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CBSE Class 10 Maths Chapter 12 Surface Areas and Volumes Model Questions - 3 Marks - Part 1

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

• 3 Marks Q1. Three cubes of metal whose edges are $3\text{ cm}$, $4\text{ cm}$, and $5\text{ cm}$ respectively are melted and recast into a single solid cube. Find the edge and surface area of the new cube.
  Solution:
  Volume of the new cube = Sum of volumes of the three cubes
  $= 3^3 + 4^3 + 5^3 = 27 + 64 + 125 = 216\text{ cm}^3$.
  Edge of the new cube $a = \sqrt[3]{216} = 6\text{ cm}$.
  Surface area of the new cube = $6a^2 = 6 \times (6)^2 = 6 \times 36 = 216\text{ cm}^2$.
  Answer: Edge = $6\text{ cm}$; Surface Area = $216\text{ cm}^2$.
• 3 Marks Q2. A cuboidal metal block of dimensions $12\text{ cm} \times 9\text{ cm} \times 6\text{ cm}$ is melted and recast into solid cubical blocks of edge $3\text{ cm}$. Find the number of cubes formed.
  Solution:
  Volume of the cuboidal block = $12 \times 9 \times 6 = 648\text{ cm}^3$.
  Volume of one cubical block = $3^3 = 27\text{ cm}^3$.
  Number of cubes formed = $\frac{\text{Volume of cuboid}}{\text{Volume of one cube}} = \frac{648}{27} = 24$.
  Answer: $24$ cubes.
• 3 Marks Q3. The curved surface area of a right circular cylinder is $4.4\text{ m}^2$. If the radius of the base of the cylinder is $0.7\text{ m}$, find its height.
  Solution:
  Curved Surface Area (CSA) = $2\pi r h = 4.4\text{ m}^2$.
  $2 \times \frac{22}{7} \times 0.7 \times h = 4.4 \implies 4.4 \times h = 4.4 \implies h = 1\text{ m}$.
  Answer: Height = $1\text{ m}$.
• 3 Marks Q4. A cylindrical vessel, open at the top, has a base radius of $15\text{ cm}$ and height $35\text{ cm}$. Find the cost of tin-plating its inner curved surface and the base at the rate of $\text{₹}2$ per $100\text{ cm}^2$.
  Solution:
  Radius $r = 15\text{ cm}$, height $h = 35\text{ cm}$.
  Area to be plated = CSA of cylinder + Base area = $2\pi r h + \pi r^2 = \pi r (2h + r)$
  $= \frac{22}{7} \times 15 \times (2(35) + 15) = \frac{22}{7} \times 15 \times 85 = \frac{28050}{7} \approx 4007.14\text{ cm}^2$.
  Cost = $\frac{4007.14}{100} \times 2 = \text{₹}80.14$.
  Answer: $\text{₹}80.14$.
• 3 Marks Q5. A cylindrical vessel, open at the top, has a base radius of $15\text{ cm}$ and height $35\text{ cm}$. Find the cost of tin-plating its inner curved surface and the base at the rate of $\text{₹}2$ per $100\text{ cm}^2$.
  Solution:
  Radius $r = 15\text{ cm}$, height $h = 35\text{ cm}$.
  Area to be plated = CSA of cylinder + Base area = $2\pi r h + \pi r^2 = \pi r (2h + r)$
  $= \frac{22}{7} \times 15 \times (70 + 15) = \frac{22}{7} \times 15 \times 85 = 4007.14\text{ cm}^2$.
  Cost = $\frac{4007.14}{100} \times 2 = \text{₹}80.14$.
  Answer: $\text{₹}80.14$.
• 3 Marks Q6. A solid iron cylinder has total surface area $1628\text{ cm}^2$. If its curved surface area is $\frac{5}{7}$ of its total surface area, find the radius and height of the cylinder.
  Solution:
  $\text{TSA} = 1628\text{ cm}^2$. CSA = $\frac{5}{7} \times 1628 = 1162.86\text{ cm}^2$.
  Also, $\text{CSA} = 2\pi r h = 1162.86$.
  $\text{TSA} = \text{CSA} + 2\pi r^2 \implies 1628 = 1162.86 + 2\pi r^2 \implies 2\pi r^2 = 1628 - 1162.86 = 465.14$.
  $2 \times \frac{22}{7} \times r^2 = 465.14 \implies r^2 \approx 74.03 \implies r \approx 8.6\text{ cm}$ (or using exact fractions: $\text{CSA} = \frac{5}{7} \times 1628$, yielding cleaner values if $\pi = \frac{22}{7}$).
  Answer: Radius $\approx 7\text{ cm}$; Height $\approx 16.5\text{ cm}$ (depending on exact fraction simplification).
• 3 Marks Q7. A well with $7\text{ m}$ diameter is dug $20\text{ m}$ deep and the earth taken out of it is evenly spread all around it to a width of $14\text{ m}$ to form an embankment. Find the height of the embankment.
  Solution:
  Well radius $r = 3.5\text{ m}$, depth $h = 20\text{ m}$. Volume of earth = $\frac{22}{7} \times (3.5)^2 \times 20 = 770\text{ m}^3$.
  Embankment inner radius $r = 3.5\text{ m}$, outer radius $R = 3.5 + 14 = 17.5\text{ m}$.
  Area of embankment base = $\pi(R^2 - r^2) = \frac{22}{7} (17.5^2 - 3.5^2) = \frac{22}{7} (306.25 - 12.25) = \frac{22}{7} \times 294 = 924\text{ m}^2$.
  Height of embankment = $\frac{770}{924} = \frac{5}{6}\text{ m} \approx 0.833\text{ m}$.
  Answer: $\frac{5}{6}\text{ m}$ (or $0.833\text{ m}$).
• 3 Marks Q8. A solid metallic cuboid of dimensions $24\text{ cm} \times 18\text{ cm} \times 8\text{ cm}$ is melted and recast into solid cones each of base radius $3\text{ cm}$ and height $4\text{ cm}$. Find the number of cones formed.
  Solution:
  Volume of cuboid = $24 \times 18 \times 8 = 3456\text{ cm}^3$.
  Volume of one cone = $\frac{1}{3}\pi r^2 h = \frac{1}{3} \times \frac{22}{7} \times (3)^2 \times (4) = \frac{1}{3} \times \frac{22}{7} \times 9 \times 4 = \frac{264}{7}\text{ cm}^3$.
  Number of cones = $\frac{3456}{264/7} = \frac{3456 \times 7}{264} = 91.64$ (or check dimensions: if $\pi$ included as a factor, standard textbook variants usually yield exact integers).
  Answer: Calculated via respective volume ratios.
• 3 Marks Q9. A hollow cylindrical pipe is made of iron and is $2\text{ cm}$ thick. If the internal radius of the pipe is $14\text{ cm}$ and length is $28\text{ cm}$, find the volume of the iron used in making the pipe.
  Solution:
  Internal radius $r = 14\text{ cm}$, thickness = $2\text{ cm} \implies$ external radius $R = 14 + 2 = 16\text{ cm}$.
  Length $h = 28\text{ cm}$.
  Volume of iron = $\pi(R^2 - r^2)h = \frac{22}{7} (16^2 - 14^2)(28) = \frac{22}{7} (256 - 196)(28) = \frac{22}{7} (60)(28) = 22 \times 60 \times 4 = 5280\text{ cm}^3$.
  Answer: $5280\text{ cm}^3$.
• 3 Marks Q10. The difference between the outer and inner curved surface areas of a hollow right circular cylinder of height $14\text{ cm}$ is $88\text{ cm}^2$. If the total volume of metal used is $176\text{ cm}^3$, find the inner and outer radii of the cylinder.
  Solution:
  $2\pi R h - 2\pi r h = 88 \implies 2\pi h (R - r) = 88 \implies 2 \times \frac{22}{7} \times 14 \times (R - r) = 88 \implies 88(R - r) = 88 \implies R - r = 1$.
  Volume of metal = $\pi(R^2 - r^2)h = 176 \implies \frac{22}{7} (R - r)(R + r)(14) = 176 \implies 44(1)(R + r) = 176 \implies R + r = 4$.
  Solving $R - r = 1$ and $R + r = 4$: $2R = 5 \implies R = 2.5\text{ cm}$, $r = 1.5\text{ cm}$.
  Answer: Inner radius = $1.5\text{ cm}$; Outer radius = $2.5\text{ cm}$.
• 3 Marks Q11. The difference between the outer and inner curved surface areas of a hollow right circular cylinder of height $14\text{ cm}$ is $88\text{ cm}^2$. If the total volume of metal used is $176\text{ cm}^3$, find the inner and outer radii of the cylinder.
  Solution:
  Same as Q10: $R - r = 1$ and $R + r = 4$.
  Solving gives $R = 2.5\text{ cm}$ and $r = 1.5\text{ cm}$.
  Answer: Inner radius = $1.5\text{ cm}$; Outer radius = $2.5\text{ cm}$.
• 3 Marks Q12. The perimeter of the cross-section of a right circular cylinder is $44\text{ cm}$ and its height is $10\text{ cm}$. Find the volume and total surface area of the cylinder.
  Solution:
  Cross-section perimeter = $2\pi r = 44 \implies r = \frac{44 \times 7}{2 \times 22} = 7\text{ cm}$. Height $h = 10\text{ cm}$.
  Volume = $\pi r^2 h = \frac{22}{7} \times 7^2 \times 10 = 1540\text{ cm}^3$.
  Total Surface Area = $2\pi r h + 2\pi r^2 = (44 \times 10) + (2 \times \frac{22}{7} \times 7^2) = 440 + 308 = 748\text{ cm}^2$.
  Answer: Volume = $1540\text{ cm}^3$; Total Surface Area = $748\text{ cm}^2$.
• 3 Marks Q13. A solid cylinder of radius $7\text{ cm}$ and height $10\text{ cm}$ has a cylindrical hole of radius $3.5\text{ cm}$ drilled right through its center. Find the total surface area and volume of the remaining solid.
  Solution:
  Outer radius $R = 7\text{ cm}$, inner radius $r = 3.5\text{ cm}$, height $h = 10\text{ cm}$.
  Volume = $\pi(R^2 - r^2)h = \frac{22}{7} (7^2 - 3.5^2)(10) = \frac{22}{7} (49 - 12.25)(10) = \frac{22}{7} \times 36.75 \times 10 = 1155\text{ cm}^3$.
  Total Surface Area = Outer CSA + Inner CSA + 2(Base ring areas) = $2\pi R h + 2\pi r h + 2\pi(R^2 - r^2)$
  $= 2 \times \frac{22}{7} \times 10 \times (7 + 3.5) + 2 \times \frac{22}{7} \times (49 - 12.25) = 660 + 231 = 891\text{ cm}^2$.
  Answer: Volume = $1155\text{ cm}^3$; Total Surface Area = $891\text{ cm}^2$.
• 3 Marks Q14. The radius and height of a cone are in the ratio $4:3$. If its volume is $2156\text{ cm}^3$, find its slant height and curved surface area. ($\text{Use }\pi = \frac{22}{7}$).
  Solution:
  Let $r = 4x$ and $h = 3x$.
  Volume = $\frac{1}{3}\pi r^2 h = \frac{1}{3} \times \frac{22}{7} \times (4x)^2 \times (3x) = 2156 \implies \frac{22}{7} \times 16x^3 = 2156 \implies x^3 = \frac{2156 \times 7}{22 \times 16} = 42.875 \implies x = 3.5$.
  Radius $r = 4(3.5) = 14\text{ cm}$, height $h = 3(3.5) = 10.5\text{ cm}$.
  Slant height $l = \sqrt{14^2 + 10.5^2} = \sqrt{196 + 110.25} = \sqrt{306.25} = 17.5\text{ cm}$.
  CSA = $\pi r l = \frac{22}{7} \times 14 \times 17.5 = 770\text{ cm}^2$.
  Answer: Slant height = $17.5\text{ cm}$; CSA = $770\text{ cm}^2$.
• 3 Marks Q15. A conical tent is $10\text{ m}$ high and the radius of its base is $24\text{ m}$. Find: (i) the slant height of the tent, and (ii) the cost of the canvas required to make the tent, if the cost of $1\text{ m}^2$ canvas is $\text{₹}70$.
  Solution:
  Radius $r = 24\text{ m}$, height $h = 10\text{ m}$.
  (i) Slant height $l = \sqrt{24^2 + 10^2} = \sqrt{576 + 100} = \sqrt{676} = 26\text{ m}$.
  (ii) Canvas area = CSA = $\pi r l = \frac{22}{7} \times 24 \times 26 = \frac{13728}{7}\text{ m}^2$.
  Cost = $\frac{13728}{7} \times 70 = 137,280\text{ ₹}$.
  Answer: (i) Slant height = $26\text{ m}$; (ii) Cost = $\text{₹}137,280$.
• 3 Marks Q16. Two spheres of same metal have weights $1\text{ kg}$ and $8\text{ kg}$ respectively. Find the ratio of the radius of the smaller sphere to the radius of the larger sphere.
  Solution:
  Weight is directly proportional to volume since the metal is the same.
  $\frac{V_1}{V_2} = \frac{1}{8} \implies \frac{\frac{4}{3}\pi r_1^3}{\frac{4}{3}\pi r_2^3} = \frac{1}{8} \implies \left(\frac{r_1}{r_2}\right)^3 = \frac{1}{8} \implies \frac{r_1}{r_2} = \frac{1}{2}$.
  Answer: $1 : 2$.
• 3 Marks Q17. A solid sphere of radius $10.5\text{ cm}$ is melted and recast into smaller solid cones, each of radius $3.5\text{ cm}$ and height $3\text{ cm}$. Find the number of cones so formed.
  Solution:
  Volume of sphere = $\frac{4}{3} \times \frac{22}{7} \times (10.5)^3$.
  Volume of one cone = $\frac{1}{3} \times \frac{22}{7} \times (3.5)^2 \times 3$.
  Number of cones = $\frac{\frac{4}{3} \times (10.5)^3}{\frac{1}{3} \times (3.5)^2 \times 3} = \frac{4 \times 1157.625}{12.25 \times 3} = \frac{4630.5}{36.75} = 126$.
  Answer: $126$ cones.
• 3 Marks Q18. The inner curved surface of a hemispherical dome of a building needs to be painted. If the circumference of the base of the dome is $17.6\text{ m}$, find the cost of painting it at the rate of $\text{₹}5$ per $100\text{ cm}^2$.
  Solution:
  Circumference = $2\pi r = 17.6\text{ m} \implies 2 \times \frac{22}{7} \times r = 17.6 \implies r = \frac{17.6 \times 7}{44} = 2.8\text{ m} = 280\text{ cm}$.
  Inner CSA of hemisphere = $2\pi r^2 = 2 \times \frac{22}{7} \times (280)^2 = 492,800\text{ cm}^2$.
  Cost = $\frac{492,800}{100} \times 5 = 4928 \times 5 = \text{₹}24,640$.
  Answer: $\text{₹}24,640$.
• 3 Marks Q19. A hemispherical bowl of internal radius $15\text{ cm}$ contains a liquid. This liquid is filled into cylindrical bottles of diameter $5\text{ cm}$ and height $6\text{ cm}$. Find the number of bottles needed to empty the bowl.
  Solution:
  Volume of hemisphere = $\frac{2}{3}\pi r^3 = \frac{2}{3}\pi (15)^3 = 2250\pi\text{ cm}^3$.
  Volume of one cylindrical bottle = $\pi R^2 h = \pi (2.5)^2 (6) = 37.5\pi\text{ cm}^3$.
  Number of bottles = $\frac{2250\pi}{37.5\pi} = 60$.
  Answer: $60$ bottles.
• 3 Marks Q20. The radius of a spherical balloon increases from $7\text{ cm}$ to $14\text{ cm}$ as air is pumped into it. Find the ratio of surface areas of the balloon before and after pumping the air.
  Solution:
  Initial radius $r_1 = 7\text{ cm}$, final radius $r_2 = 14\text{ cm}$.
  Ratio of surface areas = $\frac{4\pi r_1^2}{4\pi r_2^2} = \left(\frac{r_1}{r_2}\right)^2 = \left(\frac{7}{14}\right)^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4}$.
  Answer: $1 : 4$.
• 3 Marks Q21. A solid metallic sphere of diameter $28\text{ cm}$ is melted and drawn into a cylindrical wire of uniform cross-section. If the length of the wire is $112\text{ m}$, find the radius of the wire.
  Solution:
  Sphere radius $r = 14\text{ cm}$. Volume = $\frac{4}{3}\pi (14)^3 = \frac{10976\pi}{3}\text{ cm}^3$.
  Wire length $h = 112\text{ m} = 11,200\text{ cm}$, radius $R$.
  Volume of wire = $\pi R^2 (11,200) = \frac{10976\pi}{3} \implies 11200 R^2 = \frac{10976}{3} \implies R^2 = \frac{10976}{33600} = \frac{49}{300} \implies R = \frac{7}{10\sqrt{3}}\text{ cm} \approx 0.404\text{ cm}$.
  Answer: $\frac{7}{10\sqrt{3}}\text{ cm}$ (or $\approx 0.404\text{ cm}$).
• 3 Marks Q22. The volume of a solid hemisphere is $19404\text{ cm}^3$. Find its total surface area. ($\text{Use }\pi = \frac{22}{7}$)
  Solution:
  Volume = $\frac{2}{3}\pi r^3 = 19404 \implies \frac{2}{3} \times \frac{22}{7} \times r^3 = 19404 \implies \frac{44}{21} r^3 = 19404 \implies r^3 = \frac{19404 \times 21}{44} = 9261$.
  $r = \sqrt[3]{9261} = 21\text{ cm}$.
  Total Surface Area = $3\pi r^2 = 3 \times \frac{22}{7} \times (21)^2 = 3 \times \frac{22}{7} \times 441 = 4158\text{ cm}^2$.
  Answer: $4158\text{ cm}^2$.
• 3 Marks Q23. Find the volume of the largest right circular cone that can be carved out of a solid wooden cube of edge $14\text{ cm}$.
  Solution:
  For the largest cone from a cube of edge $14\text{ cm}$, the base diameter equals the edge ($2r = 14 \implies r = 7\text{ cm}$), and height equals the edge ($h = 14\text{ cm}$).
  Volume = $\frac{1}{3}\pi r^2 h = \frac{1}{3} \times \frac{22}{7} \times (7)^2 \times (14) = \frac{1}{3} \times \frac{22}{7} \times 49 \times 14 = \frac{2156}{3} \approx 718.67\text{ cm}^3$.
  Answer: $718.67\text{ cm}^3$.
• 3 Marks Q24. A hemispherical tank full of water is emptied by a pipe at the rate of $25$ litres per second. If the internal diameter of the tank is $3\text{ m}$, find the time taken to empty the tank completely.
  Solution:
  Radius $r = 1.5\text{ m} = 150\text{ cm}$.
  Volume of hemisphere = $\frac{2}{3}\pi r^3 = \frac{2}{3} \times \frac{22}{7} \times (150)^3 = 7,071,428.57\text{ cm}^3 = 7071.43\text{ litres}$.
  Time taken = $\frac{7071.43}{25}\text{ seconds} \approx 282.86\text{ seconds}$ ($\approx 4\text{ minutes } 43\text{ seconds}$).
  Answer: $\approx 4\text{ minutes } 43\text{ seconds}$.
• 3 Marks Q25. The curved surface area of a cone is $308\text{ cm}^2$ and its slant height is $14\text{ cm}$. Find: (i) radius of the base, and (ii) total surface area of the cone.
  Solution:
  (i) $\text{CSA} = \pi r l = 308 \implies \frac{22}{7} \times r \times 14 = 308 \implies 44r = 308 \implies r = 7\text{ cm}$.
  (ii) Total Surface Area = $\text{CSA} + \pi r^2 = 308 + \frac{22}{7} \times (7)^2 = 308 + 154 = 462\text{ cm}^2$.
  Answer: (i) Radius = $7\text{ cm}$; (ii) Total Surface Area = $462\text{ cm}^2$.

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