Surface Areas and Volumes: Class 10 Case Study Previous Year Questions with Answers

Canvas area = 1650 m².

Radius r = $\frac{\text{(}\text{30}\text{)/(}}{\text{2}\text{)}}$ = 15 m.

Cone slant height l = √(15² + 8²) = 17 m.

Canvas area = curved area of cylinder + curved area of cone.

Area = 2πrh + πrl = π × 15 × (2 × 9 + 17) = 525π m².

Using π = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$, area = 1650 m². The floor and the joined circular face are excluded.

Cost = ₹3,36,000.

Canvas to be bought = 1650 + 30 = 1680 m².

Cost = 1680 × 200 = ₹3,36,000.

Slant height = 17.5 m.

Cone height = 18.5 − 8 = 10.5 m and radius = $\frac{\text{(}\text{28}\text{)/(}}{\text{2}\text{)}}$ = 14 m.

l = √(14² + 10.5²) = √306.25 = 17.5 m.

Floor area = 616 m².

The floor is a circle of radius 14 m.

Using π = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$, area = πr² = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$ × 14² = 616 m².

Cloth area = 1474 m².

The cloth covers the cylinder’s curved side and the cone’s curved side.

Area = 2πrh + πrl, where r = 14 m, h = 8 m and l = 17.5 m.

Using π = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$, area = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$ × 14 × (16 + 17.5) = 1474 m².

Do not include the floor or the circle where the solids join.

Air volume = 7084 m³.

Air occupies both the cylindrical and conical parts.

Volume = πr²h + $\frac{\text{(}\text{1}\text{)/(}}{\text{3}\text{)}}$πr²H, with h = 8 m and H = 10.5 m.

Using π = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$, volume = 616 × (8 + $\frac{\text{(}\text{10.5}\text{)/(}}{\text{3}\text{)}}$) = 7084 m³.

Blue PVC = 1914 m²; cost = ₹1,33,980.

Cone slant height l = √(21² + 20²) = √841 = 29 m.

Using π = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$, blue area = πrl = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$ × 21 × 29 = 1914 m².

Cost = 1914 × 70 = ₹1,33,980.

White fabric = 660 m²; cost = ₹39,600.

Using π = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$, cylinder curved area = 2πrh.

Area = 2 × $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$ × 21 × 5 = 660 m².

Cost = 660 × 60 = ₹39,600.

Slant height = 2.5 m.

l = √(r² + h²) = √(1.5² + 2²) = √6.25 = 2.5 m.

Cone curved area = $\frac{\text{(}\text{165}\text{)/(}}{\text{14}\text{)}}$ m² ≈ 11.79 m².

Using π = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$, curved area = πrl = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$ × 1.5 × 2.5 = $\frac{\text{(}\text{165}\text{)/(}}{\text{14}\text{)}}$ m².

The displayed decimal is rounded from the exact fraction.

Cost = ₹1,32,000.

Using π = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$, cylinder curved area = 2πrh.

Area = 2 × $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$ × 1.5 × 7 = 66 m².

Cost = 66 × 2000 = ₹1,32,000.

Capacity = $\frac{\text{(}\text{759}\text{)/(}}{\text{14}\text{)}}$ m³ ≈ 54.21 m³.

Capacity = cylinder volume + cone volume.

V = π × 1.5² × 7 + $\frac{\text{(}\text{1}\text{)/(}}{\text{3}\text{)}}$π × 1.5² × 2 = 17.25π m³.

Using π = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$, V = $\frac{\text{(}\text{759}\text{)/(}}{\text{14}\text{)}}$ m³ ≈ 54.21 m³.

Volume of 50 toys = 16,17,000 cm³.

Volume of one toy = $\frac{\text{(}\text{2}\text{)/(}}{\text{3}\text{)}}$πr³ + $\frac{\text{(}\text{1}\text{)/(}}{\text{3}\text{)}}$πr²h.

V = $\frac{\text{(}\text{1}\text{)/(}}{\text{3}\text{)}}$π × 21² × (2 × 21 + 28) = 10290π cm³.

Using π = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$, one toy uses 32340 cm³.

Volume of 50 toys = 50 × 32340 = 16,17,000 cm³.

Hemisphere : cone = 3 : 2.

Ratio = ($\frac{\text{(}\text{2}\text{)/(}}{\text{3}\text{)}}$πr³) : ($\frac{\text{(}\text{1}\text{)/(}}{\text{3}\text{)}}$πr²h).

Cancel the common factor $\frac{\text{(}\text{1}\text{)/(}}{\text{3}\text{)}}$πr² to obtain 2r : h.

Ratio = 42 : 28 = 3 : 2.

Cylinder volume = 69.3 cm³.

V = πr²h = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$ × 2.1² × 5 = 69.3 cm³.

Hemisphere volume = 19.404 cm³.

V = $\frac{\text{(}\text{2}\text{)/(}}{\text{3}\text{)}}$πr³ = $\frac{\text{(}\text{2}\text{)/(}}{\text{3}\text{)}}$ × $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$ × 2.1³ = 19.404 cm³.

Exposed surface area = 107.58 cm².

The exposed surfaces are the cylinder’s curved side, its lower base and the hemisphere’s curved surface.

Area = 2πrh + πr² + 2πr² = 2πrh + 3πr².

Area = 2 × $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$ × 2.1 × 5 + 3 × $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$ × 2.1² = 107.58 cm².

The joined circular faces are internal and are not counted.

Cylinder total surface area = 93.72 cm².

Without the hemisphere, both circular ends of the cylinder are included.

Area = 2πrh + 2πr² = 2πr(h + r).

Area = 2 × $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$ × 2.1 × (5 + 2.1) = 93.72 cm².

Volume of both cones = 13.2 cm³.

Using π = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$, volume = 2 × $\frac{\text{(}\text{1}\text{)/(}}{\text{3}\text{)}}$πr²h.

Volume = 2 × $\frac{\text{(}\text{1}\text{)/(}}{\text{3}\text{)}}$ × $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$ × 1.5² × 2.8 = 13.2 cm³.

Cylinder volume = 49.5 cm³.

Using π = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$, V = πr²h = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$ × 1.5² × 7 = 49.5 cm³.

Wood scrapped = 36.3 cm³.

Log volume = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$ × 1.5² × 14 = 99 cm³.

Gilli volume = 13.2 + 49.5 = 62.7 cm³.

Wood scrapped = 99 − 62.7 = 36.3 cm³.

Danda surface area = $\frac{\text{(}\text{3872}\text{)/(}}{\text{7}\text{)}}$ cm² ≈ 553.14 cm².

Use Danda’s radius 2 cm and length 42 cm, rather than Gilli’s dimensions.

Total area of a closed cylinder = 2πr(h + r).

Using π = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$, area = 2 × $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$ × 2 × (42 + 2) = $\frac{\text{(}\text{3872}\text{)/(}}{\text{7}\text{)}}$ cm².

Total height = 5 cm.

A hemisphere’s height equals its radius: 3 cm.

Total height = cap height + stem height = 3 + 2 = 5 cm.

Stem volume = 3.08 cm³.

Stem radius r = $\frac{\text{(}\text{1.4}\text{)/(}}{\text{2}\text{)}}$ = 0.7 cm.

Using π = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$, V = πr²h = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$ × 0.7² × 2 = 3.08 cm³.

Volume of 7 mushrooms = 417.56 cm³.

One mushroom has volume $\frac{\text{(}\text{2}\text{)/(}}{\text{3}\text{)}}$πR³ + πr²h.

Using π = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$, one mushroom has volume $\frac{\text{(}\text{396}\text{)/(}}{\text{7}\text{)}}$ + 3.08 cm³.

Volume of 7 = 7 × ($\frac{\text{(}\text{396}\text{)/(}}{\text{7}\text{)}}$ + 3.08) = 417.56 cm³.

Total exposed area = 655.6 cm².

For one mushroom, count the cap’s curved surface, uncovered flat underside, stem’s curved side and stem’s lower base.

Area = 2πR² + (πR² − πr²) + 2πrh + πr².

Area = 3πR² + 2πrh = π(3 × 3² + 2 × 0.7 × 2) = 29.8π cm².

Using π = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$, total for 7 = 7 × 29.8 × $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$ = 655.6 cm².

Bulb surface area = 154 cm².

Bulb radius = $\frac{\text{(}\text{7}\text{)/(}}{\text{2}\text{)}}$ = 3.5 cm.

Using π = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$, area = 4πr² = 4 × $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$ × 3.5² = 154 cm².

Maximum diameter = 10 cm.

The shortest internal dimension is 12 cm.

Leave 1 cm on each of its two opposite sides.

Maximum diameter = 12 − 1 − 1 = 10 cm.

Fabric area = 1512 cm².

Fabric height before folding = 17 + 2 + 2 = 21 cm.

The fabric wraps around the four lateral faces; top and bottom stay open.

Required width around the cuboid = 2(24 + 12) = 72 cm.

Fabric area = 72 × 21 = 1512 cm².

Available space = $\frac{\text{(}\text{14149}\text{)/(}}{\text{3}\text{)}}$ cm³ ≈ 4716.33 cm³.

Available space = cuboid volume − spherical bulb volume, ignoring the rod.

Cuboid volume = 24 × 12 × 17 = 4896 cm³.

Using π = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$, bulb volume = $\frac{\text{(}\text{4}\text{)/(}}{\text{3}\text{)}}$ × $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$ × 3.5³ = $\frac{\text{(}\text{539}\text{)/(}}{\text{3}\text{)}}$ cm³.

Difference = 4896 − $\frac{\text{(}\text{539}\text{)/(}}{\text{3}\text{)}}$ = $\frac{\text{(}\text{14149}\text{)/(}}{\text{3}\text{)}}$ cm³.

Volume = 462 cm³.

Using π = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$, V = πr²h = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$ × 3.5² × 12 = 462 cm³.

Curved area = 264 cm².

Using π = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$, area = 2πrh = 2 × $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$ × 3.5 × 12 = 264 cm².

Required ratio = 40 : 3.

Total volume of small parts = 2 × $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$ × 1.05² × 5 = 34.65 cm³.

Ratio = 462 : 34.65 = 46200 : 3465 = 40 : 3.

This is equivalent to the scheme’s 3080 : 231, reduced to lowest terms.

Combined curved area = 66 cm².

Each small radius is $\frac{\text{(}\text{2.1}\text{)/(}}{\text{2}\text{)}}$ = 1.05 cm.

Using π = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$, combined curved area = 2 × 2πrh.

Area = 2 × 2 × $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$ × 1.05 × 5 = 66 cm².

Dome volume = 19404 m³.

A hemisphere’s height equals its radius, so r = 21 m.

V = $\frac{\text{(}\text{2}\text{)/(}}{\text{3}\text{)}}$πr³ = $\frac{\text{(}\text{2}\text{)/(}}{\text{3}\text{)}}$ × $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$ × 21³ = 19404 m³.

Cloth area = 1232 m².

Use the new radius 14 m for this subpart.

The cloth covers the dome’s curved surface.

Using π = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$, area = 2πr² = 2 × $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$ × 14² = 1232 m².

Cuboid total surface area = 208 m².

The scheme asks for the cuboid’s own total surface area.

Area = 2(lb + bh + hl) = 2(8 × 6 + 6 × 4 + 4 × 8) = 208 m².

This is not a calculation of the exposed area of the complete joined stupa.

Cuboid volume = 192 m³.

Volume = lbh = 8 × 6 × 4 = 192 m³.

Space diagonal = 6√3 cm.

For cube edge a, the face diagonal is a√2.

By Pythagoras, space diagonal = √((a√2)² + a²) = a√3.

With a = 6 cm, the diagonal is 6√3 cm.

Volume = 343 cm³.

Use edge 7 cm in this subpart.

Volume = a³ = 7³ = 343 cm³.

Hemisphere curved area = 308 cm².

Only the curved hemispherical surface is requested.

Using π = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$, area = 2πr² = 2 × $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$ × 7² = 308 cm².

Surface area = 160 cm².

The resulting cuboid measures 8 cm × 4 cm × 4 cm.

Area = 2(lb + bh + hl) = 2(8 × 4 + 4 × 4 + 4 × 8) = 160 cm².

The two faces at the join are hidden.

Slant height = 1.3 cm.

Radius r = $\frac{\text{(}\text{1}\text{)/(}}{\text{2}\text{)}}$ = 0.5 cm; perpendicular height h = 1.2 cm.

l = √(r² + h²) = √(0.5² + 1.2²) = √1.69 = 1.3 cm.

Curved area = 0.65π cm².

The sharpened part is a cone.

Its curved area = πrl = π × 0.5 × 1.3 = 0.65π cm².

Correct exposed surface area = 21.9π cm².

Exposed area = cylinder curved area + cone curved area + one circular base.

Cylinder curved area = 2πrh = 2π × 0.5 × 21 = 21π cm².

Cone curved area = 0.65π cm²; exposed circular base = π × 0.5² = 0.25π cm².

Total = (21 + 0.65 + 0.25)π = 21.9π cm².

Source note: the 2024–25 Basic sample scheme, PDF and printed p. 7, prints 6.15π, using π × 0.5 × 0.5 × 21 for the cylinder’s curved area. Using 2πrh with the stated radius 0.5 cm and length 21 cm gives the exposed total 21.9π cm².

Cylindrical volume = 3.2π cm³.

Following the sample’s model, the sharpened tip keeps height 1.2 cm.

Initial total height = 21 + 1.2 = 22.2 cm.

New total height = 22.2 − 8.2 = 14 cm.

New cylinder length = 14 − 1.2 = 12.8 cm.

Volume = πr²h = π × 0.5² × 12.8 = 3.2π cm³.

Wood removed = 138.6 cm³.

Each depression is a cone of radius 2.1 cm and height 6 cm.

Using π = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$, total removed volume = 5 × $\frac{\text{(}\text{1}\text{)/(}}{\text{3}\text{)}}$πr²h.

Volume = 5 × $\frac{\text{(}\text{1}\text{)/(}}{\text{3}\text{)}}$ × $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$ × 2.1² × 6 = 138.6 cm³.

Wood remaining = 661.4 cm³.

Original cuboid volume = 10 × 10 × 8 = 800 cm³.

Wood remaining = 800 − 138.6 = 661.4 cm³.

One hollow volume = $\frac{\text{(}\text{264}\text{)/(}}{\text{7}\text{)}}$ cm³ ≈ 37.71 cm³.

Using π = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$, V = πr²h = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$ × 2² × 3 = $\frac{\text{(}\text{264}\text{)/(}}{\text{7}\text{)}}$ cm³.

Cuboid lateral area = 248 cm².

Lateral area includes only the four vertical rectangular faces.

Area = 2(l + b)h = 2(14 + 17) × 4 = 248 cm².

Wood remaining = 688 cm³.

Cuboid volume = 14 × 17 × 4 = 952 cm³.

Volume of seven hollows = 7 × $\frac{\text{(}\text{264}\text{)/(}}{\text{7}\text{)}}$ = 264 cm³.

Wood remaining = 952 − 264 = 688 cm³.

Top wooden area = 150 cm².

Subtract the seven circular openings from the top rectangle.

Using π = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$, area = 14 × 17 − 7 × $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$ × 2².

Area = 238 − 88 = 150 cm².

One dimple curved area = $\frac{\text{(}\text{176}\text{)/(}}{\text{7}\text{)}}$ mm² ≈ 25.14 mm².

Convert the ball diameter: 4.2 cm = 42 mm, so its radius is 21 mm.

The dimple has radius r = 2 mm. Its exposed surface is curved.

Using π = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$, area = 2πr² = 2 × $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$ × 2² = $\frac{\text{(}\text{176}\text{)/(}}{\text{7}\text{)}}$ mm².

One dimple volume = $\frac{\text{(}\text{352}\text{)/(}}{\text{21}\text{)}}$ mm³ ≈ 16.76 mm³.

Using π = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$, hemisphere volume = $\frac{\text{(}\text{2}\text{)/(}}{\text{3}\text{)}}$πr³.

V = $\frac{\text{(}\text{2}\text{)/(}}{\text{3}\text{)}}$ × $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$ × 2³ = $\frac{\text{(}\text{352}\text{)/(}}{\text{21}\text{)}}$ mm³.

Exposed area = 9504 mm².

Area = full sphere area − circular openings + curved dimple areas.

Area = 4π × 21² − 315π × 2² + 315 × 2π × 2².

Area = (1764 − 1260 + 2520)π = 3024π mm².

Using π = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$, exposed area = 9504 mm².

The flat opening discs are not additional exposed surfaces.

Material volume = 33528 mm³.

Material volume = sphere volume − volume of 315 dimples.

V = $\frac{\text{(}\text{4}\text{)/(}}{\text{3}\text{)}}$π × 21³ − 315 × $\frac{\text{(}\text{2}\text{)/(}}{\text{3}\text{)}}$π × 2³.

Using π = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$, V = 38808 − 5280 = 33528 mm³.

Mallet volume = 125.6 cm³.

V = πr²h = 3.14 × 2² × 10 = 125.6 cm³.

Inner surface area = 157 cm².

The inner curved hemisphere has radius 5 cm.

Area = 2πr² = 2 × 3.14 × 5² = 157 cm².

Metal volume = $\frac{\text{(}\text{571.48}\text{)/(}}{\text{3}\text{)}}$ cm³ ≈ 190.49 cm³.

Subtract the inner hemisphere from the outer hemisphere.

V = $\frac{\text{(}\text{2}\text{)/(}}{\text{3}\text{)}}$π(R³ − r³).

V = $\frac{\text{(}\text{2}\text{)/(}}{\text{3}\text{)}}$ × 3.14 × (6³ − 5³) = $\frac{\text{(}\text{571.48}\text{)/(}}{\text{3}\text{)}}$ cm³.

Rounding to two decimal places gives 190.49 cm³.

Mallet total surface area = 150.72 cm².

Both circular ends and the curved side are included.

Area = 2πr(h + r) = 2 × 3.14 × 2 × (10 + 2) = 150.72 cm².

Box dimensions = 21 cm × 21 cm × 10.5 cm.

The box length and breadth equal the bowl’s outer diameter: 2 × 10.5 = 21 cm.

The box height equals the hemisphere’s outer radius: 10.5 cm.

Box surface area = 1764 cm².

Area = 2(lb + bh + hl).

Area = 2(21 × 21 + 21 × 10.5 + 10.5 × 21) = 1764 cm².

Positive volume difference = $\frac{\text{(}\text{15223}\text{)/(}}{\text{6}\text{)}}$ cm³ ≈ 2537.17 cm³.

Bowl capacity uses the inner radius 10 cm.

Capacity = $\frac{\text{(}\text{2}\text{)/(}}{\text{3}\text{)}}$ × 3.14 × 10³ = $\frac{\text{(}\text{6280}\text{)/(}}{\text{3}\text{)}}$ cm³.

Box volume = 21 × 21 × 10.5 = $\frac{\text{(}\text{9261}\text{)/(}}{\text{2}\text{)}}$ cm³.

Positive difference = $\frac{\text{(}\text{9261}\text{)/(}}{\text{2}\text{)}}$ − $\frac{\text{(}\text{6280}\text{)/(}}{\text{3}\text{)}}$ = $\frac{\text{(}\text{15223}\text{)/(}}{\text{6}\text{)}}$ cm³.

The box volume is larger; this difference is not the volume of metal in the bowl.

Painted area = $\frac{\text{(}\text{9251}\text{)/(}}{\text{14}\text{)}}$ cm² ≈ 660.79 cm².

Count the inner curved surface and the flat annular rim.

Area = 2πr² + π(R² − r²) = π(R² + r²).

The π = 3.14 instruction is printed in the volume alternative. The 2025 Basic 430/6/1 scheme, PDF pp. 17–18, uses 3.14 there and $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$ for this painting alternative.

Area = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$ × (10.5² + 10²) = $\frac{\text{(}\text{9251}\text{)/(}}{\text{14}\text{)}}$ cm².

Number of jars = 54.

Bowl capacity = $\frac{\text{(}\text{2}\text{)/(}}{\text{3}\text{)}}$π × 9³ = 486π cm³.

One jar capacity = π × 1.5² × 4 = 9π cm³.

Number of jars = $\frac{\text{(}\text{486π}\text{)/(}}{\text{9π}\text{)}}$ = 54.

Overflow volume = $\frac{\text{(}\text{66}\text{)/(}}{\text{7}\text{)}}$ cm³ ≈ 9.43 cm³.

Use the displacement model intended in the source: the immersed cone displaces its volume.

Overflow = $\frac{\text{(}\text{1}\text{)/(}}{\text{3}\text{)}}$πr²h = $\frac{\text{(}\text{1}\text{)/(}}{\text{3}\text{)}}$π × 1.5² × 4 = 3π cm³.

Using π = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$, overflow = $\frac{\text{(}\text{66}\text{)/(}}{\text{7}\text{)}}$ cm³.

Tank volume = 198 m³.

With r = 3 m, cylinder height = 6 m and cone height = 3 m.

V = πr²(2r) + $\frac{\text{(}\text{1}\text{)/(}}{\text{3}\text{)}}$πr²(r).

V = π × 3² × 6 + $\frac{\text{(}\text{1}\text{)/(}}{\text{3}\text{)}}$π × 3² × 3 = 63π m³.

Using π = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$, V = 198 m³.

Cylinder : cone = 6 : 1.

Cylinder volume = πr² × 2r = 2πr³.

Cone volume = $\frac{\text{(}\text{1}\text{)/(}}{\text{3}\text{)}}$πr³.

Ratio = 2 : $\frac{\text{(}\text{1}\text{)/(}}{\text{3}\text{)}}$ = 6 : 1.

Sphere radius = 6 cm.

Let R be the sphere radius. Since no clay is lost, the volumes are equal.

$\frac{\text{(}\text{4}\text{)/(}}{\text{3}\text{)}}$πR³ = π × 6² × 8 = 288π.

R³ = 216, so R = 6 cm.

Cylinder area : sphere area = 7 : 6.

Cylinder total area = 2π × 6 × (8 + 6) = 168π cm².

Sphere area = 4π × 6² = 144π cm².

Ratio = 168 : 144 = 7 : 6.

Base area = 38.5 cm².

Using π = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$, base area = πr² = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$ × 3.5² = 38.5 cm².

Hemispherical cup capacity = 2425.5 cm³.

Radius = $\frac{\text{(}\text{21}\text{)/(}}{\text{2}\text{)}}$ = 10.5 cm.

Using π = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$, V = $\frac{\text{(}\text{2}\text{)/(}}{\text{3}\text{)}}$πr³ = $\frac{\text{(}\text{2}\text{)/(}}{\text{3}\text{)}}$ × $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$ × 10.5³ = 2425.5 cm³.

Cylindrical cup capacity = 539 cm³.

Radius = $\frac{\text{(}\text{7}\text{)/(}}{\text{2}\text{)}}$ = 3.5 cm.

Using π = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$, V = πr²h = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$ × 3.5² × 14 = 539 cm³.

Curved area = 308 cm².

Using π = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$, area = 2πrh = 2 × $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$ × 3.5 × 14 = 308 cm².

Carton volume = 14400 cm³.

Volume = lbh = 30 × 32 × 15 = 14400 cm³.

Packet surface area = 470 cm².

Area = 2(lb + bh + hl).

Area = 2(15 × 8 + 8 × 5 + 5 × 15) = 470 cm².

Number of packets = 24.

Align the packet edges 15 cm, 8 cm and 5 cm with carton edges 30 cm, 32 cm and 15 cm.

Number along the three directions = $\frac{\text{(}\text{30}\text{)/(}}{\text{15}\text{)}}$, $\frac{\text{(}\text{32}\text{)/(}}{\text{8}\text{)}}$ and $\frac{\text{(}\text{15}\text{)/(}}{\text{5}\text{)}}$.

Total = 2 × 4 × 3 = 24 packets.

The arrangement confirms the packets fit; the 500 mL milk content is not the packet’s outside volume.

Cup capacity = 550 mL.

The diagram gives radius 5 cm, not diameter.

Using π = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$, V = πr²h = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$ × 5² × 7 = 550 cm³.

Since 1 cm³ = 1 mL, capacity = 550 mL.

One ball volume = $\frac{\text{(}\text{539}\text{)/(}}{\text{3}\text{)}}$ cm³ ≈ 179.67 cm³.

Radius = $\frac{\text{(}\text{7}\text{)/(}}{\text{2}\text{)}}$ = 3.5 cm.

Using π = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$, V = $\frac{\text{(}\text{4}\text{)/(}}{\text{3}\text{)}}$πr³ = $\frac{\text{(}\text{4}\text{)/(}}{\text{3}\text{)}}$ × $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$ × 3.5³ = $\frac{\text{(}\text{539}\text{)/(}}{\text{3}\text{)}}$ cm³.

Painted area = 1540 cm².

Area of 10 spheres = 10 × 4πr².

Using π = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$, area = 10 × 4 × $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$ × 3.5² = 1540 cm².

Empty space = 83300 cm³.

Box volume = 42 × 91 × 50 = 191100 cm³.

Volume of 600 balls = 600 × $\frac{\text{(}\text{539}\text{)/(}}{\text{3}\text{)}}$ = 107800 cm³.

Empty space = 191100 − 107800 = 83300 cm³.

Square-cell area = 49 cm²; lowermost layer = 78 balls.

The scheme uses a square grid: allocate one diameter-by-diameter cell to each ball.

Cell area = 7 × 7 = 49 cm². This is not the circular projection area or a contact area.

Balls along the two base edges = $\frac{\text{(}\text{42}\text{)/(}}{\text{7}\text{)}}$ = 6 and $\frac{\text{(}\text{91}\text{)/(}}{\text{7}\text{)}}$ = 13.

Number in the lowermost layer = 6 × 13 = 78.

Cylinder curved area = 176 cm².

Using π = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$, area = 2πrh = 2 × $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$ × 4 × 7 = 176 cm².

Hemisphere inner area = 308 cm².

Using π = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$, area = 2πr² = 2 × $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$ × 7² = 308 cm².

Capacity difference = $\frac{\text{(}\text{1100}\text{)/(}}{\text{3}\text{)}}$ cm³ ≈ 366.67 cm³.

Hemisphere capacity = $\frac{\text{(}\text{2}\text{)/(}}{\text{3}\text{)}}$ × $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$ × 7³ = $\frac{\text{(}\text{2156}\text{)/(}}{\text{3}\text{)}}$ cm³.

Cylinder capacity = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$ × 4² × 7 = 352 cm³.

Difference = $\frac{\text{(}\text{2156}\text{)/(}}{\text{3}\text{)}}$ − 352 = $\frac{\text{(}\text{1100}\text{)/(}}{\text{3}\text{)}}$ cm³.

Total coffee = $\frac{\text{(}\text{4268}\text{)/(}}{\text{3}\text{)}}$ cm³ ≈ 1422.67 cm³.

Two cylindrical cups contain 2 × 352 = 704 cm³.

One hemispherical cup contains $\frac{\text{(}\text{2156}\text{)/(}}{\text{3}\text{)}}$ cm³.

Total = 704 + $\frac{\text{(}\text{2156}\text{)/(}}{\text{3}\text{)}}$ = $\frac{\text{(}\text{4268}\text{)/(}}{\text{3}\text{)}}$ cm³.

Model height = 35 cm.

Total height = solid base height + ball diameter = 14 + 21 = 35 cm.

Ball volume = 4851 cm³.

Radius = $\frac{\text{(}\text{21}\text{)/(}}{\text{2}\text{)}}$ = 10.5 cm.

Using π = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$, V = $\frac{\text{(}\text{4}\text{)/(}}{\text{3}\text{)}}$πr³ = $\frac{\text{(}\text{4}\text{)/(}}{\text{3}\text{)}}$ × $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$ × 10.5³ = 4851 cm³.

Water volume = 7183 cm³.

The 2026 Standard 30/3/1 scheme, PDF p. 22, subtracts one-third of the ball’s volume from the inner cylinder volume. We follow that volume model here.

Cylinder volume up to water level = π × 20² × 7 = 8800 cm³, using π = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$.

Displaced volume = $\frac{\text{(}\text{4851}\text{)/(}}{\text{3}\text{)}}$ = 1617 cm³.

Water volume = 8800 − 1617 = 7183 cm³.

This uses the stated one-third-volume condition; it does not derive a spherical-cap volume from the drawing.

Sum of the named areas = 4686 cm².

The cylinder’s full height is 14 + 7 = 21 cm.

Using π = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$, outer cylinder curved area = 2π × 25 × 21 = 3300 cm².

Whole ball surface area = 4π × 10.5² = 1386 cm².

Sum = 3300 + 1386 = 4686 cm².

The question asks for these complete component areas, rather than just the ball surface above water.

Initial water volume = 770 cm³.

Water height = 21 − 1 = 20 cm.

Using π = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$, V = πr²h = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$ × 3.5² × 20 = 770 cm³.

Water remaining = 638 cm³.

One full straw holds π × 0.5² × 24 = 6π cm³.

Using π = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$, seven straws hold 7 × $\frac{\text{(}\text{132}\text{)/(}}{\text{7}\text{)}}$ = 132 cm³.

Water remaining = 770 − 132 = 638 cm³.

Inner surface area = 500.5 cm².

Count the inside curved wall and the circular bottom, without a top lid.

Area = 2πrh + πr².

Using π = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$, area = 2 × $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$ × 3.5 × 21 + $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$ × 3.5² = 500.5 cm².

Straw inner curved area = $\frac{\text{(}\text{528}\text{)/(}}{\text{7}\text{)}}$ cm² ≈ 75.43 cm².

Use the straw’s inner radius 0.5 cm and full length 24 cm.

Using π = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$, area = 2πrh = 2 × $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$ × 0.5 × 24 = $\frac{\text{(}\text{528}\text{)/(}}{\text{7}\text{)}}$ cm².

Capsule surface area = $\frac{\text{(}\text{1650}\text{)/(}}{\text{7}\text{)}}$ mm² ≈ 235.71 mm².

Radius r = $\frac{\text{(}\text{5}\text{)/(}}{\text{2}\text{)}}$ = 2.5 mm; cylinder length h = 10 mm.

Exposed area = cylinder curved area + two hemisphere curved areas.

Area = 2πrh + 4πr² = 2π × 2.5 × 10 + 4π × 2.5² = 75π mm².

Using π = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$, area = $\frac{\text{(}\text{1650}\text{)/(}}{\text{7}\text{)}}$ mm².

Capsule capacity = $\frac{\text{(}\text{5500}\text{)/(}}{\text{21}\text{)}}$ mm³ ≈ 261.90 mm³.

Capacity = cylinder volume + volume of a sphere.

V = πr²h + $\frac{\text{(}\text{4}\text{)/(}}{\text{3}\text{)}}$πr³ = π × 2.5² × 10 + $\frac{\text{(}\text{4}\text{)/(}}{\text{3}\text{)}}$π × 2.5³.

Using π = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$, V = $\frac{\text{(}\text{5500}\text{)/(}}{\text{21}\text{)}}$ mm³.

Cost ≈ ₹13.10.

Use the exact capacity before rounding.

Cost = $\frac{\text{(}\text{5500}\text{)/(}}{\text{21}\text{)}}$ × 0.05 = $\frac{\text{(}\text{275}\text{)/(}}{\text{21}\text{)}}$ rupees ≈ ₹13.10.

Scheme interpretation: extra volume = $\frac{\text{(}\text{1375}\text{)/(}}{\text{7}\text{)}}$ mm³ ≈ 196.43 mm³.

The 2026 second-board Basic schemes double the cylindrical length from 10 mm to 20 mm, keeping the radius and end caps unchanged: 430/8/1, PDF pp. 18–19; 430/8/2, p. 16; 430/8/3, p. 18.

Extra volume = πr²(20 − 10).

Using π = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$, extra = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$ × 2.5² × 10 = $\frac{\text{(}\text{1375}\text{)/(}}{\text{7}\text{)}}$ mm³.

Wording note: the original end-to-end length is 10 + 5 = 15 mm. If that full length were doubled to 30 mm, the new cylindrical length would be 25 mm, giving extra volume $\frac{\text{(}\text{4125}\text{)/(}}{\text{14}\text{)}}$ mm³ ≈ 294.64 mm³. The primary answer follows the printed scheme’s cylindrical-length interpretation.