Chapter 12 : Class 10 : Surface Areas and Volumes : Case Study Question Bank

✦ NCF 2023 • COMPETENCY-FOCUSED QUESTION BANK ✦

Chapter 12: Surface Areas and Volumes

Class 10 Standard Mathematics (Case Study Bank)

Curated By Akash Srivastva

✨ Interactive TLMs, Worksheets & Digital Study Notes: www.akashmaths.online

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Case Study 1 Medical Engineering 4 Marks

Capsule-Shaped Smart Drug Delivery Vehicle

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A smart drug-delivery capsule has a cylindrical center with two identical hemispherical ends:

• Total length: \(12\text{ mm}\)
• Capsule diameter: \(6\text{ mm}\) (radius \(r = 3\text{ mm}\))
• Parameter: Take \(\pi = 3.14\)

Q1. Cylindrical Height & CSA 1 Mark

Calculate the height (h) of the cylindrical portion and find its curved surface area in mm².

(A) h = 8 mm, CSA = 150.72 mm²

(B) h = 6 mm, CSA = 113.04 mm²

(C) h = 6 mm, CSA = 120.50 mm²

(D) h = 5 mm, CSA = 94.20 mm²

Q2. Total Surface Area 1 Mark

Determine the total outer surface area (TSA) of the drug-delivery capsule.

(A) 226.08 mm²

(B) 245.50 mm²

(C) 210.00 mm²

(D) 282.60 mm²

Q3. Capsule Volume & Redesign 2 Marks

Calculate the total volume (capacity) of the drug-delivery capsule.

(A) 250.4 mm³

(B) 300.0 mm³

(C) 282.6 mm³

(D) 275.2 mm³

OR (Alternative Q3)

If total length is reduced to 10 mm while keeping radius at 3 mm, find the percentage reduction in volume.

(A) 20%

(B) 15%

(C) 25%

(D) 18%

📋 View Step-by-Step Marking Scheme & Solution▼

Question	Expected Answer / Step-by-Step Solution	CBSE Marks Allocation
Q1	r = 6/2 = 3 mm. Cylindrical height h = 12 − 2(3) = 6 mm. CSA = 2πrh = 2 × 3.14 × 3 × 6 = 113.04 mm².	[0.5 Mark] for h = 6 mm. [0.5 Mark] for CSA = 113.04 mm².
Q2	TSA = 2πrh + 2(2πr²) = 2πr(h + 2r) = 2 × 3.14 × 3 × (6 + 6) = 226.08 mm².	[1 Mark] for TSA = 226.08 mm².
Q3	V = πr²h + 2(2/3 πr³) = πr²(h + 4/3 r) = 3.14 × 9 × (6 + 4) = 282.6 mm³.	[1 Mark] for volume formula setup. [1 Mark] for 282.6 mm³.
Q3 (OR)	New h = 10 − 6 = 4 mm. V_new = 3.14 × 9 × (4 + 4) = 226.08 mm³. Reduction = 282.6 − 226.08 = 56.52 mm³. % Reduction = (56.52 / 282.6) × 100% = 20%.	[1 Mark] for new volume. [1 Mark] for 20% reduction.

Case Study 2 Sustainable Design 4 Marks

Double-Walled Thermal Flask

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A double-walled stainless steel thermal flask combines a cylinder and a conical neck with common radius \(r = 7\text{ cm}\):

• Cylinder height: \(h_{\text{cyl}} = 10\text{ cm}\)
• Cone height: \(h_{\text{cone}} = 24\text{ cm}\)
• Parameter: Take \(\pi = \frac{22}{7}\)

Q1. Slant Height 1 Mark

Calculate the slant height (l) of the conical neck of the thermal flask.

(A) 26 cm

(B) 25 cm

(C) 24 cm

(D) 28 cm

Q2. Conical CSA 1 Mark

Calculate the curved surface area (CSA) of the conical neck portion.

(A) 550 cm²

(B) 500 cm²

(C) 600 cm²

(D) 525 cm²

Q3. Flask Capacity & Outer Area 2 Marks

Find the total capacity (volume) of the thermal flask in cm³.

(A) 2540 cm³

(B) 2650 cm³

(C) 2772 cm³

(D) 2880 cm³

OR (Alternative Q3)

Find the total outer curved surface area of the flask (excluding the flat base).

(A) 950 cm²

(B) 990 cm²

(C) 1020 cm²

(D) 1050 cm²

📋 View Step-by-Step Marking Scheme & Solution▼

Question	Expected Answer / Step-by-Step Solution	CBSE Marks Allocation
Q1	l = √(r² + h_cone²) = √(7² + 24²) = √625 = 25 cm.	[1 Mark] for l = 25 cm.
Q2	CSA_cone = πrl = (22/7) × 7 × 25 = 550 cm².	[1 Mark] for 550 cm².
Q3	V = πr²h_cyl + 1/3 πr²h_cone = (22/7) × 49 × (10 + 8) = 154 × 18 = 2772 cm³.	[1 Mark] for formula setup. [1 Mark] for 2772 cm³.
Q3 (OR)	CSA_cyl = 2 × (22/7) × 7 × 10 = 440 cm². Total CSA = 440 + 550 = 990 cm².	[1 Mark] for cylinder CSA. [1 Mark] for total CSA = 990 cm².

Case Study 3 Space Technology 4 Marks

Rocket Booster Payload Fairing

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A rocket payload fairing has a lower cylindrical booster section and a conical nose cone with common radius \(r = 5\text{ m}\):

• Cylinder height: \(h_{\text{cyl}} = 20\text{ m}\)
• Nose cone height: \(h_{\text{cone}} = 12\text{ m}\)
• Parameter: Take \(\pi = 3.14\)

Q1. Cone Slant Height 1 Mark

Calculate the slant height (l) of the conical nose cone of the fairing.

(A) 13 m

(B) 14 m

(C) 15 m

(D) 12.5 m

Q2. Thermal Tile Surface Area 1 Mark

Calculate total outer surface area requiring thermal protection tiles (excluding flat base).

(A) 815.4 m²

(B) 832.1 m²

(C) 850.0 m²

(D) 820.5 m²

Q3. Payload Volume & Redesign 2 Marks

Determine the total volume of payload space enclosed within this fairing structure.

(A) 1800 m³

(B) 1920 m³

(C) 1884 m³

(D) 1850 m³

OR (Alternative Q3)

If cone is replaced by a hemisphere of radius r = 5 m, find the difference in payload volume.

(A) 52.33 m³

(B) 48.50 m³

(C) 55.00 m³

(D) 62.25 m³

📋 View Step-by-Step Marking Scheme & Solution▼

Question	Expected Answer / Step-by-Step Solution	CBSE Marks Allocation
Q1	l = √(5² + 12²) = √169 = 13 m.	[1 Mark] for l = 13 m.
Q2	Outer Area = 2πrh_cyl + πrl = 3.14 × 5 × (2(20) + 13) = 15.7 × 53 = 832.1 m².	[1 Mark] for 832.1 m².
Q3	V = πr²h_cyl + 1/3 πr²h_cone = 3.14 × 25 × (20 + 4) = 78.5 × 24 = 1884 m³.	[1 Mark] for formula setup. [1 Mark] for 1884 m³.
Q3 (OR)	V_hemi = 3.14 × 25 × (20 + 10/3) = 1831.67 m³. Difference = 1884 − 1831.67 = 52.33 m³.	[1 Mark] for hemispherical volume. [1 Mark] for difference = 52.33 m³.

Case Study 4 Food Processing 4 Marks

Artisanal Ice Cream Treat

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A confectioner's ice cream treat consists of a hemispherical scoop atop a conical waffle sleeve:

• Common radius: \(r = 3\text{ cm}\)
• Cone height: \(h = 4\text{ cm}\)
• Parameter: Take \(\pi = 3.14\)

Q1. Slant Height & Cone CSA 1 Mark

Find the slant height (l) and curved surface area of the conical wafer sleeve.

(A) l = 5 cm, CSA = 47.1 cm²

(B) l = 6 cm, CSA = 56.52 cm²

(C) l = 5 cm, CSA = 50.0 cm²

(D) l = 4.5 cm, CSA = 42.4 cm²

Q2. Exposed Scoop Surface Area 1 Mark

Find the outer surface area of the hemispherical scoop exposed to ambient air.

(A) 47.10 cm²

(B) 56.52 cm²

(C) 62.80 cm²

(D) 50.24 cm²

Q3. Treat Volume & Ratio 2 Marks

Calculate the total volume of the ice cream treat (scoop and sleeve combined).

(A) 88.5 cm³

(B) 90.0 cm³

(C) 94.2 cm³

(D) 98.4 cm³

OR (Alternative Q3)

If cone height is doubled to 8 cm (radius unchanged), calculate the ratio V_cone : V_hemi.

(A) 4 : 3

(B) 3 : 2

(C) 2 : 1

(D) 5 : 4

📋 View Step-by-Step Marking Scheme & Solution▼

Question	Expected Answer / Step-by-Step Solution	CBSE Marks Allocation
Q1	l = √(3² + 4²) = 5 cm. CSA = πrl = 3.14 × 3 × 5 = 47.1 cm².	[0.5 Mark] for l = 5 cm. [0.5 Mark] for CSA = 47.1 cm².
Q2	Exposed Area = 2πr² = 2 × 3.14 × 9 = 56.52 cm².	[1 Mark] for 56.52 cm².
Q3	V = 1/3 πr²(2r + h) = 1/3 × 3.14 × 9 × (6 + 4) = 3.14 × 3 × 10 = 94.2 cm³.	[1 Mark] for volume setup. [1 Mark] for 94.2 cm³.
Q3 (OR)	V_cone = 1/3 π(3²)(8) = 24π cm³. V_hemi = 2/3 π(3³) = 18π cm³. Ratio = 24π / 18π = 4/3 = 4 : 3.	[1 Mark] for component volumes. [1 Mark] for ratio 4 : 3.

Case Study 5 Eco-Tourism 4 Marks

Circular Safari Glamping Tent

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A glamping tent combines a cylindrical wall with a conical roof on a common radius \(r = 7\text{ m}\):

• Cylinder height: \(h_1 = 10\text{ m}\)
• Cone roof height: \(h_2 = 24\text{ m}\)
• Parameter: Take \(\pi = \frac{22}{7}\)

Q1. Roof Slant Height 1 Mark

Calculate the exact slant height (l) of the conical canvas roof of the safari tent.

(A) 24 m

(B) 25 m

(C) 26 m

(D) 27 m

Q2. Cylindrical Wall Area 1 Mark

Find the curved surface area of the wooden cylindrical wall of the tent.

(A) 440 m²

(B) 420 m²

(C) 460 m²

(D) 480 m²

Q3. Canvas Fabric & Enclosed Volume 2 Marks

Calculate the total area of canvas fabric required (excluding the floor).

(A) 950 m²

(B) 980 m²

(C) 990 m²

(D) 1020 m²

OR (Alternative Q3)

Calculate the total volume of air enclosed within one luxury safari tent.

(A) 2,772 m³

(B) 2,650 m³

(C) 2,850 m³

(D) 2,700 m³

📋 View Step-by-Step Marking Scheme & Solution▼

Question	Expected Answer / Step-by-Step Solution	CBSE Marks Allocation
Q1	l = √(r² + h₂²) = √(7² + 24²) = √625 = 25 m.	[1 Mark] for l = 25 m.
Q2	CSA_cyl = 2πrh₁ = 2 × (22/7) × 7 × 10 = 440 m².	[1 Mark] for 440 m².
Q3	CSA_cone = πrl = (22/7) × 7 × 25 = 550 m². Total Canvas = 440 + 550 = 990 m².	[1 Mark] for cone CSA. [1 Mark] for 990 m².
Q3 (OR)	V_cyl = (22/7) × 49 × 10 = 1540 m³. V_cone = 1/3 × (22/7) × 49 × 24 = 1232 m³. Total Volume = 1540 + 1232 = 2772 m³.	[1 Mark] for component volumes. [1 Mark] for 2772 m³.

Case Study 6 Agricultural Engineering 4 Marks

Industrial Grain Storage Silo

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An industrial grain silo consists of a cylindrical tank with a bottom conical hopper on common radius \(r = 3.5\text{ m}\) (\(\frac{7}{2}\text{ m}\)):

• Cylinder tank height: \(h_1 = 20\text{ m}\)
• Hopper cone height: \(h_2 = 12\text{ m}\)
• Parameter: Take \(\pi = \frac{22}{7}\)

Q1. Hopper Slant Height 1 Mark

Find the exact slant height (l) of the bottom conical hopper of the silo.

(A) 12.0 m

(B) 12.5 m (or 25/2 m)

(C) 13.0 m

(D) 13.5 m

Q2. Hopper Capacity 1 Mark

Calculate the storage capacity (volume) of the bottom conical hopper alone.

(A) 154 m³

(B) 144 m³

(C) 160 m³

(D) 168 m³

Q3. Total Silo Volume & Steel Area 2 Marks

Calculate the total storage capacity (total volume) of one industrial silo when completely filled.

(A) 900 m³

(B) 910 m³

(C) 924 m³

(D) 950 m³

OR (Alternative Q3)

Find total internal steel surface area of the silo in contact with grain (excluding top cover).

(A) 560.0 m²

(B) 577.5 m²

(C) 585.0 m²

(D) 590.5 m²

📋 View Step-by-Step Marking Scheme & Solution▼

Question	Expected Answer / Step-by-Step Solution	CBSE Marks Allocation
Q1	l = √((3.5)² + 12²) = √(12.25 + 144) = √156.25 = 12.5 m.	[1 Mark] for l = 12.5 m.
Q2	V_hopper = 1/3 × (22/7) × (7/2)² × 12 = 22 × 7 = 154 m³.	[1 Mark] for 154 m³.
Q3	V_cyl = (22/7) × (7/2)² × 20 = 770 m³. Total Volume = 770 + 154 = 924 m³.	[1 Mark] for cylinder volume. [1 Mark] for total = 924 m³.
Q3 (OR)	CSA_cyl = 2 × (22/7) × (7/2) × 20 = 440 m². CSA_cone = (22/7) × (7/2) × 12.5 = 137.5 m². Total Area = 440 + 137.5 = 577.5 m².	[1 Mark] for area components. [1 Mark] for 577.5 m².

Case Study 7 Chemical Safety 4 Marks

Boiling Flask Calibration

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A round-bottom laboratory boiling flask combines a spherical bulb with a cylindrical neck:

• Spherical bulb radius: \(R = 10.5\text{ cm}\) (\(\frac{21}{2}\text{ cm}\))
• Cylindrical neck: radius \(r = 3.5\text{ cm}\) (\(\frac{7}{2}\text{ cm}\)), height \(h = 8\text{ cm}\)
• Parameter: Take \(\pi = \frac{22}{7}\)

Q1. Neck Surface Area 1 Mark

Calculate the curved surface area of the cylindrical neck of the boiling flask.

(A) 160 cm²

(B) 176 cm²

(C) 184 cm²

(D) 192 cm²

Q2. Neck Capacity 1 Mark

Calculate the volumetric capacity (volume) of the cylindrical neck of the flask.

(A) 308 cm³

(B) 294 cm³

(C) 316 cm³

(D) 324 cm³

Q3. Flask Total Volume & Overflow 2 Marks

Calculate total volume of liquid the boiling flask can hold when completely filled.

(A) 5000 cm³

(B) 5120 cm³

(C) 5,159 cm³

(D) 5250 cm³

OR (Alternative Q3)

If 4,500 cm³ of biofuel is poured into empty flask, determine overflow status and remaining empty volume.

(A) Will not overflow; remaining empty volume = 659 cm³

(B) Will overflow by 150 cm³

(C) Will not overflow; remaining empty volume = 550 cm³

(D) Exact fill, 0 cm³ left

📋 View Step-by-Step Marking Scheme & Solution▼

Question	Expected Answer / Step-by-Step Solution	CBSE Marks Allocation
Q1	CSA_neck = 2πrh = 2 × (22/7) × (7/2) × 8 = 176 cm².	[1 Mark] for 176 cm².
Q2	V_neck = πr²h = (22/7) × (7/2)² × 8 = 308 cm³.	[1 Mark] for 308 cm³.
Q3	V_sphere = 4/3 × (22/7) × (21/2)³ = 11 × 441 = 4851 cm³. Total = 4851 + 308 = 5159 cm³.	[1 Mark] for spherical volume. [1 Mark] for 5159 cm³.
Q3 (OR)	4500 cm³ < 5159 cm³, so it will NOT overflow. Empty volume = 5159 − 4500 = 659 cm³.	[1 Mark] for overflow verdict. [1 Mark] for 659 cm³.

Case Study 8 Heritage Conservation 4 Marks

Historic Gateway Pillar Restoration

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A heritage solid sandstone gateway pillar combines a cylinder surmounted by a hemispherical dome of radius \(r = 21\text{ cm}\):

• Cylinder height: \(h = 50\text{ cm}\)
• Parameter: Take \(\pi = \frac{22}{7}\)

Q1. Total Pillar Height 1 Mark

Find the total vertical height (H) of one solid sandstone pillar.

(A) 70 cm

(B) 71 cm

(C) 72 cm

(D) 75 cm

Q2. Dome Curved Surface Area 1 Mark

Calculate the curved surface area of the hemispherical dome surmounting the top of the pillar.

(A) 2,772 cm²

(B) 2,650 cm²

(C) 2,800 cm²

(D) 2,540 cm²

Q3. Sandstone Volume & Polishing Area 2 Marks

Calculate the total volume of solid sandstone material used in one gateway pillar.

(A) 86,500 cm³

(B) 87,400 cm³

(C) 88,704 cm³

(D) 89,200 cm³

OR (Alternative Q3)

Calculate total visible surface area to be polished (excluding bottom base on foundation).

(A) 9,250 cm²

(B) 9,372 cm²

(C) 9,480 cm²

(D) 9,550 cm²

📋 View Step-by-Step Marking Scheme & Solution▼

Question	Expected Answer / Step-by-Step Solution	CBSE Marks Allocation
Q1	Total height H = h + r = 50 + 21 = 71 cm.	[1 Mark] for H = 71 cm.
Q2	CSA_dome = 2πr² = 2 × (22/7) × 21 × 21 = 2772 cm².	[1 Mark] for 2772 cm².
Q3	V_cyl = (22/7) × 21² × 50 = 69,300 cm³. V_hemi = 2/3 × (22/7) × 21³ = 19,404 cm³. Total Volume = 69,300 + 19,404 = 88,704 cm³.	[1 Mark] for component volumes. [1 Mark] for 88,704 cm³.
Q3 (OR)	CSA_cyl = 2 × (22/7) × 21 × 50 = 6600 cm². Total Polishing Area = 6600 + 2772 = 9372 cm².	[1 Mark] for cylinder CSA. [1 Mark] for 9372 cm².

Live Practice: Chapter 12 Surface Areas and Volumes

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Case Study 1

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