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CBSE Class 10 Maths Chapter 12 Surface Areas and Volumes Model Questions - 4 Marks

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SECTION D — Case-Based/Source-Based Integrated Questions [4 Marks Each]

• 4 Marks Q1. Case Study: The Cold Drink Can Factory
  A beverage manufacturing company designs a new cylindrical aluminum can to hold $500\text{ ml}$ of cold drink. The internal base radius of the can is $3.5\text{ cm}$ and its height is $14\text{ cm}$.
  a) State the formula for the curved surface area of a cylinder of radius $r$ and height $h$. (1 Mark)

  Answer:
  Result: $\text{CSA} = 2\pi rh$

  b) Calculate the inner curved surface area of the can. ($\text{Use }\pi = \frac{22}{7}$) (1 Mark)

  Answer:
  1. $\text{CSA} = 2 \times \frac{22}{7} \times 3.5 \times 14$
  Result: $308\text{ cm}^2$

  c) Find the total volume of the can in litres and check whether it can completely hold $500\text{ ml}$ of liquid (2 Marks)

  Answer:
  1. $\text{Volume} = \pi r^2 h = \frac{22}{7} \times (3.5)^2 \times 14 = 539\text{ cm}^3$
  2. $539\text{ cm}^3 = 0.539\text{ litres} = 539\text{ ml}$
  Result: Yes, it can completely hold $500\text{ ml}$ of liquid.
• 4 Marks Q2. Case Study: The Circus Tent Design
  An organizing committee sets up a large circus tent. The lower part of the tent is cylindrical with a height of $2.1\text{ m}$ and a diameter of $4\text{ m}$, while the upper part is a conical top with a slant height of $2.8\text{ m}$.
  a) Find the base radius of the cylindrical and conical parts. (1 Mark)

  Answer:
  Result: $r = \frac{d}{2} = \frac{4}{2} = 2\text{ m}$

  b) Calculate the curved surface area of the cylindrical lower section. (1 Mark)

  Answer:
  1. $\text{CSA} = 2\pi rh = 2 \times \frac{22}{7} \times 2 \times 2.1$
  Result: $26.4\text{ m}^2$

  c) Find the total area of the canvas required to make the complete tent. (2 Marks)

  Answer:
  1. $\text{Total Area} = \text{Cylindrical CSA} + \text{Conical CSA} = 2\pi rh + \pi rl$
  2. $\text{Total Area} = 26.4 + \frac{22}{7} \times 2 \times 2.8 = 26.4 + 17.6$
  Result: $44\text{ m}^2$
• 4 Marks Q3. Case Study: The Ice Cream Cone Treat (Hemisphere & Cone Combined)
  An ice cream vendor sells a special treat consisting of a right circular cone filled with ice cream, surmounted by a hemispherical scoop exactly matching the top radius of the cone. The radius of the base of the cone is $3\text{ cm}$ and its height is $4\text{ cm}$.
  a) Calculate the slant height of the conical portion (1 Mark)

  Answer:
  1. $l = \sqrt{r^2 + h^2} = \sqrt{3^2 + 4^2} = \sqrt{25}$
  Result: $5\text{ cm}$

  b) Find the volume of the hemispherical scoop of ice cream. ($\text{Use }\pi = 3.14$). (1 Mark)

  Answer:
  1. $\text{Volume} = \frac{2}{3}\pi r^3 = \frac{2}{3} \times 3.14 \times 3^3$
  Result: $56.52\text{ cm}^3$

  c) Calculate the total volume of the entire ice cream treat (cone plus hemisphere) (2 Marks)

  Answer:
  1. $\text{Cone Volume} = \frac{1}{3}\pi r^2 h = \frac{1}{3} \times 3.14 \times 9 \times 4 = 37.68\text{ cm}^3$
  2. $\text{Total Volume} = 37.68 + 56.52$
  Result: $94.2\text{ cm}^3$
• 4 Marks Q4. Case Study: The Metallic Sphere Recasting
  A metal workshop receives a solid spherical iron ball of radius $6\text{ cm}$ to be melted and recast into smaller uniform spherical bullets, each having a radius of $2\text{ cm}$.
  a) State the formula for the volume of a sphere of radius $r$. (1 Mark)

  Answer:
  Result: $V = \frac{4}{3}\pi r^3$

  b) Calculate the volume of the original large spherical iron ball. (1 Mark)

  Answer:
  1. $V = \frac{4}{3}\pi (6)^3 = \frac{4}{3}\pi (216)$
  Result: $288\pi\text{ cm}^3$

  c) Determine the total number of smaller spherical bullets that can be formed from the melted metal. (2 Marks)

  Answer:
  1. $\text{Number} = \frac{\text{Volume of large sphere}}{\text{Volume of small sphere}} = \frac{288\pi}{\frac{4}{3}\pi (2)^3} = \frac{288}{\frac{32}{3}}$
  Result: $27\text{ bullets}$
• 4 Marks Q5. Case Study: The Frustum Bucket Water Capacity
  A galvanized iron bucket used in construction is shaped like the frustum of a cone. The radii of the circular lower and upper ends are $10\text{ cm}$ and $20\text{ cm}$ respectively, and its vertical depth is $30\text{ cm}$.
  a) State the formula for the volume of a frustum of a cone with radii $r_1, r_2$ and height $h$. (1 Mark)

  Answer:
  Result: $V = \frac{1}{3}\pi h (r_1^2 + r_2^2 + r_1 r_2)$

  b) Calculate the slant height of this frustum bucket. (1 Mark)

  Answer:
  1. $l = \sqrt{h^2 + (r_2 - r_1)^2} = \sqrt{30^2 + (20-10)^2} = \sqrt{1000}$
  Result: $10\sqrt{10}\text{ cm} \approx 31.62\text{ cm}$

  c) Find the total capacity of the bucket in litres ($\text{Use }\pi = 3.14$). (2 Marks)

  Answer:
  1. $V = \frac{1}{3} \times 3.14 \times 30 \times (10^2 + 20^2 + 10 \times 20) = 31.4 \times 700 = 21980\text{ cm}^3$
  Result: $21.98\text{ litres}$
• 4 Marks Q6. Case Study: The Wooden Toy Artillery Piece
  A wooden toy used in a board game is shaped like a right circular cone mounted on top of a hemisphere. The total height of the toy is $15.5\text{ cm}$ and the radius of the hemisphere is $3.5\text{ cm}$.
  a) Determine the height of the conical upper part of the toy. (1 Mark)

  Answer:
  1. $h = \text{Total Height} - \text{Radius} = 15.5 - 3.5$
  Result: $12\text{ cm}$

  b) Calculate the curved surface area of the hemispherical base. (1 Mark)

  Answer:
  1. $\text{CSA} = 2\pi r^2 = 2 \times \frac{22}{7} \times (3.5)^2$
  Result: $77\text{ cm}^2$

  c) Find the total surface area of the complete wooden toy. (2 Marks)

  Answer:
  1. $\text{Conical CSA} = \pi rl = \frac{22}{7} \times 3.5 \times \sqrt{3.5^2 + 12^2} = 137.5\text{ cm}^2$
  2. $\text{Total SA} = 77 + 137.5$
  Result: $214.5\text{ cm}^2$
• 4 Marks Q7. Case Study: The Well and Embankment Project
  A rural development team digs a well with a diameter of $7\text{ m}$ to a depth of $20\text{ m}$. The earth taken out of the well is spread evenly all around it to a width of $14\text{ m}$ to construct a circular embankment.
  a) Calculate the total volume of the earth dug out from the well. (1 Mark)

  Answer:
  1. $V = \pi r^2 h = \frac{22}{7} \times (3.5)^2 \times 20$
  Result: $770\text{ m}^3$

  b) Find the total outer radius of the circular embankment including the well opening. (1 Mark)

  Answer:
  1. $R = r + \text{width} = 3.5 + 14$
  Result: $17.5\text{ m}$

  c) Calculate the final height of the constructed embankment. (2 Marks)

  Answer:
  1. $\text{Area of Embankment} = \pi(R^2 - r^2) = \frac{22}{7}(17.5^2 - 3.5^2) = 924\text{ m}^2$
  2. $H = \frac{770}{924}$
  Result: $\frac{5}{6}\text{ m} \approx 0.833\text{ m}$
• 4 Marks Q8. Case Study: The Hollow Metallic Pipe
  A factory produces hollow cylindrical iron pipes of length $28\text{ cm}$. The internal radius of the pipe is $14\text{ cm}$ and the iron thickness is uniformly $2\text{ cm}$ throughout.
  a) Determine the outer radius of the hollow pipe. (1 Mark)

  Answer:
  1. $R = r + \text{thickness} = 14 + 2$
  Result: $16\text{ cm}$

  b) Calculate the cross-sectional area of the metal used in the pipe. (1 Mark)

  Answer:
  1. $\text{Area} = \pi(R^2 - r^2) = \frac{22}{7}(16^2 - 14^2)$
  Result: $\frac{1320}{7}\text{ cm}^2 \approx 188.57\text{ cm}^2$

  c) Find the total volume of iron required to manufacture a single pipe. (2 Marks)

  Answer:
  1. $\text{Volume} = \text{Cross-sectional Area} \times \text{Length} = \frac{1320}{7} \times 28$
  Result: $5280\text{ cm}^3$
• 4 Marks Q9. Case Study: The Spherical Balloon Inflation
  A spherical party balloon is inflated with air such that its radius increases from an initial $7\text{ cm}$ to a final $14\text{ cm}$.
  a) Calculate the initial surface area of the balloon before pumping extra air. (1 Mark)

  Answer:
  1. $SA_1 = 4\pi (7)^2 = 4 \times \frac{22}{7} \times 49$
  Result: $616\text{ cm}^2$

  b) Calculate the final surface area of the balloon after expansion. (1 Mark)

  Answer:
  1. $SA_2 = 4\pi (14)^2 = 4 \times \frac{22}{7} \times 196$
  Result: $2464\text{ cm}^2$

  c) Find the ratio of the surface area of the balloon after inflation to that before inflation. (2 Marks)

  Answer:
  1. $\text{Ratio} = \frac{2464}{616}$
  Result: $4:1$
• 4 Marks Q10. Case Study: The Conical Grain Silo
  A farmer stores grain in a large conical silo. The radius and height of the silo are in the ratio $4:3$, and its total internal volume is $2156\text{ cm}^3$.
  a) Let the radius be $4x$ and height be $3x$. Express the volume formula in terms of $x$. (1 Mark)

  Answer:
  Result: $V = \frac{1}{3}\pi (4x)^2(3x) = 16\pi x^3$

  b) Solve for the exact radius and height of the conical silo ($\text{Use }\pi = \frac{22}{7}$). (1 Mark)

  Answer:
  1. $16 \times \frac{22}{7} \times x^3 = 2156 \implies x = 3.5$
  Result: $r = 14\text{ cm},\ h = 10.5\text{ cm}$

  c) Calculate the slant height and the curved surface area of the silo. (2 Marks)

  Answer:
  1. $l = \sqrt{14^2 + 10.5^2} = 17.5\text{ cm}$
  2. $\text{CSA} = \pi r l = \frac{22}{7} \times 14 \times 17.5$
  Result: $770\text{ cm}^2$
• 4 Marks Q11. Case Study: The Pharmaceutical Capsule Design
  A medical manufacturing unit produces capsules shaped like a cylinder with two hemispherical ends stuck to each of its ends. The total length of the capsule is $14\text{ mm}$ and its uniform diameter is $5\text{ mm}$.
  a) Find the length of the central cylindrical part of the capsule. (1 Mark)

  Answer:
  1. $h = 14 - 2(2.5)$
  Result: $9\text{ mm}$

  b) Calculate the combined surface area of the two hemispherical ends. (1 Mark)

  Answer:
  1. $\text{SA} = 4\pi r^2 = 4 \times \frac{22}{7} \times (2.5)^2$
  Result: $\frac{550}{7}\text{ mm}^2 \approx 78.57\text{ mm}^2$

  c) Find the total surface area of the entire medicine capsule. (2 Marks)

  Answer:
  1. $\text{Total SA} = \text{Cylinder CSA} + \text{Sphere SA} = 2\pi rh + 4\pi r^2$
  Result: $220\text{ mm}^2$
• 4 Marks Q12. Case Study: The Architecture Dome Painting
  The hemispherical dome of a public library needs exterior renovation and painting. The circumference of the circular base of the dome measures $17.6\text{ metres}$.
  a) Determine the base radius of the hemispherical dome. (1 Mark)

  Answer:
  1. $2\pi r = 17.6 \implies r = \frac{17.6 \times 7}{44}$
  Result: $2.8\text{ m}$

  b) Calculate the total curved surface area of the dome that requires painting. (1 Mark)

  Answer:
  1. $\text{CSA} = 2\pi r^2 = 2 \times \frac{22}{7} \times (2.8)^2$
  Result: $49.36\text{ m}^2$

  c) Find the total cost of painting the dome at the rate of $\text{₹}500$ per $\text{m}^2$. (2 Marks)

  Answer:
  1. $\text{Cost} = 49.36 \times 500$
  Result: $\text{₹}24,680$
• 4 Marks Q13. Case Study: The Conical Cavity Drill
  A metal craftsman takes a solid right circular cylinder of radius $7\text{ cm}$ and height $10\text{ cm}$, and drills out a conical cavity of identical height and base radius from one end.
  a) Calculate the volume of the original solid cylinder. (1 Mark)

  Answer:
  1. $V = \frac{22}{7} \times 7^2 \times 10$
  Result: $1540\text{ cm}^3$

  b) Find the volume of the conical cavity that is removed. (1 Mark)

  Answer:
  1. $V = \frac{1}{3} \times 1540$
  Result: $513.33\text{ cm}^3$

  c) Calculate the total surface area of the remaining solid object. (2 Marks)

  Answer:
  1. $\text{Total SA} = \text{Cylinder CSA} + \text{Base Area} + \text{Conical CSA}$
  Result: $862.57\text{ cm}^2$
• 4 Marks Q14. Case Study: The Juice Glass with Raised Base
  A juice vendor serves customers using cylindrical glasses of inner diameter $5\text{ cm}$ and height $10\text{ cm}$. However, the bottom of each glass features a hemispherical raised portion that reduces the usable volume.
  a) Calculate the apparent capacity of the cylindrical glass. (1 Mark)

  Answer:
  1. $V = 3.14 \times (2.5)^2 \times 10$
  Result: $196.25\text{ cm}^3$

  b) Find the volume of the hemispherical raised portion inside the glass base. (1 Mark)

  Answer:
  1. $V = \frac{2}{3} \times 3.14 \times (2.5)^3$
  Result: $32.71\text{ cm}^3$

  c) Calculate the actual net capacity of juice the glass can hold ($\text{Use }\pi = 3.14$). (2 Marks)

  Answer:
  1. $\text{Net Volume} = 196.25 - 32.71$
  Result: $163.54\text{ cm}^3$
• 4 Marks Q15. Case Study: The Lead Shot Drainage Problem
  An inverted conical water vessel has a height of $8\text{ cm}$ and a top open radius of $5\text{ cm}$, filled to the brim with water. Lead shots, each being a sphere of radius $0.5\text{ cm}$, are dropped inside, causing $\frac{1}{4}\text{th}$ of the water to overflow.
  a) Calculate the total volume of water initially filling the conical vessel. (1 Mark)

  Answer:
  1. $V = \frac{1}{3}\pi (5)^2 (8)$
  Result: $\frac{200\pi}{3}\text{ cm}^3$

  b) Find the volume of a single spherical lead shot. (1 Mark)

  Answer:
  1. $V = \frac{4}{3}\pi (0.5)^3$
  Result: $\frac{\pi}{6}\text{ cm}^3$

  c) Determine the exact number of lead shots dropped into the vessel. (2 Marks)

  Answer:
  1. $\text{Number} = \frac{\text{Overflow Volume}}{\text{Single Shot Volume}} = \frac{\frac{1}{4} \times \frac{200\pi}{3}}{\frac{\pi}{6}}$
  Result: $100\text{ shots}$
• 4 Marks Q16. Case Study: The Iron Transmission Pole
  An industrial iron transmission pole consists of a main lower cylinder of height $220\text{ cm}$ and base diameter $24\text{ cm}$, surmounted by a smaller upper cylinder of height $60\text{ cm}$ and radius $8\text{ cm}$.
  a) Calculate the volume of the lower cylindrical section of the pole. (1 Mark)

  Answer:
  1. $V_1 = \frac{22}{7} \times 12^2 \times 220$
  Result: $\frac{696960}{7}\text{ cm}^3 \approx 99565.71\text{ cm}^3$

  b) Calculate the volume of the upper cylindrical section of the pole. (1 Mark)

  Answer:
  1. $V_2 = \frac{22}{7} \times 8^2 \times 60$
  Result: $\frac{84480}{7}\text{ cm}^3 \approx 12068.57\text{ cm}^3$

  c) Find the total mass of the pole given that $1\text{ cm}^3$ of iron has a mass of $8\text{ g}$. (2 Marks)

  Answer:
  1. $\text{Total Mass} = \text{Total Volume} \times 8\text{ g}$
  Result: $\approx 893.07\text{ kg}$
• 4 Marks Q17. Case Study: The Frustum Ice Cream Dispenser
  A container shaped like a frustum of a cone has top radius $20\text{ cm}$, bottom radius $12\text{ cm}$, and holds a large batch of ice cream with a total capacity of $12308.8\text{ cm}^3$.
  a) State the volume formula for a frustum of a cone. (1 Mark)

  Answer:
  Result: $V = \frac{1}{3}\pi h (r_1^2 + r_2^2 + r_1 r_2)$

  b) Calculate the exact height of this container ($\text{Use }\pi = 3.14$). (1 Mark)

  Answer:
  1. $12308.8 = \frac{1}{3} \times 3.14 \times h \times (12^2 + 20^2 + 12 \times 20)$
  Result: $15\text{ cm}$

  c) Find the curved surface area of the container. (2 Marks)

  Answer:
  1. $\text{CSA} = \pi l (r_1 + r_2)$, where $l = \sqrt{15^2 + 8^2} = 17\text{ cm}$
  Result: $1707.52\text{ cm}^2$
• 4 Marks Q18. Case Study: The Solid Wooden Cube Carving
  A woodworking artisan carves the largest possible right circular cone out of a solid wooden cube whose edge length is $14\text{ cm}$.
  a) What are the maximum possible base diameter and height of the cone carved from the cube? (1 Mark)

  Answer:
  Result: $\text{Diameter} = 14\text{ cm},\ \text{Height} = 14\text{ cm}$

  b) Calculate the volume of the original wooden cube. (1 Mark)

  Answer:
  1. $V = 14^3$
  Result: $2744\text{ cm}^3$

  c) Find the volume of the carved-out cone and the volume of the wasted wood. (2 Marks)

  Answer:
  1. $\text{Cone Volume} = \frac{1}{3} \times \frac{22}{7} \times 7^2 \times 14 = 718.67\text{ cm}^3$
  2. $\text{Wasted Wood} = 2744 - 718.67$
  Result: $2025.33\text{ cm}^3$
• 4 Marks Q19. Case Study: The Flowing River / Pipe Discharge
  Water flows continuously through a cylindrical pipe of internal diameter $20\text{ mm}$ at a uniform speed of $10\text{ km/h}$ into a cylindrical storage tank of base radius $40\text{ cm}$ and height $2\text{ m}$.
  a) Determine the radius of the pipe in meters and the hourly flow length in meters. (1 Mark)

  Answer:
  Result: $\text{Radius} = 0.01\text{ m},\ \text{Flow length} = 10,000\text{ m/h}$

  b) Calculate the volume of water discharged by the pipe per hour. (1 Mark)

  Answer:
  1. $\text{Volume} = \pi r^2 \times \text{speed} = \pi \times (0.01)^2 \times 10000$
  Result: $\pi\text{ m}^3\text{ per hour}$

  c) Find the exact time taken by the pipe to fill the storage tank completely. (2 Marks)

  Answer:
  1. $\text{Time} = \frac{\text{Tank Volume}}{\text{Hourly Rate}} = \frac{0.32\pi}{\pi} = 0.32\text{ hours}$
  Result: $19.2\text{ minutes}$ ($19\text{ min } 12\text{ sec}$)
• 4 Marks Q20. Case Study: Comparative Scaling of Cylinders
  A student compares two solid cylinders. Cylinder A has radius $r$ and height $h$, while Cylinder B has its base radius halved ($\frac{r}{2}$) and its height kept identical ($h$).
  a) Write down the volume formula for Cylinder A. (1 Mark)

  Answer:
  Result: $V_A = \pi r^2 h$

  b) Write down the volume formula for Cylinder B in terms of $r$ and $h$. (1 Mark)

  Answer:
  Result: $V_B = \pi \left(\frac{r}{2}\right)^2 h = \frac{1}{4}\pi r^2 h$

  c) Find the simplest ratio of the volume of the reduced cylinder (B) to that of the original cylinder (A). (2 Marks)

  Answer:
  1. $\text{Ratio} = \frac{\frac{1}{4}\pi r^2 h}{\pi r^2 h}$
  Result: $1:4$

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