SURFACE AREAS AND VOLUMES - Q&A
EXERCISE 12.1
1. 2 cubes each of volume 64 cm³ are joined end to end. Find the surface area of the resulting cuboid.
Solution:
Let the side of each cube be 'a'.
Given, Volume of cube = 64 cm³
a³ = 64
a = ∛64 = 4 cm.
When two cubes are joined end to end, a cuboid is formed.
Dimensions of the resulting cuboid:
Length (l) = 4 + 4 = 8 cm
Breadth (b) = 4 cm
Height (h) = 4 cm
Surface Area of cuboid = 2(lb + bh + hl)
= 2(8×4 + 4×4 + 4×8)
= 2(32 + 16 + 32)
= 2(80)
= 160 cm²
Answer: The surface area of the resulting cuboid is 160 cm².
2. A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel.
Solution:
Given:
Diameter of hemisphere = 14 cm, so Radius (r) = 7 cm.
Total height of vessel = 13 cm.
Height of the cylindrical part (h) = Total height - Radius of hemisphere
h = 13 - 7 = 6 cm.
Inner Surface Area = Curved Surface Area of Cylinder + Curved Surface Area of Hemisphere
= 2πrh + 2πr²
= 2πr(h + r)
= 2 × (22/7) × 7 × (6 + 7)
= 44 × 13
= 572 cm²
Answer: The inner surface area of the vessel is 572 cm².
3. A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy.
Solution:
Radius (r) = 3.5 cm = 7/2 cm.
Total height of toy = 15.5 cm.
Height of conical part (h) = Total height - Radius of hemisphere
h = 15.5 - 3.5 = 12 cm.
Slant height of cone (l) = √(h² + r²)
l = √(12² + (3.5)²)
l = √(144 + 12.25)
l = √156.25 = 12.5 cm.
Total Surface Area of toy = CSA of Cone + CSA of Hemisphere
= πrl + 2πr²
= πr(l + 2r)
= (22/7) × 3.5 × (12.5 + 2×3.5)
= (22/7) × (7/2) × (12.5 + 7)
= 11 × 19.5
= 214.5 cm²
Answer: The total surface area of the toy is 214.5 cm².
4. A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have? Find the surface area of the solid.
Solution:
The greatest diameter the hemisphere can have is equal to the side of the cube.
Greatest diameter = 7 cm.
Radius (r) = 7/2 = 3.5 cm.
Side of cube (a) = 7 cm.
Surface Area of solid = TSA of Cube - Area of base of Hemisphere + CSA of Hemisphere
= 6a² - πr² + 2πr²
= 6a² + πr²
= 6(7)² + (22/7) × (3.5)²
= 6(49) + (22/7) × 12.25
= 294 + 38.5
= 332.5 cm²
Answer: The greatest diameter is 7 cm and the surface area of the solid is 332.5 cm².
5. A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter l of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid.
Solution:
Edge of the cube = l.
Diameter of hemisphere = l, so Radius (r) = l/2.
Surface Area of remaining solid = TSA of Cube - Area of circular base of hemisphere + CSA of hemisphere
= 6l² - πr² + 2πr²
= 6l² + πr²
= 6l² + π(l/2)²
= 6l² + πl²/4
= (l²/4)(24 + π) unit²
Answer: The surface area is (l²/4)(24 + π) sq. units.
6. A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends (see Fig. 12.10). The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area.
Solution:
Diameter = 5 mm, so Radius (r) = 2.5 mm.
Total length of capsule = 14 mm.
Height of cylindrical part (h) = Total length - 2 × Radius
h = 14 - (2.5 + 2.5) = 14 - 5 = 9 mm.
Surface Area = CSA of Cylinder + 2 × CSA of Hemisphere
= 2πrh + 2 × (2πr²)
= 2πr(h + 2r)
= 2 × (22/7) × 2.5 × (9 + 5)
= (44/7) × 2.5 × 14
= 44 × 2.5 × 2
= 220 mm²
Answer: The surface area of the capsule is 220 mm².
7. A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 2.1 m and 4 m respectively, and the slant height of the top is 2.8 m, find the area of the canvas used for making the tent. Also, find the cost of the canvas of the tent at the rate of ₹ 500 per m². (Note that the base of the tent will not be covered with canvas.)
Solution:
For cylindrical part: Radius (r) = 4/2 = 2 m, Height (h) = 2.1 m.
For conical part: Radius (r) = 2 m, Slant height (l) = 2.8 m.
Area of canvas = CSA of Cylinder + CSA of Cone
= 2πrh + πrl
= πr(2h + l)
= (22/7) × 2 × (2 × 2.1 + 2.8)
= (44/7) × (4.2 + 2.8)
= (44/7) × 7
= 44 m²
Cost of canvas = Area × Rate
= 44 × 500
= ₹ 22,000
Answer: The area of the canvas is 44 m² and the cost is ₹ 22,000.
8. From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest cm².
Solution:
Radius (r) = 1.4/2 = 0.7 cm.
Height (h) = 2.4 cm.
Slant height of cone (l) = √(h² + r²) = √((2.4)² + (0.7)²)
l = √(5.76 + 0.49) = √6.25 = 2.5 cm.
Total Surface Area = CSA of Cylinder + Area of cylindrical base + CSA of Cone cavity
= 2πrh + πr² + πrl
= πr(2h + r + l)
= (22/7) × 0.7 × (2 × 2.4 + 0.7 + 2.5)
= 2.2 × (4.8 + 0.7 + 2.5)
= 2.2 × 8.0
= 17.6 cm²
Rounding to the nearest cm²:
Answer: The total surface area of the remaining solid is approximately 18 cm².
9. A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in Fig. 12.11. If the height of the cylinder is 10 cm, and its base is of radius 3.5 cm, find the total surface area of the article.
Solution:
Height of cylinder (h) = 10 cm.
Radius (r) = 3.5 cm.
Total Surface Area = CSA of Cylinder + 2 × CSA of Hemisphere
= 2πrh + 2 × (2πr²)
= 2πr(h + 2r)
= 2 × (22/7) × 3.5 × (10 + 2×3.5)
= 22 × (10 + 7)
= 22 × 17
= 374 cm²
Answer: The total surface area of the article is 374 cm².
EXERCISE 12.2
1. A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1 cm and the height of the cone is equal to its radius. Find the volume of the solid in terms of π.
Solution:
Radius of hemisphere (r) = 1 cm.
Radius of cone (r) = 1 cm.
Height of cone (h) = r = 1 cm.
Volume of solid = Volume of Cone + Volume of Hemisphere
= (1/3)πr²h + (2/3)πr³
= (1/3)π(1)²(1) + (2/3)π(1)³
= (1/3)π + (2/3)π
= π cm³
Answer: The volume of the solid is π cm³.
2. Rachel, an engineering student, was asked to make a model shaped like a cylinder with two cones attached at its two ends by using a thin aluminium sheet. The diameter of the model is 3 cm and its length is 12 cm. If each cone has a height of 2 cm, find the volume of air contained in the model that Rachel made. (Assume the outer and inner dimensions of the model to be nearly the same.)
Solution:
Diameter = 3 cm, so Radius (r) = 1.5 cm.
Total length of model = 12 cm.
Height of each cone (h_cone) = 2 cm.
Height of cylinder (h_cyl) = Total length - 2 × Height of cone
h_cyl = 12 - (2 + 2) = 8 cm.
Total Volume = Volume of Cylinder + 2 × Volume of Cone
= πr²h_cyl + 2 × (1/3)πr²h_cone
= πr² (h_cyl + (2/3)h_cone)
= (22/7) × (1.5)² × (8 + (2/3)×2)
= (22/7) × 2.25 × (8 + 4/3)
= (22/7) × 2.25 × (28/3)
= 22 × 0.75 × 4
= 66 cm³
Answer: The volume of air contained in the model is 66 cm³.
3. A gulab jamun, contains sugar syrup up to about 30% of its volume. Find approximately how much syrup would be found in 45 gulab jamuns, each shaped like a cylinder with two hemispherical ends with length 5 cm and diameter 2.8 cm (see Fig. 12.15).
Solution:
Diameter = 2.8 cm, so Radius (r) = 1.4 cm.
Total length of gulab jamun = 5 cm.
Height of cylindrical part (h) = 5 - (1.4 + 1.4) = 5 - 2.8 = 2.2 cm.
Volume of one gulab jamun = Volume of Cylinder + 2 × Volume of Hemisphere
= πr²h + 2 × (2/3)πr³
= πr² (h + 4/3 r)
= (22/7) × (1.4)² × (2.2 + 4/3 × 1.4)
= (22/7) × 1.96 × (2.2 + 1.866...)
= 6.16 × 4.066... ≈ 25.05 cm³.
Volume of 45 gulab jamuns = 45 × 25.05 = 1127.25 cm³.
Quantity of syrup = 30% of Total Volume
= (30/100) × 1127.25
= 338.175 cm³
Answer: Approximately 338 cm³ of syrup would be found.
4. A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm. The radius of each of the depressions is 0.5 cm and the depth is 1.4 cm. Find the volume of wood in the entire stand (see Fig. 12.16).
Solution:
Volume of cuboid = L × B × H = 15 × 10 × 3.5 = 525 cm³.
For conical depression: Radius (r) = 0.5 cm, Depth (h) = 1.4 cm.
Volume of one cone = (1/3)πr²h
= (1/3) × (22/7) × (0.5)² × 1.4
= (1/3) × (22/7) × 0.25 × 1.4
= (1/3) × 22 × 0.25 × 0.2
= 1.1 / 3 = 0.366... cm³.
Volume of 4 conical depressions = 4 × 0.366... = 1.47 cm³ (approx).
Volume of wood = Volume of cuboid - Volume of 4 cones
= 525 - 1.47
= 523.53 cm³
Answer: The volume of wood in the entire stand is 523.53 cm³.
5. A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of its top, which is open, is 5 cm. It is filled with water up to the brim. When lead shots, each of which is a sphere of radius 0.5 cm are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped in the vessel.
Solution:
For Cone: Radius (R) = 5 cm, Height (H) = 8 cm.
Volume of water in cone = (1/3)πR²H = (1/3)π(5)²(8) = 200π/3 cm³.
Volume of water flowed out = 1/4 × Volume of cone = 1/4 × (200π/3) = 50π/3 cm³.
For Lead Shot (Sphere): Radius (r) = 0.5 cm.
Volume of one lead shot = (4/3)πr³ = (4/3)π(0.5)³ = (4/3)π(0.125) = π/6 cm³.
Let 'n' be the number of lead shots.
n × Volume of one shot = Volume of water flowed out
n × (π/6) = 50π/3
n = (50π/3) × (6/π)
n = 50 × 2 = 100.
Answer: The number of lead shots dropped is 100.
6. A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, which is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the pole, given that 1 cm³ of iron has approximately 8g mass. (Use π = 3.14)
Solution:
First Cylinder (Lower): Radius (R) = 24/2 = 12 cm, Height (H) = 220 cm.
Volume V1 = πR²H = 3.14 × 12² × 220 = 3.14 × 144 × 220 = 99475.2 cm³.
Second Cylinder (Upper): Radius (r) = 8 cm, Height (h) = 60 cm.
Volume V2 = πr²h = 3.14 × 8² × 60 = 3.14 × 64 × 60 = 12057.6 cm³.
Total Volume = V1 + V2 = 99475.2 + 12057.6 = 111532.8 cm³.
Mass = Volume × Density
= 111532.8 × 8 grams
= 892262.4 grams
= 892.26 kg.
Answer: The mass of the pole is 892.26 kg.
7. A solid consisting of a right circular cone of height 120 cm and radius 60 cm standing on a hemisphere of radius 60 cm is placed upright in a right circular cylinder full of water such that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is 60 cm and its height is 180 cm.
Solution:
Radius (r) is same for cone, hemisphere, and cylinder = 60 cm.
Height of Cone (h_cone) = 120 cm.
Height of Cylinder (h_cyl) = 180 cm.
Volume of Cylinder = πr²h_cyl = π(60)²(180) = 648000π cm³.
Volume of Solid = Volume of Cone + Volume of Hemisphere
= (1/3)πr²h_cone + (2/3)πr³
= (1/3)π(60)²(120) + (2/3)π(60)³
= 144000π + 144000π
= 288000π cm³.
Volume of water left = Volume of Cylinder - Volume of Solid
= 648000π - 288000π
= 360000π cm³
= 360000 × (22/7)
= 1131428.57 cm³
= 1.131 m³ (approx).
Answer: The volume of water left in the cylinder is 1.131 m³.
8. A spherical glass vessel has a cylindrical neck 8 cm long, 2 cm in diameter; the diameter of the spherical part is 8.5 cm. By measuring the amount of water it holds, a child finds its volume to be 345 cm³. Check whether she is correct, taking the above as the inside measurements, and π = 3.14.
Solution:
Cylindrical neck: Length (h) = 8 cm, Diameter = 2 cm, so Radius (r) = 1 cm.
Volume of neck = πr²h = 3.14 × 1² × 8 = 25.12 cm³.
Spherical part: Diameter = 8.5 cm, so Radius (R) = 4.25 cm.
Volume of sphere = (4/3)πR³ = (4/3) × 3.14 × (4.25)³
= 1.333... × 3.14 × 76.765...
≈ 321.39 cm³.
Total Volume = Volume of sphere + Volume of neck
= 321.39 + 25.12
= 346.51 cm³.
The child found the volume to be 345 cm³.
Since 346.51 ≠ 345, the child is incorrect.
Answer: She is not correct. The correct volume is approximately 346.51 cm³.