Chapter 12: Surface Areas and Volumes Exercise Questions & Answers
12 cubes each of volume 64 cm³ are joined end to end. Find the surface area of the resulting cuboid.
Answer & Reference:
Volume of each cube = 64 cm³
Side of each cube = ∛64 = 4 cm
When joined end to end, dimensions of cuboid:
Length = 4 + 4 = 8 cm, Breadth = 4 cm, Height = 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 Reference: Exercise 12.1, Question 1 (Page 166)
2A 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.
Answer & Reference:
Radius of hemisphere = 14/2 = 7 cm
Height of cylinder = Total height - Radius of hemisphere = 13 - 7 = 6 cm
Inner surface area = CSA of hemisphere + CSA of cylinder
= 2πr² + 2πrh = 2πr(r + h)
= 2 × (22/7) × 7 × (7 + 6) = 44 × 13 = 572 cm² Reference: Exercise 12.1, Question 2 (Page 166)
3A 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.
Answer & Reference:
Radius of cone and hemisphere = 3.5 cm
Height of cone = Total height - Radius of hemisphere = 15.5 - 3.5 = 12 cm
Slant height of cone (l) = √(r² + h²) = √(3.5² + 12²) = √(12.25 + 144) = √156.25 = 12.5 cm
TSA of toy = CSA of cone + CSA of hemisphere
= πrl + 2πr² = πr(l + 2r)
= (22/7) × 3.5 × (12.5 + 7) = 11 × 19.5 = 214.5 cm² Reference: Exercise 12.1, Question 3 (Page 166)
4A 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.
Answer & Reference:
Greatest diameter of hemisphere = Side of cube = 7 cm
Radius of hemisphere = 7/2 = 3.5 cm
Surface area of solid = TSA of cube - Base area of hemisphere + CSA of hemisphere
= 6a² - πr² + 2πr² = 6a² + πr²
= 6×7² + (22/7)×(3.5)² = 294 + 38.5 = 332.5 cm² Reference: Exercise 12.1, Question 4 (Page 166)
5A 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.
Answer & Reference:
Let edge of cube = l
Radius of hemisphere = l/2
Surface area of remaining solid = TSA of cube - Area of base of hemisphere + CSA of hemisphere
= 6l² - π(l/2)² + 2π(l/2)² = 6l² + π(l/2)²
= 6l² + (πl²)/4 = l²(6 + π/4) = l²(24 + π)/4 Reference: Exercise 12.1, Question 5 (Page 166)
6A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends. The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area.
Answer & Reference:
Diameter = 5 mm, so radius = 2.5 mm
Length of cylindrical part = Total length - 2×radius = 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) = (110/7) × 14 = 220 mm² Reference: Exercise 12.1, Question 6 (Page 166)
7A 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.)
Answer & Reference:
Radius of cylinder and cone = 4/2 = 2 m
Height of cylinder = 2.1 m, Slant height of cone = 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 = 44 × 500 = ₹22,000 Reference: Exercise 12.1, Question 7 (Page 167)
8From 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.
Answer & Reference:
Radius = 1.4/2 = 0.7 cm, Height = 2.4 cm
Slant height of cone (l) = √(r² + h²) = √(0.7² + 2.4²) = √(0.49 + 5.76) = √6.25 = 2.5 cm
TSA of remaining solid = CSA of cylinder + CSA of cone + Area of base of cylinder
= 2πrh + πrl + πr² = πr(2h + l + r)
= (22/7) × 0.7 × (2×2.4 + 2.5 + 0.7) = 2.2 × (4.8 + 2.5 + 0.7) = 2.2 × 8 = 17.6 cm² ≈ 18 cm² Reference: Exercise 12.1, Question 8 (Page 167)
9A 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.
Answer & Reference:
Radius = 3.5 cm, Height of cylinder = 10 cm
TSA of article = CSA of cylinder + 2×CSA of hemisphere
= 2πrh + 2×2πr² = 2πr(h + 2r)
= 2 × (22/7) × 3.5 × (10 + 7) = 22 × 17 = 374 cm² Reference: Exercise 12.1, Question 9 (Page 167)
10A 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.
Answer & Reference:
Radius of cone and hemisphere = 1 cm
Height of cone = 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 Reference: Exercise 12.2, Question 1 (Page 169)
11Rachel, 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.)
Answer & Reference:
Radius = 3/2 = 1.5 cm
Height of each cone = 2 cm
Height of cylinder = Total length - 2×height of cone = 12 - 4 = 8 cm
Volume of air = Volume of cylinder + 2×Volume of cone
= πr²h + 2×(1/3)πr²h_cone = πr²(h + (2/3)h_cone)
= π(1.5)²(8 + (2/3)×2) = π×2.25×(8 + 4/3) = π×2.25×(28/3) = π×21 = 21π cm³ ≈ 66 cm³ Reference: Exercise 12.2, Question 2 (Page 169)
12A 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.
Answer & Reference:
Radius = 2.8/2 = 1.4 cm
Length of cylindrical part = Total length - 2×radius = 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 + 1.867) ≈ 6.16×4.067 ≈ 25.05 cm³
Volume of 45 gulab jamuns = 45×25.05 ≈ 1127.25 cm³
Syrup volume = 30% of 1127.25 ≈ 338.175 cm³ Reference: Exercise 12.2, Question 3 (Page 170)
13A 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.
Answer & Reference:
Volume of cuboid = 15×10×3.5 = 525 cm³
Volume of one conical depression = (1/3)πr²h = (1/3)×(22/7)×(0.5)²×1.4 ≈ 0.367 cm³
Volume of 4 depressions = 4×0.367 ≈ 1.47 cm³
Volume of wood = Volume of cuboid - Volume of depressions = 525 - 1.47 ≈ 523.53 cm³ Reference: Exercise 12.2, Question 4 (Page 170)
14A 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.
Answer & Reference:
Volume of water in vessel = (1/3)πr²h = (1/3)×(22/7)×5²×8 ≈ 209.33 cm³
Volume of water that flows out = (1/4)×209.33 ≈ 52.33 cm³
Volume of one lead shot = (4/3)πr³ = (4/3)×(22/7)×(0.5)³ ≈ 0.5236 cm³
Number of lead shots = 52.33 / 0.5236 ≈ 100 Reference: Exercise 12.2, Question 5 (Page 170)
15A 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)
Answer & Reference:
Radius of larger cylinder = 24/2 = 12 cm
Volume of larger cylinder = πr²h = 3.14×12²×220 = 3.14×144×220 = 99475.2 cm³
Volume of smaller cylinder = πr²h = 3.14×8²×60 = 3.14×64×60 = 12057.6 cm³
Total volume = 99475.2 + 12057.6 = 111532.8 cm³
Mass = 111532.8 × 8 = 892262.4 g = 892.262 kg Reference: Exercise 12.2, Question 6 (Page 170)
16A 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.
Answer & Reference:
Volume of cylinder = πr²h = π×60²×180 = 648000π cm³
Volume of cone = (1/3)πr²h = (1/3)π×60²×120 = 144000π cm³
Volume of hemisphere = (2/3)πr³ = (2/3)π×60³ = 144000π cm³
Volume of solid = 144000π + 144000π = 288000π cm³
Volume of water left = Volume of cylinder - Volume of solid = 648000π - 288000π = 360000π cm³
= 360000 × (22/7) ≈ 1131428.57 cm³ Reference: Exercise 12.2, Question 7 (Page 170)
17A 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.
Answer & Reference:
Radius of spherical part = 8.5/2 = 4.25 cm
Volume of sphere = (4/3)πr³ = (4/3)×3.14×(4.25)³ ≈ 321.39 cm³
Radius of cylindrical neck = 2/2 = 1 cm
Volume of cylinder = πr²h = 3.14×1²×8 = 25.12 cm³
Total volume = 321.39 + 25.12 = 346.51 cm³
The child's measurement of 345 cm³ is approximately correct. Reference: Exercise 12.2, Question 8 (Page 170)