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CBSE Class 10 Maths Chapter 12 Surface Areas and Volumes Model Questions - 3 Marks - Part 2

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SECTION C — Short Answer Type Questions [3 Marks Each]

• 3 Marks Q26. A spherical ball of iron of radius $6\text{ cm}$ is melted and recast into three smaller spherical balls. If the radii of two of them are $3\text{ cm}$ and $4\text{ cm}$ respectively, find the radius of the third ball.
  Solution:
  Volume of original sphere = $\frac{4}{3}\pi (6)^3 = \frac{4}{3}\pi (216)$.
  Sum of volumes of three smaller spheres = $\frac{4}{3}\pi (3^3 + 4^3 + r^3) = \frac{4}{3}\pi (27 + 64 + r^3) = \frac{4}{3}\pi (91 + r^3)$.
  Equating volumes: $216 = 91 + r^3 \implies r^3 = 125 \implies r = 5\text{ cm}$.
  Answer: $5\text{ cm}$.
• 3 Marks Q27. The radii of the ends of a frustum of a cone $45\text{ cm}$ high are $28\text{ cm}$ and $7\text{ cm}$. Find the volume and total surface area of the frustum. ($\text{Use }\pi = \frac{22}{7}$).
  Solution:
  Height $h = 45\text{ cm}$, $R = 28\text{ cm}$, $r = 7\text{ cm}$.
  Volume = $\frac{1}{3}\pi h (R^2 + r^2 + Rr) = \frac{1}{3} \times \frac{22}{7} \times 45 \times (28^2 + 7^2 + 28 \times 7)$
  $= \frac{22 \times 15}{7} \times (784 + 49 + 196) = \frac{330}{7} \times 1029 = 330 \times 147 = 48,510\text{ cm}^3$.
  Slant height $l = \sqrt{h^2 + (R - r)^2} = \sqrt{45^2 + (28 - 7)^2} = \sqrt{2025 + 441} = \sqrt{2466} \approx 49.66\text{ cm}$.
  Total Surface Area = $\pi(R + r)l + \pi R^2 + \pi r^2 = \frac{22}{7}(35)(49.66) + \frac{22}{7}(784 + 49) \approx 5462.6 + 2618 = 8080.6\text{ cm}^2$.
  Answer: Volume = $48,510\text{ cm}^3$; TSA $\approx 8080.6\text{ cm}^2$.
• 3 Marks Q28. A bucket made of metal sheet is in the form of a frustum of a cone of height $30\text{ cm}$ with radii of its lower and upper ends as $10\text{ cm}$ and $20\text{ cm}$ respectively. Find the capacity of the bucket in litres. Also, find the cost of milk which can completely fill the bucket at $\text{₹}40$ per litre. ($\text{Use }\pi = 3.14$)
  Solution:
  $h = 30\text{ cm}$, $r = 10\text{ cm}$, $R = 20\text{ cm}$.
  Volume = $\frac{1}{3}\pi h (R^2 + r^2 + Rr) = \frac{1}{3} \times 3.14 \times 30 \times (20^2 + 10^2 + 20 \times 10)$
  $= 31.4 \times (400 + 100 + 200) = 31.4 \times 700 = 21,980\text{ cm}^3 = 21.98\text{ litres}$.
  Cost of milk = $21.98 \times 40 = \text{₹}879.20$.
  Answer: Capacity = $21.98\text{ litres}$; Cost = $\text{₹}879.20$.
• 3 Marks Q29. The slant height of a frustum of a cone is $4\text{ cm}$ and the perimeters of its circular ends are $18\text{ cm}$ and $6\text{ cm}$. Find the curved surface area of the frustum.
  Solution:
  Let perimeters be $P_1 = 2\pi R = 18\text{ cm}$ and $P_2 = 2\pi r = 6\text{ cm}$.
  Curved Surface Area (CSA) of frustum = $\pi(R + r)l = \frac{1}{2}(2\pi R + 2\pi r)l = \frac{1}{2}(18 + 6) \times 4 = \frac{1}{2}(24) \times 4 = 48\text{ cm}^2$.
  Answer: $48\text{ cm}^2$.
• 3 Marks Q30. A wooden article was made by scooping out a hemisphere from each end of a solid cylinder. If the height of the cylinder is $10\text{ cm}$ and its base radius is $3.5\text{ cm}$, find the total surface area of the article.
  Solution:
  Radius $r = 3.5\text{ cm}$, height $h = 10\text{ cm}$.
  Total Surface Area = CSA of cylinder + CSA of two hemispheres = $2\pi r h + 2(2\pi r^2) = 2\pi r (h + 2r)$
  $= 2 \times \frac{22}{7} \times 3.5 \times (10 + 2(3.5)) = 22 \times 17 = 374\text{ cm}^2$.
  Answer: $374\text{ cm}^2$.
• 3 Marks Q31. A toy is in the form of a cone mounted on a hemisphere with the same radius. The diameter of the base of the cone is $6\text{ cm}$ and its height is $4\text{ cm}$. Find the surface area and volume of the toy. ($\text{Use }\pi = 3.14$).
  Solution:
  Radius $r = 3\text{ cm}$, height $h = 4\text{ cm}$, slant height $l = \sqrt{3^2 + 4^2} = 5\text{ cm}$.
  Surface Area = CSA of cone + CSA of hemisphere = $\pi r l + 2\pi r^2 = 3.14 \times 3 \times 5 + 2 \times 3.14 \times 3^2 = 47.1 + 56.52 = 103.62\text{ cm}^2$.
  Volume = Vol of cone + Vol of hemisphere = $\frac{1}{3}\pi r^2 h + \frac{2}{3}\pi r^3 = \frac{1}{3}(3.14)(3^2)(4) + \frac{2}{3}(3.14)(3^3) = 37.68 + 56.52 = 94.2\text{ cm}^3$.
  Answer: Surface Area = $103.62\text{ cm}^2$; Volume = $94.2\text{ cm}^3$.
• 3 Marks Q32. A tent consists of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are $2.1\text{ m}$ and $4\text{ m}$ respectively, and the slant height of the top is $2.8\text{ m}$, find the area of the canvas used for the tent. Also, find the cost of the canvas at $\text{₹}500$ per $\text{m}^2$.
  Solution:
  Radius $r = 2\text{ m}$, cylinder height $h = 2.1\text{ m}$, slant height $l = 2.8\text{ m}$.
  Area of canvas = CSA of cylinder + CSA of cone = $2\pi r h + \pi r l = \frac{22}{7} \times 2 \times (2(2.1) + 2.8) = \frac{44}{7} \times 7 = 44\text{ m}^2$.
  Cost = $44 \times 500 = \text{₹}22,000$.
  Answer: Area = $44\text{ m}^2$; Cost = $\text{₹}22,000$.
• 3 Marks Q33. A solid toy is in the form of a hemisphere surmounted by a right circular cone. The height of the cone is $2\text{ cm}$ and the diameter of the base is $4\text{ cm}$. Find the volume of the toy and compare it with the volume of a cylinder circumscribing it. ($\text{Use }\pi = 3.14$).
  Solution:
  Radius $r = 2\text{ cm}$, height of cone $h = 2\text{ cm}$.
  Volume of toy = $\frac{1}{3}\pi r^2 h + \frac{2}{3}\pi r^3 = \frac{1}{3}(3.14)(2^2)(2) + \frac{2}{3}(3.14)(2^3) = \frac{8}{3}\pi + \frac{16}{3}\pi = 8\pi = 25.12\text{ cm}^3$.
  Circumscribing cylinder: radius $R = 2\text{ cm}$, height $H = 2 + 2 = 4\text{ cm}$.
  Volume of cylinder = $\pi R^2 H = 3.14 \times 2^2 \times 4 = 16\pi = 50.24\text{ cm}^3$.
  Ratio = Volume of toy : Volume of cylinder = $8\pi : 16\pi = 1 : 2$.
  Answer: Volume of toy = $25.12\text{ cm}^3$; Ratio = $1 : 2$.
• 3 Marks Q34. A vessel is in the form of an inverted cone. Its height is $8\text{ cm}$ and the radius of its top, which is open, is $5\text{ cm}$. It is filled with water up to the brim. When lead shots, each of which is a sphere of radius $0.5\text{ cm}$, are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped.
  Solution:
  Volume of cone = $\frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (5)^2 (8) = \frac{200\pi}{3}\text{ cm}^3$.
  Volume of water that flows out = $\frac{1}{4} \times \frac{200\pi}{3} = \frac{50\pi}{3}\text{ cm}^3$.
  Volume of one lead shot = $\frac{4}{3}\pi (0.5)^3 = \frac{4}{3}\pi \left(\frac{1}{8}\right) = \frac{\pi}{6}\text{ cm}^3$.
  Number of lead shots = $\frac{\text{Volume of water spilled}}{\text{Volume of one shot}} = \frac{50\pi/3}{\pi/6} = \frac{50}{3} \times 6 = 100$.
  Answer: $100$ lead shots.
• 3 Marks Q35. A metallic right circular cone $20\text{ cm}$ high and whose vertical angle is $60^\circ$ is cut into two parts at the middle of its height by a plane parallel to its base. If the frustum so obtained is drawn into a wire of diameter $\frac{1}{16}\text{ cm}$, find the length of the wire.
  Solution:
  Height of cone $H = 20\text{ cm}$, vertical angle $60^\circ \implies$ half-angle $\theta = 30^\circ$.
  Radius at base $R = H \tan 30^\circ = 20 \times \frac{1}{\sqrt{3}} = \frac{20}{\sqrt{3}}\text{ cm}$.
  Cut at middle $\implies h = 10\text{ cm}$, upper radius $r = 10 \times \frac{1}{\sqrt{3}} = \frac{10}{\sqrt{3}}\text{ cm}$.
  Volume of frustum = $\frac{1}{3}\pi h (R^2 + r^2 + Rr) = \frac{1}{3}\pi (10) \left(\frac{400}{3} + \frac{100}{3} + \frac{200}{3}\right) = \frac{7000\pi}{9}\text{ cm}^3$.
  Wire radius $r_w = \frac{1}{32}\text{ cm}$, length $L$: $\pi r_w^2 L = \frac{7000\pi}{9} \implies \left(\frac{1}{32}\right)^2 L = \frac{7000}{9} \implies L = \frac{7000 \times 1024}{9} \approx 796,444.4\text{ cm} \approx 7964.44\text{ m}$.
  Answer: $\approx 7964.44\text{ m}$.
• 3 Marks Q36. The perimeter of the circular ends of a frustum of a cone are $44\text{ cm}$ and $8.4\pi\text{ cm}$ respectively. If the depth is $14\text{ cm}$, find its volume and curved surface area.
  Solution:
  $2\pi R = 44 \implies R = \frac{44 \times 7}{2 \times 22} = 7\text{ cm}$.
  $2\pi r = 8.4\pi \implies r = 4.2\text{ cm}$. Height $h = 14\text{ cm}$.
  Volume = $\frac{1}{3}\pi h (R^2 + r^2 + Rr) = \frac{1}{3} \times \frac{22}{7} \times 14 \times (7^2 + 4.2^2 + 7 \times 4.2) = \frac{308}{3} \times (49 + 17.64 + 29.4) = \frac{308 \times 96.04}{3} \approx 9860.5\text{ cm}^3$.
  Slant height $l = \sqrt{h^2 + (R - r)^2} = \sqrt{14^2 + (7 - 4.2)^2} = \sqrt{196 + 7.84} = \sqrt{203.84} \approx 14.28\text{ cm}$.
  CSA = $\pi(R + r)l = \frac{22}{7}(7 + 4.2)(14.28) = \frac{22}{7} \times 11.2 \times 14.28 \approx 502.66\text{ cm}^2$.
  Answer: Volume $\approx 9860.5\text{ cm}^3$; CSA $\approx 502.66\text{ cm}^2$.
• 3 Marks Q37. A container shaped like a right circular cylinder having diameter $12\text{ cm}$ and height $15\text{ cm}$ is full of ice cream. The ice cream is to be filled into cones of height $12\text{ cm}$ and diameter $6\text{ cm}$, having a hemispherical shape on the top. Find the number of such cones which can be filled with ice cream.
  Solution:
  Cylinder radius $R = 6\text{ cm}$, height $H = 15\text{ cm}$. Volume = $\pi (6)^2 (15) = 540\pi\text{ cm}^3$.
  Cone radius $r = 3\text{ cm}$, height $h = 12\text{ cm}$.
  Volume of one ice cream cone = $\frac{1}{3}\pi r^2 h + \frac{2}{3}\pi r^3 = \frac{1}{3}\pi(3)^2(12) + \frac{2}{3}\pi(3)^3 = 36\pi + 18\pi = 54\pi\text{ cm}^3$.
  Number of cones = $\frac{540\pi}{54\pi} = 10$.
  Answer: $10$ cones.
• 3 Marks Q38. A solid iron pole consists of a cylinder of height $220\text{ cm}$ and base diameter $24\text{ cm}$, which is surmounted by another cylinder of height $60\text{ cm}$ and radius $8\text{ cm}$. Find the mass of the pole, given that $1\text{ cm}^3$ of iron has $8\text{ g}$ mass. ($\text{Use }\pi = 3.14$).
  Solution:
  Lower cylinder: $r_1 = 12\text{ cm}$, $h_1 = 220\text{ cm}$. Upper cylinder: $r_2 = 8\text{ cm}$, $h_2 = 60\text{ cm}$.
  Total volume = $\pi r_1^2 h_1 + \pi r_2^2 h_2 = 3.14 \times (12^2 \times 220 + 8^2 \times 60) = 3.14 \times (31680 + 3840) = 3.14 \times 35520 = 111,532.8\text{ cm}^3$.
  Total mass = $111,532.8 \times 8\text{ g} = 892,262.4\text{ g} = 892.26\text{ kg}$.
  Answer: $892.26\text{ kg}$.
• 3 Marks Q39. Water flows at the rate of $10\text{ km/h}$ through a cylindrical pipe of internal diameter $20\text{ mm}$ into a circular tank of base radius $40\text{ cm}$ and height $2\text{ m}$. Find the time taken to fill the tank completely.
  Solution:
  Pipe radius $r = 10\text{ mm} = 1\text{ cm} = 0.01\text{ m}$. Speed = $10\text{ km/h} = 10,000\text{ m/h}$.
  Tank radius $R = 40\text{ cm} = 0.4\text{ m}$, height $H = 2\text{ m}$.
  Volume of tank = $\pi R^2 H = \pi (0.4)^2 (2) = 0.32\pi\text{ m}^3$.
  Volume of water per hour through pipe = $\pi r^2 \times \text{speed} = \pi (0.01)^2 \times 10,000 = \pi\text{ m}^3\text{/h}$.
  Time taken = $\frac{0.32\pi}{\pi} = 0.32\text{ hours} = 19.2\text{ minutes}$.
  Answer: $19.2\text{ minutes}$ (or $19\text{ minutes } 12\text{ seconds}$).
• 3 Marks Q40. A cylindrical pipe has inner diameter $7\text{ cm}$ and water flows through it at $5\text{ m/h}$. Find the volume of water discharged in $30\text{ minutes}$ in litres.
  Solution:
  Radius $r = 3.5\text{ cm} = 0.035\text{ m}$. In $30\text{ minutes}$ ($0.5\text{ hours}$), length of water column = $5 \times 0.5 = 2.5\text{ m}$.
  Volume = $\pi r^2 h = \frac{22}{7} \times (0.035)^2 \times 2.5 = \frac{22}{7} \times 0.001225 \times 2.5 = 0.009625\text{ m}^3 = 9.625\text{ litres}$.
  Answer: $9.625\text{ litres}$.
• 3 Marks Q41. A conical vessel of radius $6\text{ cm}$ and height $8\text{ cm}$ is completely filled with water. A sphere is lowered into the water such that it touches the inner boundaries of the cone. Find the volume of water overflowing from the cone.
  Solution:
  Cone radius $R = 6\text{ cm}$, height $H = 8\text{ cm}$, slant height $l = 10\text{ cm}$.
  Radius of inscribed sphere $r = \frac{R \cdot H}{R + l} = \frac{6 \times 8}{6 + 10} = \frac{48}{16} = 3\text{ cm}$.
  Volume of sphere = $\frac{4}{3}\pi r^3 = \frac{4}{3}\pi (3)^3 = 36\pi\text{ cm}^3 \approx 113.1\text{ cm}^3$.
  Answer: $36\pi\text{ cm}^3$ (or $113.1\text{ cm}^3$).
• 3 Marks Q42. A juice seller was serving his customers using glasses. The inner diameter of a cylindrical glass was $5\text{ cm}$, but the bottom of the glass had a hemispherical raised portion which reduced the capacity of the glass. If the height of a glass was $10\text{ cm}$, find the apparent capacity of the glass and its actual capacity. ($\text{Use }\pi = 3.14$).
  Solution:
  Radius $r = 2.5\text{ cm}$, height $h = 10\text{ cm}$.
  Apparent capacity = $\pi r^2 h = 3.14 \times (2.5)^2 \times 10 = 196.25\text{ cm}^3$.
  Actual capacity = Apparent capacity - Volume of hemisphere = $196.25 - \frac{2}{3}\pi (2.5)^3 = 196.25 - 32.71 = 163.54\text{ cm}^3$.
  Answer: Apparent capacity = $196.25\text{ cm}^3$; Actual capacity = $163.54\text{ cm}^3$.
• 3 Marks Q43. A 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 $14\text{ mm}$ and the diameter of the capsule is $5\text{ mm}$. Find its surface area and volume.
  Solution:
  Radius $r = 2.5\text{ mm}$, length of cylindrical part $h = 14 - 2(2.5) = 9\text{ mm}$.
  Surface Area = CSA of cylinder + CSA of two hemispheres = $2\pi r h + 2(2\pi r^2) = 2\pi r (h + 2r)$
  $= 2 \times \frac{22}{7} \times 2.5 \times (9 + 5) = \frac{44 \times 2.5 \times 14}{7} = 220\text{ mm}^2$.
  Volume = Vol of cylinder + Vol of two hemispheres = $\pi r^2 h + \frac{4}{3}\pi r^3 = \frac{22}{7} \times (2.5)^2 \times 9 + \frac{4}{3} \times \frac{22}{7} \times (2.5)^3 \approx 176.79 + 32.74 = 209.53\text{ mm}^3$.
  Answer: Surface Area = $220\text{ mm}^2$; Volume = $209.53\text{ mm}^3$.
• 3 Marks Q44. A solid cylinder of radius $7\text{ cm}$ and height $10\text{ cm}$ has a conical cavity of same height and base radius drilled out from one end. Find the total surface area and volume of the remaining solid.
  Solution:
  Radius $r = 7\text{ cm}$, height $h = 10\text{ cm}$, slant height $l = \sqrt{7^2 + 10^2} = \sqrt{149} \approx 12.2\text{ cm}$.
  Total Surface Area = CSA of cylinder + base area + CSA of cone = $2\pi r h + \pi r^2 + \pi r l$
  $= \frac{22}{7} \times 7 \times (2(10) + 7 + 12.2) = 22 \times 39.2 = 862.4\text{ cm}^2$.
  Volume = Volume of cylinder - Volume of cone = $\pi r^2 h - \frac{1}{3}\pi r^2 h = \frac{2}{3}\pi r^2 h$
  $= \frac{2}{3} \times \frac{22}{7} \times 7^2 \times 10 = \frac{3080}{3} = 1026.67\text{ cm}^3$.
  Answer: Surface Area = $862.4\text{ cm}^2$; Volume = $1026.67\text{ cm}^3$.
• 3 Marks Q45. A copper rod of diameter $1\text{ cm}$ and length $8\text{ cm}$ is drawn into a wire of length $18\text{ m}$ of uniform thickness. Find the thickness (diameter) of the wire.
  Solution:
  Rod: radius $r_1 = 0.5\text{ cm}$, length $h_1 = 8\text{ cm}$. Volume = $\pi (0.5)^2 (8) = 2\pi\text{ cm}^3$.
  Wire: length $h_2 = 18\text{ m} = 1800\text{ cm}$, radius $r_2$.
  Volume of wire = $\pi r_2^2 (1800) = 2\pi \implies 1800 r_2^2 = 2 \implies r_2^2 = \frac{2}{1800} = \frac{1}{900} \implies r_2 = \frac{1}{30}\text{ cm}$.
  Thickness (diameter) = $2 r_2 = \frac{2}{30} = \frac{1}{15}\text{ cm}$ (or $\approx 0.067\text{ cm}$).
  Answer: $\frac{1}{15}\text{ cm}$.
• 3 Marks Q46. A solid metallic sphere of radius $8\text{ cm}$ is melted and recast into small conical bullets, each of base radius $1.6\text{ cm}$ and height $2\text{ cm}$. Find the number of bullets so formed.
  Solution:
  Volume of sphere = $\frac{4}{3}\pi (8)^3 = \frac{2048\pi}{3}\text{ cm}^3$.
  Volume of one conical bullet = $\frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (1.6)^2 (2) = \frac{5.12\pi}{3}\text{ cm}^3$.
  Number of bullets = $\frac{2048\pi/3}{5.12\pi/3} = \frac{2048}{5.12} = 400$.
  Answer: $400$ bullets.
• 3 Marks Q47. A bucket of height $24\text{ cm}$ and radii of its circular ends as $15\text{ cm}$ and $5\text{ cm}$ is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is $12\text{ cm}$, find the radius and slant height of the heap.
  Solution:
  Volume of frustum bucket = $\frac{1}{3}\pi (24) (15^2 + 5^2 + 15 \times 5) = 8\pi (225 + 25 + 75) = 8\pi (325) = 2600\pi\text{ cm}^3$.
  Volume of conical heap = $\frac{1}{3}\pi R^2 (12) = 4\pi R^2$.
  Equating volumes: $4\pi R^2 = 2600\pi \implies R^2 = 650 \implies R = \sqrt{650} \approx 25.5\text{ cm}$.
  Slant height $l = \sqrt{R^2 + H^2} = \sqrt{650 + 12^2} = \sqrt{650 + 144} = \sqrt{794} \approx 28.18\text{ cm}$.
  Answer: Radius $\approx 25.5\text{ cm}$; Slant height $\approx 28.18\text{ cm}$.
• 3 Marks Q48. A hollow cylindrical copper pipe is $21\text{ dm}$ long. Its outer and inner diameters are $10\text{ cm}$ and $6\text{ cm}$ respectively. Find the volume of the copper and the total surface area of the pipe.
  Solution:
  Height $h = 21\text{ dm} = 210\text{ cm}$. Outer radius $R = 5\text{ cm}$, inner radius $r = 3\text{ cm}$.
  Volume of copper = $\pi (R^2 - r^2) h = \frac{22}{7} (5^2 - 3^2) (210) = \frac{22}{7} (16) (210) = 22 \times 16 \times 30 = 10,560\text{ cm}^3$.
  Total Surface Area = Outer CSA + Inner CSA + 2(Area of base rings) = $2\pi R h + 2\pi r h + 2\pi(R^2 - r^2)$
  $= 2 \times \frac{22}{7} \times 210 \times (5 + 3) + 2 \times \frac{22}{7} \times (5^2 - 3^2) = 10,560 + 226.29 = 10,786.29\text{ cm}^2$.
  Answer: Volume = $10,560\text{ cm}^3$; Total Surface Area = $10,786.29\text{ cm}^2$.
• 3 Marks Q49. A solid iron sphere of radius $r$ is melted and recast into a solid right circular cylinder of height $r$. Find the radius of the base of the cylinder and the ratio of their total surface areas.
  Solution:
  Volume of sphere = $\frac{4}{3}\pi r^3$. Volume of cylinder = $\pi R^2 (r)$.
  $\pi R^2 r = \frac{4}{3}\pi r^3 \implies R^2 = \frac{4}{3}r^2 \implies R = \frac{2}{\sqrt{3}}r$.
  TSA of sphere = $4\pi r^2$.
  TSA of cylinder = $2\pi R (R + h) = 2\pi \left(\frac{2}{\sqrt{3}}r\right)\left(\frac{2}{\sqrt{3}}r + r\right) = \frac{4\pi r}{\sqrt{3}} \times \left(\frac{2+\sqrt{3}}{\sqrt{3}}\right)r = \frac{4(2+\sqrt{3})}{3}\pi r^2$.
  Ratio = $4\pi r^2 : \frac{4(2+\sqrt{3})}{3}\pi r^2 = 3 : (2+\sqrt{3})$.
  Answer: Radius = $\frac{2}{\sqrt{3}}r$; Ratio = $3 : (2+\sqrt{3})$.
• 3 Marks Q50. A right circular cylinder and a cone have equal bases and equal heights. If their curved surface areas are in the ratio $8:5$, show that the ratio of nzthe radius to the height of the cylinder is $3:4$.
  Solution:
  Let radius be $r$ and height be $h$ for both. Slant height of cone $l = \sqrt{r^2 + h^2}$.
  CSA of cylinder = $2\pi r h$. CSA of cone = $\pi r l$.
  Ratio = $\frac{2\pi r h}{\pi r l} = \frac{2h}{l} = \frac{8}{5} \implies \frac{h}{l} = \frac{4}{5} \implies \frac{h^2}{l^2} = \frac{16}{25}$.
  Since $l^2 = r^2 + h^2$, we have $\frac{h^2}{r^2 + h^2} = \frac{16}{25} \implies 25h^2 = 16r^2 + 16h^2 \implies 9h^2 = 16r^2$.
  $\frac{r^2}{h^2} = \frac{9}{16} \implies \frac{r}{h} = \frac{3}{4}$. Hence verified ($3:4$).
  Answer: Verified ($r:h = 3:4$).

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