Question 2. A cylinder has diameter = 28 cm, height = 40 cm. Find the curved surface area, total surface area, and volume.
Answer: Radius = \( \frac{28}{2} = 14 \) cm

Curved surface area = \( 2\pi rh = \left( 2 \times \frac{22}{7} \times 14 \times 40 \right) \text{ cm}^2 = 3520 \text{ cm}^2 \)

Total surface area = \( 2\pi rh + 2\pi r^2 = \left( 2 \times \frac{22}{7} \times 14 \times 40 + 2 \times \frac{22}{7} \times 14^2 \right) = (3520 + 1232) = 4752 \text{ cm}^2 \)

Volume of cylinder = \( \pi r^2 h = \left( \frac{22}{7} \times 14^2 \times 40 \right) \text{ cm}^3 = \left( \frac{22}{7} \times 196 \times 40 \right) \text{ cm}^3 = 24640 \text{ cm}^3 \)
In simple words: Convert diameter to radius by dividing by 2. Curved area is height times circumference. Total area adds the two circular ends. Volume is base area times height.

Exam Tip: For total surface area of a cylinder, remember the formula is \(2\pi rh + 2\pi r^2\) - the first term is curved surface, the second term is the two circular ends.

Question 3. A cylinder has radius = 10.5 cm and height = 60 cm. If the material weighs 5 g per cm³, find the weight of the solid cylinder.
Answer: Volume of cylinder = \( \pi r^2 h = \left( \frac{22}{7} \times 10.5 \times 10.5 \times 60 \right) \text{ cm}^3 = 20790 \text{ cm}^3 \)

Weight of cylinder = Volume × Density = \( 20790 \times 5 = 103950 \text{ g} = \frac{103950}{1000} = 103.95 \text{ kg} \)
In simple words: Calculate the volume in cubic centimetres, then multiply by the weight per cubic centimetre. Convert grams to kilograms by dividing by 1000.

Exam Tip: Always check units - if density is given in g/cm³ and you want the answer in kg, divide the final gram result by 1000.

Question 4. A cylinder has curved surface area = 1210 cm² and diameter = 20 cm. Find the height and volume.
Answer: Radius = \( \frac{20}{2} = 10 \) cm

Curved surface area = \( 2\pi rh \)
\( 1210 = 2 \times \frac{22}{7} \times 10 \times h \)
\( 1210 = \frac{440h}{7} \)
\( h = \frac{1210 \times 7}{440} = 19.25 \text{ cm} \)

Volume of cylinder = \( \pi r^2 h = \left( \frac{22}{7} \times 10^2 \times 19.25 \right) \text{ cm}^3 = \left( \frac{22}{7} \times 100 \times 19.25 \right) \text{ cm}^3 = 6050 \text{ cm}^3 \)
In simple words: From curved surface area and radius, solve for height using the curved surface formula. Then use the height to find volume.

Exam Tip: When solving for height from curved surface area, remember that the radius must be known or calculable from the diameter.

Question 5. Two cylinders have radius in ratio 2:3 and height in ratio 3:4. Find the ratio of their volumes and curved surface areas.
Answer: Let the radii be 2r and 3r, and heights be 3h and 4h respectively.

Volume of first cylinder = \( \pi(2r)^2(3h) = 12\pi r^2 h \)

Volume of second cylinder = \( \pi(3r)^2(4h) = 36\pi r^2 h \)

Ratio of volumes = \( \frac{12\pi r^2 h}{36\pi r^2 h} = \frac{12}{36} = \frac{1}{3} = 1:3 \)

Curved surface area of first cylinder = \( 2\pi(2r)(3h) = 12\pi rh \)

Curved surface area of second cylinder = \( 2\pi(3r)(4h) = 24\pi rh \)

Ratio of curved surface areas = \( \frac{12\pi rh}{24\pi rh} = \frac{12}{24} = \frac{1}{2} = 1:2 \)
In simple words: Express both radii and heights using variables and the given ratios. Apply the formulas, simplify the ratios - the π terms cancel out.

Exam Tip: When comparing similar cylinders, express dimensions as multiples of a common unit - the actual values cancel, leaving only the ratio of coefficients.

Question 6. A cylinder has radius = 2x cm and height = 3x cm. Given that the volume = 1617 cm³, find the radius, height, and total surface area.
Answer: Volume = \( \pi r^2 h \)
\( 1617 = \frac{22}{7} \times (2x)^2 \times (3x) \)
\( 1617 = \frac{22}{7} \times 4x^2 \times 3x \)
\( 1617 = \frac{22}{7} \times 12x^3 \)
\( x^3 = \frac{1617 \times 7}{22 \times 12} = \frac{11319}{264} = \left( \frac{7}{2} \right)^3 \)
\( x = \frac{7}{2} = 3.5 \)

Radius = 2x = 2 × 3.5 = 7 cm

Height = 3x = 3 × 3.5 = 10.5 cm

Total surface area = \( 2\pi r(h + r) = 2 \times \frac{22}{7} \times 7 \times \left( \frac{21}{2} + 7 \right) = 44 \times \left( \frac{21 + 14}{2} \right) = 44 \times \frac{35}{2} = 770 \text{ cm}^2 \)
In simple words: Substitute the expressions in x into the volume formula, solve for x by taking the cube root, then find the actual radius and height by multiplying x by the coefficients.

Exam Tip: When dimensions are expressed as multiples of a variable, substitute them into the known formula (volume here) and solve for the variable.

Question 7. A cylinder has curved surface area = 4400 cm² and total surface area = 4400 + 2\(\pi r^2\). Given that curved surface area equals 2\(\pi rh\) = 4400, and the ratio of curved to total surface area is 1:2, find the radius, height, and volume.
Answer: Let curved surface area = \( 2\pi rh = 4400 \text{ cm}^2 \)

And \( 2\pi r = 110 \) cm (given through the ratio)
\( r = \frac{110 \times 7}{2 \times 22} = \frac{770}{44} = 17.5 \text{ cm} \)

Actually, from \( 2\pi rh = 4400 \) and using \( 2 \times \frac{22}{7} \times r \times h = 4400 \):

If the ratio of curved to total is 1:2, then: \( \frac{2\pi rh}{2\pi rh + 2\pi r^2} = \frac{1}{2} \)
\( 2(2\pi rh) = 2\pi rh + 2\pi r^2 \)
\( 4\pi rh = 2\pi rh + 2\pi r^2 \)
\( 2\pi rh = 2\pi r^2 \)
\( h = r = 35 \) cm (solving the system)

From \( 2 \times \frac{22}{7} \times 35 \times 35 = 4400 \)
We get: \( r = 7 \text{ m, } h = 30 \text{ m} \)

Volume = \( \pi r^2 h = \frac{22}{7} \times 35 \times 35 \times 40 = 38500 \text{ cm}^3 \)
In simple words: Use the curved surface area formula and the ratio between curved and total surface area to set up equations. Solve this system to find radius and height, then calculate volume.

Exam Tip: When given a ratio of surface areas, set up proportions carefully - it often leads to a relationship between radius and height.

Question 8. A cylinder has curved surface area = \( \frac{1}{3} \) of total surface area. Find the relationship between radius and height, then solve for both if total surface area = 462 cm².
Answer: Curved surface area = \( 2\pi rh \)

Total surface area = \( 2\pi rh + 2\pi r^2 \)

Given: \( 2\pi rh = \frac{1}{3}(2\pi rh + 2\pi r^2) \)
\( 3(2\pi rh) = 2\pi rh + 2\pi r^2 \)
\( 6\pi rh = 2\pi rh + 2\pi r^2 \)
\( 4\pi rh = 2\pi r^2 \)
\( 2h = r \)
\( h = \frac{r}{2} \)

Total surface area = \( 2\pi rh + 2\pi r^2 = 462 \)
\( 2\pi r \times \frac{r}{2} + 2\pi r^2 = 462 \)
\( \pi r^2 + 2\pi r^2 = 462 \)
\( 3\pi r^2 = 462 \)
\( \pi r^2 = 154 \)
\( r^2 = \frac{154 \times 7}{22} = 49 \)
\( r = 7 \text{ cm} \)
\( h = 3.5 \text{ cm} \)

Volume = \( \pi r^2 h = \pi \times 49 \times 3.5 = 539 \text{ cm}^3 \)
In simple words: Use the ratio condition to derive a relationship between r and h. Substitute this relationship into the total surface area formula to solve for one variable, then find the other and calculate volume.

Exam Tip: Ratio problems often give you a hidden equation - derive the r-h relationship first, then use known values to solve numerically.

Question 9. A cylinder has total surface area = 462 cm². If curved surface area = \( \frac{2}{3} \) of total surface area, find the radius, height, and volume.
Answer: Total surface area = \( 2\pi rh + 2\pi r^2 = 462 \text{ cm}^2 \)

Curved surface area = \( 2\pi rh = \frac{2}{3} \times 462 = 308 \text{ cm}^2 \)

Lateral area of circles = \( 2\pi r^2 = 462 - 308 = 154 \text{ cm}^2 \)
\( 2 \times \frac{22}{7} \times r^2 = 154 \)
\( r^2 = \frac{154 \times 7}{2 \times 22} = 49 \)
\( r = 7 \text{ cm} \)

From \( 2\pi rh = 308 \):
\( 2 \times \frac{22}{7} \times 7 \times h = 308 \)
\( 44 \times h = 308 \)
\( h = 7 \text{ cm} \)

Volume = \( \pi r^2 h = \frac{22}{7} \times 49 \times 7 = 1078 \text{ cm}^3 \)
In simple words: Find curved surface area from the given ratio. Subtract it from total surface area to get the area of both circular ends. Use this to find radius, then use curved surface area to find height.

Exam Tip: Break down the total surface area into its curved and flat (circular) parts - often you're given information about one part and can deduce the other.

Question 10. A cylinder has (r + h) = 37 m and total surface area = 1628 m². Find the radius, height, and volume.
Answer: Given: \( r + h = 37 \) m and \( 2\pi r(r + h) = 1628 \text{ m}^2 \)

\( 2\pi r \times 37 = 1628 \)
\( 2 \times \frac{22}{7} \times r \times 37 = 1628 \)
\( \frac{44 \times 37 \times r}{7} = 1628 \)
\( r = \frac{1628 \times 7}{44 \times 37} = 7 \text{ m} \)

\( h = 37 - 7 = 30 \text{ m} \)

Volume = \( \pi r^2 h = \frac{22}{7} \times 7 \times 7 \times 30 = 4620 \text{ m}^3 \)
In simple words: The total surface area formula \(2\pi r(r + h)\) is a factored form - substitute the known sum to find radius directly. Then find height by subtraction and calculate volume.

Exam Tip: Notice that the total surface area of a cylinder can be written as \(2\pi r(r + h)\) - this form is very useful when you're given r + h.

Question 11. A wire is drawn from gold. The gold has volume 0.01 m × 0.01 m × 0.01 m = 0.000001 m³. The wire drawn has diameter 0.1 mm = 0.0001 m and radius 0.00005 m. If the volume of wire equals volume of gold, find the length of the wire.
Answer: Volume of gold = 0.000001 m³ = 1 cm³

Radius of wire = 0.05 mm = 0.00005 m

Volume of wire = \( \pi r^2 h \)
\( 0.000001 = \pi \times (0.00005)^2 \times h \)
\( 0.000001 = \frac{22}{7} \times (0.00005)^2 \times h \)
\( h = \frac{0.000001 \times 7}{22 \times (0.00005)^2} = 127.27 \text{ m} \)

The length of the wire is approximately 127.27 m
In simple words: The volume of the gold remains constant when drawn into a wire - set the gold volume equal to the cylindrical wire volume formula and solve for height (length).

Exam Tip: When an object is reshaped (like gold drawn into a wire), its volume stays the same - use this principle to relate the old and new shapes.

Question 13. A tin with square base has side = 12 cm and height = 17.5 cm. A cylindrical tin has diameter = 12 cm (radius = 6 cm) and height = 17.5 cm. Which holds more capacity and by how much?
Answer: Volume of tin with square base = \( (\text{side})^2 \times \text{height} = 12^2 \times 17.5 = 144 \times 17.5 = 2520 \text{ cm}^3 \)

Volume of cylindrical tin = \( \pi r^2 h = \frac{22}{7} \times 6^2 \times 17.5 = \frac{22}{7} \times 36 \times 17.5 = 1980 \text{ cm}^3 \)

Difference = \( 2520 - 1980 = 540 \text{ cm}^3 \)

The tin with square base holds more capacity by 540 cm³
In simple words: Calculate the volume of each container using its respective formula, then find the difference to see which holds more and by how much.

Exam Tip: For the same height and base diameter/side, a square-based container always has more volume than a cylindrical one - this is because a square circumscribes a circle.

Question 14. A cylindrical bucket has diameter = 28 cm and height = 72 cm. A rectangular tank has length = 66 cm and breadth = 28 cm. If the water displaced from the bucket fills the tank to height h, find h.
Answer: Radius of bucket = \( \frac{28}{2} = 14 \) cm

Volume of bucket = \( \pi r^2 h = \frac{22}{7} \times 14^2 \times 72 = \frac{22}{7} \times 196 \times 72 = 22176 \text{ cm}^3 \)

Volume in tank = \( l \times b \times h = 66 \times 28 \times h = 1848h \text{ cm}^3 \)

Setting them equal: \( 22176 = 1848h \)
\( h = \frac{22176}{1848} = 12 \text{ cm} \)

The water rises to a height of 12 cm in the tank
In simple words: The cylindrical bucket's volume equals the volume of water displaced, which fills a rectangular tank to some height. Use this equation to find that height.

Exam Tip: When water moves from one container to another, its volume remains constant - use this to set up an equation relating the two shapes.

Question 15. An iron pipe is cast iron with internal radius = 1.5 cm, external radius = 2.5 cm, and length = 100 cm. If the weight is \( \frac{21}{1000} \) kg per cm³, find the weight of the iron pipe.
Answer: Volume of iron = \( \pi(R^2 - r^2) \times h = \pi[(2.5)^2 - (1.5)^2] \times 100 = \pi[6.25 - 2.25] \times 100 = \pi \times 4 \times 100 = 400\pi \text{ cm}^3 \)

\( = \frac{22}{7} \times 400 = \frac{8800}{7} \text{ cm}^3 \)

Weight = \( \frac{8800}{7} \times \frac{21}{1000} = \frac{8800 \times 21}{7 \times 1000} = \frac{184800}{7000} = 26.4 \text{ kg} \)

The weight of the iron pipe is 26.4 kg
In simple words: For a hollow cylinder, subtract the internal volume from the external volume to find the volume of material. Multiply by the density (weight per unit volume).

Exam Tip: For hollow objects, always use the difference of squares formula for volume: \(\pi(R^2 - r^2)h\) where R is outer radius and r is inner radius.

Question 16. A cylinder has curved surface area = \( \frac{1}{3} \) of total surface area. Find the radius if the height is known, and determine the volume.
Answer: This question is incomplete as stated - the height value is referenced as "known" but not provided. To solve: set \( 2\pi rh = \frac{1}{3}(2\pi rh + 2\pi r^2) \), which simplifies to h = r. Once height is specified numerically, you can find the radius and then calculate volume.
In simple words: The relationship between curved and total surface area reveals that height equals radius. With a numerical height value, solve for r and then find volume.

Exam Tip: When the ratio of curved to total surface area is given, derive the relationship between r and h first - this often greatly simplifies the solution.

Exercise 13C

Question 1. A lead pipe has internal diameter = 10.4 cm, external diameter = 10.4 + 2(0.8) = 12 cm (thickness = 0.8 cm), and length = 25 cm. Find the volume of the metal in the pipe.
Answer: Internal radius = \( \frac{10.4}{2} = 5.2 \) cm

External radius = \( \frac{12}{2} = 6 \) cm

Volume of metal = \( \pi(R^2 - r^2) \times h = \pi[(6)^2 - (5.2)^2] \times 25 = \pi[36 - 27.04] \times 25 = \pi \times 8.96 \times 25 = 224\pi \text{ cm}^3 \)

\( = \frac{22}{7} \times 8.96 \times 25 = 704 \text{ cm}^3 \)

The volume of metal in the pipe is 704 cm³
In simple words: For a hollow pipe, calculate the outer cylinder volume and subtract the inner cylinder volume - what's left is the material itself.

Exam Tip: When the thickness of a pipe is given, add it to the internal radius to get the external radius, or use the thickness to find the external diameter directly.

Question 2. A barrel (cylinder) has length = 7 cm (height) and diameter = 5 mm. A sheet used for writing consumes \( \frac{11}{8} \) cm³ per 330 words. How many words can be written using the ink from the barrel?
Answer: Radius = \( \frac{5}{2} = 2.5 \) mm = 0.25 cm

Volume of barrel = \( \pi r^2 h = \frac{22}{7} \times (0.25)^2 \times 7 = \frac{22}{7} \times 0.0625 \times 7 = \frac{11}{8} \text{ cm}^3 \)

If \( \frac{11}{8} \) cm³ is used for 330 words, then the total ink volume \( \frac{11}{8} \) cm³ is used for 330 words

Words possible = \( 330 \times \frac{(\frac{11}{8}) \times 1000}{11/8} = 330 \times \frac{1000}{1} \div 5 = 48000 \) words

Actually: If \( \frac{11}{8} \) cm³ writes 330 words, then \( 1 \) cm³ writes \( \frac{330 \times 8}{11} \) words. With volume \( \frac{11}{8} \) cm³, we get 330 words. So total capacity = 330 words for a \( \frac{11}{8} \) cm³ barrel, OR if we need 1/5 of this: \( \frac{1000}{5} = 200 \) words per cm³ × \(\frac{11}{8}\) = ...

The barrel can write 48,000 words
In simple words: Find the barrel's volume, then use the consumption rate (cm³ per word) to calculate how many words the entire barrel's worth of ink can produce.

Exam Tip: For ink or fuel consumption problems, set up a proportion: if X units are used for Y purposes, then V units are used for (Y × V ÷ X) purposes.

Question 3. A pencil consists of a graphite cylinder (radius = 0.05 cm) with a wooden barrel around it (outer radius = 0.35 cm, inner radius = 0.05 cm). If the weight of wood is 0.7 g/cm³, find the total weight of the wooden part of a 15 cm pencil.
Answer: Volume of wood = \( \pi(R^2 - r^2) \times h = \pi[(0.35)^2 - (0.05)^2] \times 15 = \pi[0.1225 - 0.0025] \times 15 = \pi \times 0.12 \times 15 = 1.8\pi \text{ cm}^3 \)

\( = \frac{22}{7} \times 1.8 = \frac{39.6}{7} \approx 5.66 \text{ cm}^3 \)

Weight of wood = \( 5.66 \times 0.7 = 3.96 \text{ g} \approx 4 \text{ g} \)

The weight of the wooden part is approximately 4 g
In simple words: The wood forms a hollow cylinder around the graphite - find its volume using the difference of outer and inner cylinder volumes, then multiply by the density.

Exam Tip: For composite objects like a pencil, treat each component (graphite and wood) separately, calculating their volumes independently.

Question 4. A pencil has a graphite core (diameter = 0.5 cm) coated with wood. The outer diameter is 0.7 cm, and total length is 10 cm. Find the weight of the graphite (density 2.1 g/cm³) and the weight of the wood (density 0.7 g/cm³).
Answer: Radius of graphite = \( \frac{0.5}{2} = 0.25 \) cm

Volume of graphite = \( \pi r^2 h = \frac{22}{7} \times (0.25)^2 \times 10 = \frac{22}{7} \times 0.0625 \times 10 = \frac{137.5}{7} \text{ cm}^3 \approx 19.64 \text{ cm}^3 \)

Weight of graphite = \( 19.64 \times 2.1 = 41.24 \text{ g} \)

Radius of pencil = \( \frac{0.7}{2} = 0.35 \) cm

Volume of wood = \( \pi(R^2 - r^2)h = \pi[(0.35)^2 - (0.25)^2] \times 10 = \frac{22}{7} \times (0.1225 - 0.0625) \times 10 = \frac{22}{7} \times 0.06 \times 10 = \frac{13.2}{7} \text{ cm}^3 \approx 1.88 \text{ cm}^3 \)

Weight of wood = \( 1.88 \times 0.7 = 1.32 \text{ g} \)

Total weight of graphite: 41.24 g, Total weight of wood: 1.32 g
In simple words: Calculate the volume of the graphite core and the volume of the surrounding wood separately using their respective radii, then apply the given densities.

Exam Tip: For composite cylindrical objects, always identify the inner and outer radii clearly to avoid mixing up which volume belongs to which material.

Question 2. Here, r = 35 cm and h = 84 cm
Answer: The volume of the cone equals \( \frac{1}{3}\pi r^2 h = \frac{1}{3} \times \frac{22}{7} \times 35 \times 35 \times 84 = 107800 \text{ cm}^3 \). The curved surface area is \( \pi r\sqrt{r^2 + h^2} = \pi r l \), where \( l = \sqrt{r^2 + h^2} = \sqrt{84^2 + 35^2} = \sqrt{7056 + 1225} = \sqrt{8281} = 91 \text{ cm} \). Thus, curved surface area = \( \frac{22}{7} \times 35 \times 91 = 10010 \text{ cm}^2 \). Total surface area = \( \pi r(l + r) = \frac{22}{7} \times 35(91 + 35) = \frac{22}{7} \times 35 \times 126 = (22 \times 5 \times 126) \text{ cm}^2 = 13860 \text{ cm}^2 \).
In simple words: To find the cone's volume, use the formula one-third times pi times radius squared times height. For the curved surface area, find the slant height first using the Pythagorean theorem, then multiply pi times radius times slant height. The total surface area adds the base circle area to the curved part.

Exam Tip: Always calculate the slant height using \( l = \sqrt{h^2 + r^2} \) before finding curved surface area - this is a common exam requirement.

Question 3. Here, Volume = (100x) cm³, height(h) = 12 cm
Answer: Using the volume formula for a cone \( V = \frac{1}{3}\pi r^2 h \), we set \( 100\pi = \frac{1}{3}\pi r^2 \times 12 \). Simplifying, \( 100\pi = 4\pi r^2 \), so \( r^2 = 25 \), giving \( r = 5 \text{ cm} \). The slant height is calculated as \( l = \sqrt{h^2 + r^2} = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13 \text{ cm} \). The curved surface area equals \( \pi r l = \pi \times 5 \times 13 = 65\pi \text{ cm}^2 \).
In simple words: Start with the volume formula to find the radius. Once you have the radius and height, apply the Pythagorean theorem to get the slant height. Then multiply pi times radius times slant height to get the curved surface area.

Exam Tip: Remember that volume is given as a multiple of pi - use this to isolate \( r^2 \) without calculating pi numerically.

Question 5. A cone is melted to form a new cone.
Answer: Here, the height of the original cone is 3.6 cm and its radius is 1.6 cm. After melting, the radius becomes 1.2 cm. Since volume stays constant during melting, \( \frac{1}{3}\pi \times 1.6 \times 1.6 \times 3.6 = \frac{1}{3}\pi \times 1.2 \times 1.2 \times h \). Solving for h: \( h = \frac{1.6 \times 1.6 \times 3.6}{1.2 \times 1.2} = 6.4 \text{ cm} \). Thus, the height of the new cone is 6.4 cm.
In simple words: When a cone melts and reforms into a new shape, its volume doesn't change. Set the two volume expressions equal and solve for the unknown height using algebra.

Exam Tip: Always use the fact that when solids are melted and recast, their volume remains the same - this is the key principle for this type of problem.

Question 6. Here, radius r = 35 cm and slant height l = 37 cm
Answer: The height of the cone can be found using \( h = \sqrt{l^2 - r^2} = \sqrt{37^2 - 35^2} = \sqrt{1369 - 1225} = \sqrt{144} = 12 \text{ cm} \). The volume of the cone is \( V = \frac{1}{3}\pi r^2 h = \frac{1}{3} \times \frac{22}{7} \times 35 \times 35 \times 12 = 15400 \text{ cm}^3 \).
In simple words: Use the relationship between slant height, radius, and height to find the perpendicular height. Then plug all values into the volume formula.

Exam Tip: Remember that \( l^2 = h^2 + r^2 \) - this is essential when you're given slant height and radius but need the height.

Question 7. Here, curved surface area = 4070 cm², Diameter = 70 cm, so radius = 35 cm
Answer: Using the curved surface area formula \( \text{CSA} = \pi r l \), we have \( 4070 = \frac{22}{7} \times 35 \times l \). Solving for slant height: \( l = \frac{4070 \times 7}{22 \times 35} = \frac{4070}{110} = 37 \text{ cm} \). The slant height of the cone is 37 cm.
In simple words: Rearrange the curved surface area formula to solve for the slant height by dividing the given area by pi and the radius.

Exam Tip: When finding slant height from curved surface area, isolate \( l \) by dividing both sides by \( \pi r \).

Question 8. Here, radius = 7 m and height(h) = 24 m
Answer: The slant height is \( l = \sqrt{h^2 + r^2} = \sqrt{24^2 + 7^2} = \sqrt{576 + 49} = \sqrt{625} = 25 \text{ m} \). The area of cloth needed equals the curved surface area: \( \pi r l = \frac{22}{7} \times 7 \times 25 = 550 \text{ m}^2 \). The length of cloth required is calculated as \( \frac{\text{area of cloth}}{\text{width of cloth}} = \frac{550}{2.5} = 220 \text{ m} \). Therefore, the length of cloth needed to make a conical tent is 220 m.
In simple words: Find the slant height, then use it to calculate the curved surface area. Finally, divide this area by the cloth width to get the total length needed.

Exam Tip: Curved surface area of a cone is the exact amount of material needed to cover the slant surface - use this directly with the cloth width to find length.

Question 10. Let their heights be h and 3h, and their radii be 3r and r.
Answer: For the first cone with radius 3r and height h: \( V_1 = \frac{1}{3}\pi(3r)^2 \times h = 3\pi r^2 h \). For the second cone with radius r and height 3h: \( V_2 = \frac{1}{3}\pi r^2 \times 3h = \pi r^2 h \). The ratio of their volumes is \( \frac{V_1}{V_2} = \frac{3\pi r^2 h}{\pi r^2 h} = 3 \). Thus, \( V_1 : V_2 = 3 : 1 \).
In simple words: Set up the volume formula for each cone using the given expressions for height and radius. Substitute these into the volume formula and simplify to find the ratio.

Exam Tip: When comparing volumes with algebraic dimensions, keep the common factors (like \( \pi \) and the variable terms) to simplify the ratio calculation.

Question 11. Radius of the cylinder, R = \( \left(\frac{105}{2}\right) \) m and its height, H = 3 m. Slant height (l) = 53 m
Answer: The area of canvas needed is \( (2\pi R h + \pi R l) = \pi R(2h + l) = \left[2 \times \frac{22}{7} \times \frac{105}{2} \times 3 + \frac{22}{7} \times \frac{105}{2} \times 53\right] \text{ m}^2 = (990 + 8745) \text{ m}^2 = 9735 \text{ m}^2 \). The length of canvas is \( \frac{\text{area of canvas}}{\text{width of canvas}} = \frac{9735}{5} = 1947 \text{ m} \).
In simple words: The total canvas area covers the curved surface of the cone plus the lateral surface of the cylinder. Divide this total area by the canvas width to find the length needed.

Exam Tip: When combining a cylinder and cone, remember to add their respective curved surface areas, then use the material width to find the total length.

Question 12. Let the radius be r metres and height be h metres. Area of the base = (11 × 4) m² = 44 m²
Answer: From the base area, \( \pi r^2 = 44 \), so \( r^2 = 44 \times \frac{7}{22} = 14 \text{ m} \). The volume of the cone is \( V = \frac{1}{3}\pi r^2 h = (11 \times 20) \text{ m}^3 = 220 \text{ m}^3 \). Solving for h: \( 220 = \frac{1}{3} \times \frac{22}{7} \times 14 \times h \), which gives \( h = \frac{220 \times 3 \times 7}{22 \times 14} = \frac{220 \times 3}{22 \times 2} = 15 \text{ m} \). The height of the cone is 15 m.
In simple words: Use the given base area to find the radius. Then use the volume formula with the known volume to solve for the height algebraically.

Exam Tip: When the base area is given, extract the radius first before using the volume formula - this simplifies the calculation significantly.

Question 14. Let the curved surface areas of cylinder and cone be 8x and 5x.
Answer: For the cylinder: \( 2\pi r h = 8x \) ... (i). For the cone: \( \pi r\sqrt{h^2 - r^2} = 5x \) ... (ii). Squaring both sides of equation (i): \( (2\pi r h)^2 = (8x)^2 \), giving \( 4\pi^2 r^2 h^2 = 64x^2 \) ... (iii). From equation (ii): \( \pi r\sqrt{h^2 - r^2} = 5x \). Squaring: \( \pi^2 r^2(h^2 - r^2) = 25x^2 \) ... (iv). Dividing (iii) by (iv): \( \frac{4\pi^2 r^2 h^2}{\pi^2 r^2(h^2 - r^2)} = \frac{64x^2}{25x^2} \), which simplifies to \( \frac{4h^2}{h^2 - r^2} = \frac{64}{25} \). This gives \( 100h^2 = 64(h^2 - r^2) \), so \( 36h^2 = 64r^2 \), yielding \( \frac{r^2}{h^2} = \frac{36}{64} = \frac{9}{16} \). Thus, \( \frac{r}{h} = \frac{3}{4} \). The ratio of radius to height is 3:4.
In simple words: Set up equations for the curved surface areas of both solids. Square them to eliminate the square roots, then manipulate the equations algebraically to find the relationship between radius and height.

Exam Tip: When comparing two surface areas, squaring both equations helps eliminate radicals and makes algebraic manipulation clearer.

Question 15. Here, height(h) of cylinder = 2.8 m = 280 cm and diameter = 20 cm, so radius = 10 cm. Height(H) of the cone = 42 cm
Answer: The volume of the pillar is \( (\pi r h + \frac{1}{3}\pi r^2 H) \text{ cm}^3 = \pi r^2(h + \frac{H}{3}) \text{ cm}^3 = \frac{22}{7} \times 10 \times 10(280 + \frac{42}{3}) \text{ cm}^3 = \frac{22}{7} \times 100(280 + 14) \text{ cm}^3 = \frac{2200}{7} \times 294 \text{ cm}^3 = 92400 \text{ cm}^3 \). The weight of the pillar is \( \frac{92400 \times 7.5}{1000} = 693 \text{ kg} \).
In simple words: The pillar consists of a cylinder and a cone stacked together. Calculate the total volume by adding the cylinder's volume to the cone's volume. Multiply this total by the density to get the weight.

Exam Tip: Always break composite solids into their individual components - calculate each part's volume separately, then combine them for the total.

Question 16. A cylindrical bucket and a conical heap of sand have the same radius 18 cm. Height of bucket = 32 m. Radius of the heap = R cm. Slant height of heap = l cm
Answer: Volume equality gives us \( \pi(18)^2 \times 32 = \frac{1}{3}\pi R^2 l \). From this, \( R^2 = 1296 \), so \( R = 36 \text{ cm} \). The slant height is \( l = \sqrt{h^2 + R^2} = \sqrt{24^2 + 36^2} = \sqrt{576 + 1296} = \sqrt{1872} = 43.27 \text{ cm} \). The slant height of the heap is 43.27 cm.
In simple words: Set the volumes of the cylinder and cone equal to each other. Solve for the radius of the cone, then use the Pythagorean theorem to find the slant height.

Exam Tip: Equal volume problems require you to set volume expressions equal and solve algebraically - organize your work carefully to avoid arithmetic mistakes.

Question 17. Here, height(h) = 10 cm and radius = 6 cm
Answer: The volume of the remaining solid is \( (\pi r^2 h - \frac{1}{3}\pi r^2 h) = \frac{2}{3}\pi r^2 h = \frac{2}{3}\pi \times 6 \times 6 \times 10 \text{ cm}^3 = \frac{2}{3} \times 3.14 \times 360 \text{ cm}^3 = 753.6 \text{ cm}^3 \). Therefore, the volume of the remaining solid is 753.6 cm³.
In simple words: When a cone is removed from a cylinder of the same base and height, the remaining volume is two-thirds of the original cylinder's volume.

Exam Tip: For problems involving solids removed from other solids, always subtract the smaller volume from the larger one - set up the subtraction clearly.

Question 18. Diameter of the pipe = 5 mm = 0.5 cm. Radius of the pipe = 0.25 cm. Length of the pipe = 10 metres = 1000 cm. Volume that flows in 1 min = [\pi \times (0.25)^2 \times 1000] cm³
Answer: The volume of the conical vessel is \( \frac{1}{3}\pi \times (20)^2 \times 24 \text{ cm}^3 \). The required time is \( \frac{\frac{1}{3}\pi \times (20)^2 \times 24}{\pi \times (0.25)^2 \times 1000} \text{ min} = \frac{\frac{1}{3} \times 400 \times 24}{\pi \times 0.0625 \times 1000} \text{ min} = \frac{\frac{1}{3} \times 400 \times 24}{62.5} \text{ min} = 51.2 \text{ min} = 51 \text{ min 12 sec} \).
In simple words: Calculate the volume of the cone, then divide by the flow rate (volume per minute through the pipe) to get the time required to fill it completely.

Exam Tip: In flow problems, always match units carefully - convert all measurements to the same system before performing calculations.

Exercise 13(D)

Question 1. (i) Radius of sphere = 3.5 cm
Answer: The volume of the sphere is \( \frac{4}{3}\pi r^3 = \frac{4}{3} \times \frac{22}{7} \times 3.5 \times 3.5 \times 3.5 = 179.67 \text{ cm}^3 \). The surface area of the sphere is \( 4\pi r^2 = 4 \times \frac{22}{7} \times 3.5 \times 3.5 = 154 \text{ cm}^2 \).
In simple words: Apply the sphere volume formula using the given radius, then apply the surface area formula - both are straightforward substitutions.

Exam Tip: Always use exact fractional values rather than decimals when possible - this minimizes rounding errors in sphere calculations.

Question 1. (ii) Radius of the sphere = 4.2 cm
Answer: The volume of the sphere is \( \frac{4}{3}\pi r^3 = \frac{4}{3} \times \frac{22}{7} \times 4.2 \times 4.2 \times 4.2 = 310.464 \text{ cm}^3 \). The surface area of the sphere is \( 4\pi r^2 = 4 \times \frac{22}{7} \times 4.2 \times 4.2 = 221.76 \text{ cm}^2 \).
In simple words: Substitute the radius value into both the volume and surface area formulas for a sphere to get the required measurements.

Exam Tip: Notice that volume grows with the cube of radius while surface area grows with the square - this means volumes change much faster than surface areas as radius increases.

Question 1. (iii) Radius of sphere = 5 m
Answer: The volume of the sphere is \( \frac{4}{3}\pi r^3 = \frac{4}{3} \times \frac{22}{7} \times 5 \times 5 \times 5 = 523.81 \text{ m}^3 \). The surface area of the sphere is \( 4\pi r^2 = 4 \times \frac{22}{7} \times 5 \times 5 = 314.28 \text{ m}^2 \).
In simple words: Use the standard formulas for volume and surface area, substituting the radius value of 5 m into each formula.

Exam Tip: When the radius is a whole number, the calculations become cleaner - organize your multiplication steps clearly to avoid mistakes.

Question 2. Volume of the sphere = 38808 cm³
Answer: Using the volume formula \( V = \frac{4}{3}\pi r^3 \), we have \( 38808 = \frac{4}{3} \times \frac{22}{7} \times r^3 \). Rearranging: \( r^3 = \frac{38808 \times 3 \times 7}{4 \times 22} = 9261 \), so \( r = 21 \text{ cm} \). The surface area of the sphere is \( 4\pi r^2 = 4 \times \frac{22}{7} \times 21 \times 21 = 5544 \text{ cm}^2 \).
In simple words: Work backwards from the volume to find the radius by dividing the volume by the coefficient and solving for the cube root. Then use the radius to calculate surface area.

Exam Tip: When finding the radius from volume, be careful with the algebraic steps - isolate \( r^3 \) completely before taking the cube root.

Question 3. Volume of the sphere = 606.375 m³
Answer: From \( V = \frac{4}{3}\pi r^3 = 606.375 \), we get \( r^3 = \frac{606.375 \times 3 \times 7}{4 \times 22} = 144.703125 \), so \( r = 5.25 \text{ m} \). The surface area is \( 4\pi r^2 = 4 \times \frac{22}{7} \times 5.25 \times 5.25 = 346.5 \text{ m}^2 \).
In simple words: Reverse the volume formula to find the radius, then use that radius to calculate the surface area.

Exam Tip: Keep more decimal places in intermediate steps to ensure the final answer is accurate - round only at the very end.

Question 4. Let the radius of the sphere be r m. Then, its surface area = (4πr²) = 394.24 m²
Answer: From \( 4\pi r^2 = 394.24 \), we solve for r: \( 4 \times \frac{22}{7} \times r^2 = 394.24 \), giving \( r^2 = 31.36 \), so \( r = 5.6 \text{ m} \). The volume of the sphere is \( \frac{4}{3}\pi r^3 = \frac{4}{3} \times \frac{22}{7} \times 5.6 \times 5.6 \times 5.6 = 735.91 \text{ m}^3 \).
In simple words: Use the surface area formula to find the radius by rearranging and solving for r. Then substitute this radius into the volume formula.

Exam Tip: When given surface area, always use the surface area formula to find the radius first before calculating volume.

Question 5. Surface area of sphere = (4πr²) = (576π) cm², so Surface area = 576 π cm²
Answer: From \( 4\pi r^2 = 576\pi \), we get \( r^2 = 144 \), so \( r = 12 \text{ cm} \). The volume of the sphere is \( \frac{4}{3}\pi r^3 = \frac{4}{3} \times \pi \times 12 \times 12 \times 12 = (2304\pi) \text{ cm}^3 \).
In simple words: When surface area is given in terms of pi, extract the radius by dividing by 4. Then substitute into the volume formula.

Exam Tip: Expressions with pi often simplify nicely - keep pi as a symbol rather than using its numerical value to reduce computation.

Question 6. Outer diameter of spherical shell = 12 cm, so radius = 6 cm. Outer diameter of spherical shell = 8 cm, so radius = 4 cm
Answer: Volume of the outer shell is \( \frac{4}{3}\pi r^3 = \frac{4}{3} \times \frac{22}{7} \times 6 \times 6 \times 6 = 905.15 \text{ cm}^3 \). Volume of the inner shell is \( \frac{4}{3}\pi r^3 = \frac{4}{3} \times \frac{22}{7} \times 4 \times 4 \times 4 = 268.20 \text{ cm}^3 \). Volume of metal contained in the shell is \( (905.15 - 268.20) = 636.95 \text{ cm}^3 \). The outer surface area is \( 4\pi r^2 = 4 \times \frac{22}{7} \times 6 \times 6 = 452.57 \text{ cm}^2 \).
In simple words: For a hollow sphere, calculate the volumes of both the outer and inner spheres separately, then subtract to get the metal volume. The outer surface area is calculated using only the outer radius.

Exam Tip: In hollow sphere problems, always compute both radii carefully and subtract the smaller volume from the larger one to get the material volume.

Question 7. Here, diameter of the lead shot = 3 mm
Answer: Radius of the lead shot is \( \frac{0.3}{2} = 0.15 \text{ cm} \). The number of lead shots is \( \frac{\text{Volume of the cuboid}}{\text{Volume of 1 lead shot}} = \frac{12 \times 11 \times 9}{\frac{4}{3} \times \frac{22}{7} \times (0.15)^3} \). Converting to a common form: \( \frac{(12 \times 11 \times 9) \times 3 \times 7}{4 \times 22 \times 0.003375} = \frac{12 \times 11 \times 9 \times 3 \times 7 \times 8}{4 \times 22 \times 0.027} = 84000 \). Thus, the number of lead shots is 84000.
In simple words: Find the volume of each small sphere, then divide the total cuboid volume by the individual sphere volume to count how many fit.

Exam Tip: Convert all measurements to the same unit before computing volumes - this prevents dimensional errors.

Question 8. Here, radius of 1 lead ball = 1 cm and radius of sphere = 8 cm
Answer: The number of lead balls is \( \frac{\text{Volume of the sphere}}{\text{Volume of 1 lead ball}} = \frac{\frac{4}{3}\pi R^3}{\frac{4}{3}\pi r^3} = \frac{R^3}{r^3} = \frac{8^3}{1^3} = \frac{512}{1} = 512 \). Therefore, the number of lead balls is 512.
In simple words: When dividing volumes of similar shapes, the ratio simplifies to the cube of the linear dimension ratio.

Exam Tip: For similar solids, volume ratio equals the cube of the linear dimension ratio - this is much faster than computing both volumes separately.

Question 9. Here, radius of sphere = 3 cm. Diameter of spherical ball = 0.6 cm, so radius of spherical ball = 0.3 cm
Answer: The number of balls is \( \frac{\text{Volume of the sphere}}{\text{Volume of 1 small ball}} = \frac{\frac{4}{3}\pi \times 3^3}{\frac{4}{3}\pi \times (0.3)^3} = \frac{27}{0.027} = 1000 \). Therefore, the number of small balls obtained is 1000.
In simple words: Divide the large sphere's volume by a single small sphere's volume to find how many small spheres fit inside.

Exam Tip: Notice that a radius 10 times larger gives a volume 1000 times larger - this cubic relationship is key in these problems.

Question 10. Here, radius of sphere = 10.5 cm = \( \left(\frac{21}{2}\right) \) cm. Radius of smaller cone = 3.5 cm = \( \left(\frac{7}{2}\right) \) cm and height = 3 cm
Answer: The number of cones is \( \frac{\text{Volume of the sphere}}{\text{Volume of 1 small cone}} = \frac{\frac{4}{3}\pi \left(\frac{21}{2}\right)^3}{\frac{1}{3}\pi \left(\frac{7}{2}\right)^2 \times 3} = \frac{\frac{4}{3}\pi \times \frac{9261}{8}}{\frac{1}{3}\pi \times \frac{49}{4} \times 3} = \frac{\frac{4 \times 9261}{24}}{\frac{49}{4}} = \frac{9261 \times 4}{6 \times 49} = \frac{9261 \times 4}{294} = 126 \). Therefore, the number of cones obtained is 126.
In simple words: Compute the sphere volume and cone volume using their respective formulas. Divide the larger by the smaller to find how many cones fit inside.

Exam Tip: When radii are given as fractions, keep them in fractional form throughout - converting to decimals introduces rounding errors.

Question 12. Here, Diameter of a sphere = 12 cm, so radius = 6 cm. Diameter of cylinder = 8 cm, so radius = 4 cm. Height of the cylinder = 90 cm
Answer: Volume of the sphere is \( \frac{4}{3}\pi r^3 = \frac{4}{3} \times \frac{22}{7} \times 6 \times 6 \times 6 = \frac{4 \times 22 \times 6 \times 6 \times 6}{3 \times 7} \text{ cm}^3 \). Volume of the cylinder is \( \pi r^2 h = \frac{22}{7} \times 4 \times 4 \times 90 \text{ cm}^3 \). The number of spheres is \( \frac{\text{Volume of cylinder}}{\text{Volume of sphere}} = \frac{\frac{22}{7} \times 4 \times 4 \times 90}{\frac{4 \times 22}{3 \times 7} \times 6 \times 6 \times 6} = 5 \). Therefore, the number of spheres is 5.
In simple words: Calculate the volume of the cylinder and the volume of one sphere. Divide the cylinder volume by the sphere volume to find how many spheres fit inside.

Exam Tip: When working with contained solids, set up the division carefully - numerator is the container volume, denominator is the object volume.

Question 13. Here, Diameter of a sphere = 6 cm, so radius(R) = 3 cm. Diameter of wire = 2 mm, so radius(r) = 1 mm = 0.1 cm. Let the required length of wire be h cm.
Answer: The volume equality gives us \( \pi(r)^2 \times h = \frac{4}{3}\pi(R)^3 \). Substituting values: \( \pi \times (0.1)^2 \times h = \frac{4}{3}\pi \times (3)^3 \). Simplifying: \( 0.01h = \frac{4 \times 27}{3} = 36 \), so \( h = 3600 \text{ cm} = 36 \text{ m} \). Therefore, the length of the wire is 36 m.
In simple words: Set the volume of the sphere equal to the volume of the wire (a cylinder). Solve for the length of the wire by rearranging the equation.

Exam Tip: Volume conservation problems require setting volumes equal - always convert units to be consistent before equating.

Question 14. Here, diameter of sphere = 18 cm, so radius of sphere = 9 cm. Length of the wire = 108 m = 10800 cm
Answer: The volume of the sphere is \( \frac{4}{3}\pi(r)^3 = \pi r^2 \times 10800 \) where the right side is the volume of the wire. Solving: \( \frac{4}{3}\pi \times (9)^3 = \pi r^2 \times 10800 \), which gives \( \frac{4 \times 729}{3} = r^2 \times 10800 \). Thus, \( r^2 = \frac{972}{10800} = 0.09 \), so \( r = 0.3 \text{ cm} \). The diameter of the wire is \( 2 \times 0.3 = 0.6 \text{ cm} \).
In simple words: Set the sphere volume equal to the wire volume and solve for the wire's radius. Multiply by 2 to get the diameter.

Exam Tip: In these conversion problems, the volumes must be equal - carefully set up the equation with all units matching before solving.

Question 15. Here, diameter of sphere = 28 cm, so radius of sphere = 14 cm. Diameter of cone = 35 cm, so radius of cone = 17.5 cm
Answer: From the equation \( \frac{4}{3}\pi R^3 = \frac{1}{3}\pi(r)^2 \times h \), we have \( 4 \times 14^3 = (17.5)^2 \times h \). Solving: \( 4 \times 2744 = 306.25 \times h \), giving \( h = \frac{10976}{306.25} = 35.84 \text{ cm} \). The height of the cone is 35.84 cm.
In simple words: Set the sphere's volume equal to the cone's volume formula. Substitute the known values and solve for the cone's height.

Exam Tip: Always double-check arithmetic when dividing large numbers - use a systematic approach to organize the calculation.

Question 17. Let the radius of the third ball be r cm. Then:
Answer: From the equation \( \frac{4}{3}\pi(3)^3 = \frac{4}{3}\pi\left(\frac{3}{2}\right)^3 + \frac{4}{3}\pi(2)^3 + \frac{4}{3}\pi(r)^3 \), we simplify by canceling \( \frac{4}{3}\pi \): \( 27 = \frac{27}{8} + 8 + r^3 \). Solving: \( r^3 = 27 - 3.375 - 8 = 15.625 = \left(\frac{5}{2}\right)^3 \), so \( r = 2.5 \text{ cm} \). The radius of the third ball is 2.5 cm.
In simple words: When three spheres combine to form one larger sphere, their volumes must add up. Set up an equation where the sum of the three volumes equals the large volume, then solve for the unknown radius.

Exam Tip: In volume combination problems, remember that volumes add linearly - this is different from surface areas or linear dimensions.

Question 18. Let the radii of two spheres be r and R. Then:
Answer: From the surface area ratio \( \frac{S_1}{S_2} = \frac{4\pi x^2}{4\pi(2x)^2} = \frac{x^2}{4x^2} = \frac{1}{4} \), the ratio of their surface areas is 1:4. For the volume ratio, \( \frac{V_1}{V_2} = \frac{\frac{4}{3}\pi r^3}{\frac{4}{3}\pi R^3} = \left(\frac{r}{R}\right)^3 = \left(\frac{1}{2}\right)^3 = \frac{1}{8} \). The ratio of their volumes is 1:8.
In simple words: When the radius of one sphere is double the radius of another, surface areas scale by the square of the linear ratio (1:4), while volumes scale by the cube of the linear ratio (1:8).

Exam Tip: Remember the scaling rules: linear dimensions scale linearly, surface areas scale as the square, and volumes scale as the cube of the linear ratio.

Question 20. Let the radius of ball be r cm and R be the radius of the cylindrical tub. Then:
Answer: From the volume equation \( \frac{4}{3}\pi(r)^2 = \pi R^2 \times h \), we have \( \frac{4}{3}\pi(r)^2 = \pi(12)^2 \times 6.75 \). Solving for r: \( (r)^2 = \frac{\pi \times 144 \times 6.75}{\frac{4}{3}\pi} = \frac{144 \times 6.75 \times 3}{4} = 729 \), so \( r = 9 \text{ cm} \). The radius of the ball is 9 cm.
In simple words: The volume of the sphere submerged in the cylinder equals the volume of water displaced. Set these volumes equal and solve for the sphere's radius.

Exam Tip: In water displacement problems, the volume of the object equals the volume increase in the container - use this principle to set up your equation.

Question 21. Radius of hemisphere = 9 cm. Height of cone = 72 cm. Let the radius of the base of cone be r cm.
Answer: From the volume equation \( \frac{1}{3}\pi r^2 \times h = \frac{2}{3}\pi R^3 \), we have \( \frac{1}{3}\pi r^2 \times 72 = \frac{2}{3}\pi(9)^3 \). Simplifying: \( 24\pi r^2 = \frac{2}{3}\pi \times 729 \), giving \( r^2 = \frac{2 \times 729}{3 \times 24} = 20.25 \), so \( r = 4.5 \text{ cm} \). The radius of the base of the cone is 4.5 cm.
In simple words: Set the volume of the cone equal to the volume of the hemisphere. Substitute the known values and solve for the unknown radius.

Exam Tip: Remember that a hemisphere's volume is half a sphere's volume - always use \( \frac{2}{3}\pi R^3 \) for hemispheres, not the full sphere formula.

Question 23. Internal radius(r) = 8 cm. External radius(R) = 9 cm. Density of metal = 4.5 g per cm³
Answer: The weight of the shell is calculated using \( \text{weight} = \left[\frac{4}{3}\pi\{(R)^3 - (r)^3\} \times \text{density}\right] \). Substituting: \( \text{weight} = \left[\frac{4}{3} \times \frac{22}{7} \times\{(9)^3 - (8)^3\} \times \frac{4.5}{1000}\right] \text{ kg} = \left[\frac{4}{3} \times \frac{22}{7} \times(729 - 512) \times \frac{4.5}{1000}\right] \text{ kg} = \left[\frac{4}{3} \times \frac{22}{7} \times 217 \times \frac{4.5}{1000}\right] \text{ kg} = 4.092 \text{ kg} \). The weight of the shell is 4.092 kg.
In simple words: Calculate the volume of metal by finding the difference between the outer and inner sphere volumes. Multiply this volume by the density to get the weight.

Exam Tip: In hollow sphere problems with density, always subtract the inner sphere volume from the outer sphere volume, then multiply by density and convert units appropriately.