Let outer radius = $R$ cm, inner radius = $r$ cm.

Given:

• $R - r = 1$ … (i)
• Volume of metal = $\pi(R^2 - r^2) \times h = 176$ cm³

Substituting $h = 14$ cm and $\pi = \dfrac{22}{7}$:

$$\frac{22}{7} \times (R^2 - r^2) \times 14 = 176$$

$$44(R^2 - r^2) = 176$$

$$R^2 - r^2 = 4$$

$$(R+r)(R-r) = 4$$

Since $R - r = 1$:

$$R + r = 4 \quad \text{…(ii)}$$

Adding (i) and (ii): $2R = 5 \Rightarrow R = 2.5$ cm

Subtracting (i) from (ii): $2r = 3 \Rightarrow r = 1.5$ cm

Outer radius = 2.5 cm, Inner radius = 1.5 cm

Source: Chapter 12, Section 12.3 — Volume of a Combination of Solids

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• The key formula is: Volume of hollow cylinder = $\pi(R^2 - r^2)h$, which uses the difference of squares.
• Examiners expect you to set up two equations from the two given conditions and solve simultaneously — show both equations clearly.
• Use $\pi = \dfrac{22}{7}$ as instructed and simplify step by step to avoid errors.

(d) $\dfrac{2\pi}{3}$ cu cm

For a cube of edge 2 cm, the largest cone has radius $r = 1$ cm and height $h = 2$ cm.

$$V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (1)^2(2) = \frac{2\pi}{3} \text{ cu cm}$$

The largest cone carved from a cube of edge $a$ has its base circle fitting the face of the cube, so radius $r = a/2$, and height = edge = $a$. Here $a = 2$ cm, giving $r = 1$ cm, $h = 2$ cm. Apply the cone volume formula directly. A common mistake is taking $r = 2$ cm (the full edge) instead of half the edge.

(c) 2 : 3

Surface area of sphere = $4\pi r^2$. Each hemisphere has CSA $= 2\pi r^2$ and a flat circular face $= \pi r^2$, so TSA of one hemisphere $= 3\pi r^2$. Total for two hemispheres $= 6\pi r^2$. Ratio $= 4\pi r^2 : 6\pi r^2 = 2:3$.

When the sphere is cut, each hemisphere gains a flat circular face ($\pi r^2$). So the two hemispheres together have surface area $2(2\pi r^2 + \pi r^2) = 6\pi r^2$, not $4\pi r^2$. The sphere's surface area stays $4\pi r^2$. The key mistake students make is forgetting to add the two flat faces. Always account for newly exposed surfaces when a solid is cut.

(i) Surface area of the bulb:

Diameter = 7 cm, so radius = 3.5 cm

Surface area = $4\pi r^2 = 4 \times \dfrac{22}{7} \times 3.5 \times 3.5 = 154 \text{ cm}^2$

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(ii) Maximum diameter of bulb leaving 1 cm from each side:

The smallest dimension of the base = 12 cm.
Maximum diameter = 12 − 1 − 1 = 10 cm

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(iii) Area of fabric used (with 2 cm fold on top and bottom):

The fold adds 2 cm on each of the two open ends, so effective height = 17 + 2 + 2 = 21 cm.

Lateral surface area of cuboid (open top & bottom) = Perimeter of base × height
= 2(24 + 12) × 21
= 2 × 36 × 21
= 1512 cm²

OR

Space available inside the lamp:

Volume of cuboid = 24 × 12 × 17 = 4896 cm³

Volume of bulb = $\dfrac{4}{3}\pi r^3 = \dfrac{4}{3} \times \dfrac{22}{7} \times 3.5^3 = \dfrac{4}{3} \times \dfrac{22}{7} \times 42.875 = 179.67 \approx 179.67 \text{ cm}^3$

Space available = 4896 − 179.67 = 4716.33 cm³

Source: Mensuration, Surface Areas and Volumes

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• (i) Standard formula $4\pi r^2$; remember to halve the diameter for radius.
• (ii) Use the smallest base dimension (12 cm) to restrict bulb size; subtract 1 cm from each side.
• (iii) Main option: The fold increases the fabric height by 2 cm on both top and bottom (total +4 cm). Since the lamp is open top and bottom, only lateral faces are covered — use Perimeter × new height.
• OR option: Subtract bulb volume from cuboid volume. Use $\frac{4}{3}\pi r^3$ for sphere. Examiners award full marks for either option attempted correctly.

(A) 11.78 m²

CSA of hemisphere = $2\pi r^2 = 2 \times \dfrac{22}{7} \times 1.4 \times 1.4 = 12.32 \text{ m}^2$

Outer surface area = $12.32 - 0.50 = \mathbf{11.82}$ — wait:

$2 \times \frac{22}{7} \times 1.96 = 12.32\ \text{m}^2$; subtracting door area: $12.32 - 0.50 = 11.82\ \text{m}^2$

(C) 11.82 m²

The tent is hemispherical, so its curved surface area = $2\pi r^2 = 2 \times \frac{22}{7} \times (1.4)^2 = 12.32\ \text{m}^2$. The door opening (0.50 m²) is cut out from this surface, so outer surface area = $12.32 - 0.50 = 11.82\ \text{m}^2$. Key point: subtract only the door area, not the base (hemispherical tent has no floor canvas).

For maximum volume, the cone carved from a solid hemisphere has radius = 10 cm and height = 10 cm.

Slant height $l = \sqrt{r^2 + h^2} = \sqrt{10^2 + 10^2} = 10\sqrt{2}$ cm

CSA of conical cavity $= \pi r l = 3.14 \times 10 \times 10\sqrt{2} = 314\sqrt{2}$ cm²

(A) $314\sqrt{2}$ cm²

Source: Chapter 12, Section 12.2

• The largest cone that fits inside a hemisphere of radius $R$ has base radius $r = R$ and height $h = R$ (the apex touches the curved surface and the base coincides with the flat face).
• Only the curved (lateral) surface area of the cone is asked — use $\pi r l$, not total surface area.
• Slant height $l = \sqrt{R^2 + R^2} = R\sqrt{2}$, giving CSA $= \pi R \cdot R\sqrt{2} = \pi R^2\sqrt{2} = 3.14 \times 100 \times \sqrt{2} = 314\sqrt{2}$ cm².

(a)

Volume of hemispherical bowl = $\dfrac{2}{3}\pi r^3 = \dfrac{2}{3}\pi (9)^3 = \dfrac{2}{3}\pi \times 729 = 486\pi$ cm³

Volume of one cylindrical jar = $\pi r^2 h = \pi (1.5)^2 \times 4 = \pi \times 2.25 \times 4 = 9\pi$ cm³

Number of jars required $= \dfrac{486\pi}{9\pi} = \mathbf{54}$

∴ 54 cylindrical jars are required.

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(b)

The conical funnel has the same height ($h = 4$ cm) and same diameter ($r = 1.5$ cm) as the cylindrical jar.

Volume of cone $= \dfrac{1}{3}\pi r^2 h = \dfrac{1}{3} \times 9\pi = 3\pi$ cm³

Volume of cylinder $= 9\pi$ cm³

Water that flows out = Volume of cone $= 3\pi = 3 \times \dfrac{22}{7} \approx \mathbf{9.43}$ cm³

(When the cone is immersed, it displaces a volume equal to the cone's own volume, so that much water flows out.)

---

• (a): Divide total volume of hemisphere by volume of one cylinder. Key formula: hemisphere volume = $\frac{2}{3}\pi r^3$.
• (b): A solid cone immersed in a full jar displaces water equal to its own volume = $\frac{1}{3}$ of cylinder's volume. Examiners expect the reasoning that water flowing out = volume of cone. Don't forget to show the formula and substitution clearly for full marks.

For the largest sphere carved from a cube of side 21 cm, the diameter of the sphere = side of cube = 21 cm, so radius $r = \dfrac{21}{2} = 10.5$ cm.

$$\text{Volume of sphere} = \frac{4}{3}\pi r^3 = \frac{4}{3} \times \frac{22}{7} \times (10.5)^3$$

$$= \frac{4}{3} \times \frac{22}{7} \times 1157.625 = 4851 \text{ cm}^3$$

The key step is recognising that the largest sphere that fits inside a cube has its diameter equal to the side of the cube. Then apply the standard formula $V = \frac{4}{3}\pi r^3$ with $\pi = \frac{22}{7}$. Examiners award one mark for identifying $r = 10.5$ cm and one mark for the correct calculation.

Given: CSA of cylinder = 176 cm², Volume = 1232 cm³

CSA = $2\pi r h = 176$ ... (i)

Volume = $\pi r^2 h = 1232$ ... (ii)

Dividing (ii) by (i):

$$\frac{\pi r^2 h}{2\pi r h} = \frac{1232}{176}$$

$$\frac{r}{2} = 7 \implies r = 14 \text{ cm}$$

Substituting in (i):

$$2 \times \frac{22}{7} \times 14 \times h = 176$$

$$88h = 176 \implies h = 2 \text{ cm}$$

The height of the cylinder is 2 cm.

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• This question tests your ability to use two formulas simultaneously: CSA = $2\pi rh$ and Volume = $\pi r^2 h$.
• The key step is dividing Volume by CSA to eliminate $h$ and find $r$ first, then back-substitute to get $h$.
• Show both steps clearly for full marks — examiners award 1 mark for finding $r$ and 1 mark for finding $h$.

Given: Radius (r) = 15 m, Height of cylinder (H) = 9 m, Height of cone (h) = 8 m

(1) Area of canvas used:

Slant height of cone: $l = \sqrt{r^2 + h^2} = \sqrt{15^2 + 8^2} = \sqrt{225 + 64} = \sqrt{289} = 17$ m

Curved Surface Area of cylinder = $2\pi r H = 2 \times \frac{22}{7} \times 15 \times 9 = \frac{5940}{7}$ m²

Curved Surface Area of cone = $\pi r l = \frac{22}{7} \times 15 \times 17 = \frac{5610}{7}$ m²

Total canvas area = $\frac{5940 + 5610}{7} = \frac{11550}{7} = \mathbf{1650 \ m^2}$

(2) Cost of canvas:

Total canvas needed = 1650 + 30 = 1680 m²

Cost = 1680 × ₹200 = ₹3,36,000

• The tent has no base, so only the curved surfaces are counted — CSA of cylinder + CSA of cone.
• The slant height $l = \sqrt{r^2+h^2}$ is a must-calculate step; examiners deduct marks if skipped.
• In part (2), remember to add the wasted canvas (30 m²) before multiplying by rate — a common error students make.
• Use $\pi = \frac{22}{7}$ unless told otherwise; it gives clean numbers here.

Volume of cuboid = $11 \times 7 \times 7 = 539 \text{ cm}^3$

Volume of one sphere $= \dfrac{4}{3}\pi r^3 = \dfrac{4}{3} \times \dfrac{22}{7} \times \dfrac{7}{2} \times \dfrac{7}{2} \times \dfrac{7}{2} = \dfrac{4}{3} \times \dfrac{22}{7} \times \dfrac{343}{8} = \dfrac{539}{3} \text{ cm}^3$

Since metal is melted and recast:

$$n = \frac{\text{Volume of cuboid}}{\text{Volume of one sphere}} = \frac{539}{\dfrac{539}{3}} = 3$$

$$\boxed{n = 3}$$

Source: Chapter 12, Section 12.3 (Volume of a Combination of Solids)

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• The key principle: when a solid is melted and recast, total volume is conserved.
• Equate volume of cuboid = $n \times$ volume of one sphere, then solve for $n$.
• Use $\pi = \frac{22}{7}$ and $r = \frac{7}{2}$ cm; the numbers cancel cleanly to give $n = 3$.
• Show both volume calculations clearly for full marks (1 mark each step typically).

Given:

• Radius of cone = Radius of hemisphere = r = 7 cm
• Height of cone = diameter = 2r = 14 cm

Volume of the solid = Volume of cone + Volume of hemisphere

$$V = \frac{1}{3}\pi r^2 h + \frac{2}{3}\pi r^3$$

$$= \frac{1}{3}\pi r^2(h + 2r)$$

$$= \frac{1}{3} \times \frac{22}{7} \times 7 \times 7 \times (14 + 2 \times 7)$$

$$= \frac{1}{3} \times \frac{22}{7} \times 49 \times 28$$

$$= \frac{1}{3} \times 22 \times 7 \times 28$$

$$= \frac{4312}{3}$$

$$= \mathbf{1437.33 \ cm^3} \text{ (approx.)}$$

Source: Chapter 12, Section 12.3 – Volume of a Combination of Solids

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• The key step is recognising that volume adds directly for combined solids (unlike surface area, where overlapping parts are excluded).
• Height of cone = diameter = 2r = 14 cm — don't use radius here.
• Factor out $\frac{1}{3}\pi r^2$ to simplify calculation.
• Examiners award marks for: correct formula (1 mark), substitution (1 mark), simplification (2 marks), final answer with unit (1 mark).
• Use $\pi = \frac{22}{7}$ unless told otherwise.

Option D: 416 cm³

Volume of cone = $\frac{1}{3} \times \text{Base Area} \times h = \frac{1}{3} \times 156 \times 8 = \frac{1248}{3} = 416 \text{ cm}^3$

The formula for volume of a cone is $V = \frac{1}{3} \times \pi r^2 \times h$. Since base area ($\pi r^2$) is directly given as 156 cm², substitute directly: $\frac{1}{3} \times 156 \times 8 = 416$ cm³. No need to find radius separately.

Given: Cone radius $r = 3$ cm, height $h = 12$ cm.

Volume of cone:
$$V_{cone} = \frac{1}{3}\pi r^2 h = \frac{1}{3} \times 3.14 \times 9 \times 12 = 113.04 \text{ cm}^3$$

Volume of ice-cream in cone (cone is $\frac{1}{6}$ unfilled):
$$V_{cone\,filled} = \frac{5}{6} \times 113.04 = 94.2 \text{ cm}^3$$

Volume of hemisphere (radius = 3 cm):
$$V_{hemi} = \frac{2}{3}\pi r^3 = \frac{2}{3} \times 3.14 \times 27 = 56.52 \text{ cm}^3$$

Total volume of ice-cream:
$$= 94.2 + 56.52 = \boxed{150.72 \text{ cm}^3}$$

Source: Volume of a Combination of Solids, Chapter 12

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• The cone is $\frac{1}{6}$ unfilled, so ice-cream fills $\frac{5}{6}$ of the cone's volume.
• The hemisphere on top has the same radius as the cone's open end (3 cm).
• Add the two volumes: filled cone part + hemisphere.
• Examiners expect clear formula, substitution, and final answer with units. Don't forget to multiply by $\frac{5}{6}$ — a common error is using the full cone volume.

Let the base radius of hemisphere = $r$, then height of cylinder $h = 2r$ (given $r = \frac{h}{2}$).

Total volume of air = Volume of cylinder + Volume of hemisphere

$$\frac{1408}{21} = \pi r^2 h + \frac{2}{3}\pi r^3$$

$$\frac{1408}{21} = \pi r^2(2r) + \frac{2}{3}\pi r^3 = 2\pi r^3 + \frac{2}{3}\pi r^3 = \frac{8}{3}\pi r^3$$

$$\frac{1408}{21} = \frac{8}{3} \times \frac{22}{7} \times r^3$$

$$r^3 = \frac{1408}{21} \times \frac{3 \times 7}{8 \times 22} = \frac{1408 \times 21}{21 \times 176} = \frac{1408}{176} = 8$$

$$r = 2 \text{ m}$$

Total height = $h + r = 2r + r = 3r = 3 \times 2 = \boxed{6 \text{ m}}$

Source: Chapter 12, Section 12.3 – Volume of a Combination of Solids

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• The key condition is: radius of hemisphere = half the height of cylinder, i.e., $r = h/2$, so $h = 2r$.
• Total volume = cylinder + hemisphere (not half-sphere surface — full volumes add up).
• Simplify to a single variable $r$, substitute $\pi = 22/7$, and solve for $r$.
• Total height = height of cylinder + radius of hemisphere (since the dome sits on top) = $2r + r = 3r$.
• Examiners expect the setup equation, substitution, and final answer clearly labelled.

Slant height $l = \sqrt{r^2 + h^2} = \sqrt{7^2 + 24^2} = \sqrt{49 + 576} = \sqrt{625} = 25$ cm

CSA of cone $= \pi r l = \dfrac{22}{7} \times 7 \times 25 = 550$ cm²

Answer: C — 550 cm²

The key step is finding slant height $l$ using $l = \sqrt{r^2 + h^2}$ before applying the CSA formula $\pi r l$. Students often mistakenly use height instead of slant height — always compute $l$ first. Here $r = 7$, $h = 24$ gives the Pythagorean triple 7-24-25.

Using CSA of cylinder = $2\pi rh$:

$94.2 = 2 \times 3.14 \times r \times 5$

$r = \dfrac{94.2}{31.4} = 3$ cm

Answer: B) 3 cm

Substitute the CSA formula $2\pi rh = 94.2$, plug in $\pi = 3.14$ and $h = 5$, then solve for $r$. The denominator $2 \times 3.14 \times 5 = 31.4$, giving $r = 3$ cm exactly. Watch units and don't confuse CSA with TSA (which adds the two circular bases).

Given: Cube side = 14 cm; largest cone carved from one face has radius r = 7 cm, height h = 14 cm.

Volume of remaining solid:

Volume of cube = $14^3 = 2744 \text{ cm}^3$

Volume of cone = $\dfrac{1}{3}\pi r^2 h = \dfrac{1}{3} \times \dfrac{22}{7} \times 7^2 \times 14 = \dfrac{1}{3} \times \dfrac{22}{7} \times 49 \times 14 = \dfrac{1}{3} \times 2156 = 718.67 \text{ cm}^3$

Volume of remaining solid = 2744 − 718.67 = 2025.33 cm³

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Surface area of remaining solid:

Slant height of cone: $l = \sqrt{r^2 + h^2} = \sqrt{49 + 196} = \sqrt{245} = 7\sqrt{5} = 7 \times 2.2 = 15.4 \text{ cm}$

Surface area = 5 faces of cube + base of cube (with circular hole) + CSA of cone

$$= 5 \times 14^2 + (14^2 - \pi r^2) + \pi r l$$

$$= 5 \times 196 + (196 - \tfrac{22}{7} \times 49) + \tfrac{22}{7} \times 7 \times 15.4$$

$$= 980 + (196 - 154) + 338.8$$

$$= 980 + 42 + 338.8$$

Surface area of remaining solid = 1360.8 cm²

Source: Chapter 12, Surface Areas and Volumes (Sections 12.2 & 12.3)

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• The largest cone carved from a face of a cube of side 14 cm has r = 7 cm (half the face) and h = 14 cm (full side).
• Volume = Volume of cube − Volume of cone. Straightforward subtraction.
• Surface area is trickier: the 5 untouched faces contribute fully; the 6th face loses the circular base of the cone, so add $(14^2 - \pi r^2)$; then add the curved surface area of the cone ($\pi r l$). The circular base of the cone is not added separately because it is open (carved out).
• Examiners award marks at each step: identifying r & h, slant height, each area component, and final addition — show all steps clearly.

Option C: 2 : 1

The cone of greatest volume inside a cylinder has the same base radius and height as the cylinder.
Volume of cylinder = πr²h; Volume of cone = ⅓πr²h.
Remaining wood = πr²h − ⅓πr²h = ⅔πr²h.
Ratio = ⅔πr²h : ⅓πr²h = 2 : 1.

The key is recognising that the largest cone carved from a cylinder shares the same radius (r) and height (h). The remaining wood is cylinder minus cone. Examiners expect the ratio simplified to lowest terms: 2:1. Don't confuse this with the cone-to-cylinder ratio (1:3).

Given: Cylinder: height = 24 cm, radius = 5 cm. Two cones hollowed out: height = 12 cm each, radius = 5 cm each.

Volume of remaining solid:

Volume of cylinder = $\pi r^2 h = \dfrac{22}{7} \times 5^2 \times 24 = \dfrac{22 \times 25 \times 24}{7} = \dfrac{13200}{7}$ cm³

Volume of one cone = $\dfrac{1}{3}\pi r^2 h = \dfrac{1}{3} \times \dfrac{22}{7} \times 25 \times 12 = \dfrac{2200}{7}$ cm³

Volume of two cones = $2 \times \dfrac{2200}{7} = \dfrac{4400}{7}$ cm³

Volume of remaining solid = $\dfrac{13200 - 4400}{7} = \dfrac{8800}{7} \approx \mathbf{1257.14}$ cm³

Surface Area of remaining solid:

The visible surfaces are: curved surface of cylinder + curved surfaces of two cones (the circular ends of the cylinder remain open as cone bases coincide).

CSA of cylinder = $2\pi r h = 2 \times \dfrac{22}{7} \times 5 \times 24 = \dfrac{5280}{7}$ cm²

Slant height of cone: $l = \sqrt{r^2 + h^2} = \sqrt{25 + 144} = \sqrt{169} = 13$ cm

CSA of two cones = $2 \times \pi r l = 2 \times \dfrac{22}{7} \times 5 \times 13 = \dfrac{2860}{7}$ cm²

Total Surface Area = $\dfrac{5280 + 2860}{7} = \dfrac{8140}{7} \approx \mathbf{1162.86}$ cm²

Source: Chapter 12, Sections 12.2 and 12.3

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• Volume: Subtract both cone volumes from the cylinder volume. Since the two cones are hollowed out internally, their base circles are inside — no circular flat areas remain on the ends of the cylinder.
• Surface Area: Only the curved surfaces count. The top and bottom circular faces of the cylinder are entirely removed (each cone's base = the cylinder's circular end), so no $\pi r^2$ terms are added. The slant height $l = 13$ cm is key — calculate it via Pythagoras.
• A common mistake is adding $2\pi r^2$ (circular ends of cylinder); avoid it here since both ends are fully hollowed by the cones.

Step 1: Number of hemispheres

Side of cube = 14 cm, diameter of hemisphere = 1.4 cm, so radius $r$ = 0.7 cm.

Number of hemispheres along one edge = $\dfrac{14}{1.4} = 10$

Total hemispheres on one face = $10 \times 10 = \mathbf{100}$

Step 2: Total Surface Area of remaining solid

TSA = (Surface area of cube) – (Area of 100 circular bases on one face) + (CSA of 100 hemispheres)

• TSA of cube $= 6 \times 14^2 = 1176 \text{ cm}^2$
• Area of 100 circular holes $= 100 \times \pi r^2 = 100 \times \dfrac{22}{7} \times (0.7)^2 = 100 \times 1.54 = 154 \text{ cm}^2$
• CSA of 100 hemispheres $= 100 \times 2\pi r^2 = 2 \times 154 = 308 \text{ cm}^2$

$$\text{TSA} = 1176 - 154 + 308 = \mathbf{1330 \text{ cm}^2}$$

Source: Surface Area of a Combination of Solids, Chapter 12

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• The key idea is that scooping removes the flat circular area from the face but adds the curved surface of each hemisphere.
• Formula used: TSA of remaining solid = TSA of cube − (100 × πr²) + (100 × 2πr²)
• Examiners award marks for: correct count (100), correct formula setup, and accurate arithmetic. Show each step clearly.

Radius of cone: $r = \sqrt{l^2 - h^2} = \sqrt{13^2 - 12^2} = \sqrt{169-144} = 5$ cm.

Total height = height of cone + radius of hemisphere = 12 + 5 = 17 cm. → Option A

The hemisphere's radius equals the cone's base radius (r = 5 cm). The hemisphere adds its radius (not diameter) to the cone's height, giving 12 + 5 = 17 cm. A common mistake is adding the diameter (10 cm) instead.

Given:

• Edge of cube = 5 cm
• Diameter of each hemispherical depression = 1.4 cm, so radius r = 0.7 cm
• Opposite faces sum to 7: pairs are (1,6), (2,5), (3,4) → total depressions = 1+2+3+4+5+6 = 21

Total Surface Area of the die:

TSA = TSA of cube − (area of 21 circular bases) + (CSA of 21 hemispheres)

Each circular base area = πr² and CSA of one hemisphere = 2πr², so each depression contributes: 2πr² − πr² = πr² (net addition per depression).

TSA of cube = 6 × (5)² = 150 cm²

Net area added per depression = πr² = $\dfrac{22}{7} × 0.7 × 0.7$ = 1.54 cm²

Total change = 21 × 1.54 = 32.34 cm²

$$\text{Total Surface Area} = 150 + 32.34 = \boxed{182.34 \text{ cm}^2}$$

Source: Surface Area of a Combination of Solids, Chapter 12

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• The key idea (from Chapter 12) is: when a depression is made, you subtract the flat circular base (πr²) but add the curved surface of the hemisphere (2πr²). Net gain per depression = πr².
• Opposite faces summing to 7 is just standard die convention — use it only to confirm total = 21 (sum 1 to 6).
• Don't forget to count ALL 21 depressions across all 6 faces.
• Examiners award marks for: correct formula setup, correct r value, correct count (21), and final answer.

The top face of the cube has area = 7 × 7 = 49 cm².
Each depression has diameter = 2 × 0.35 = 0.70 cm, so each occupies a square of side 0.70 cm.
Number of depressions = $\dfrac{49}{0.70 \times 0.70} = \dfrac{49}{0.49} = 100$

Answer: (b) 100

Each hemispherical depression occupies a circular area of diameter 0.70 cm. For packing purposes, treat each as fitting in a 0.70 cm × 0.70 cm square on the face. Divide total face area (49 cm²) by area per depression (0.49 cm²) to get the maximum count of 100. This is a straightforward area-division problem linked to Chapter 12 concepts of combining/removing solids.

(i) Total height of the fountain:

Total height = Height of solid base + Height of water filled + Radius of ball
= 14 + 7 + 10.5 = 31.5 cm

(ii) Volume of the ball:

Radius of ball = 21/2 = 10.5 cm

$$V = \frac{4}{3}\pi r^3 = \frac{4}{3} \times \frac{22}{7} \times (10.5)^3 = \frac{4}{3} \times \frac{22}{7} \times 1157.625 = \textbf{4851 cm}^3$$

(iii) Volume of water filled in the pool:

Volume of water = Volume of cylindrical annular region – Volume of submerged part of ball

Inner radius = 20 cm, Outer radius = 25 cm → Inner pool radius = 20 cm

$$\text{Volume of water region (cylinder)} = \pi r^2 h = \frac{22}{7} \times (20)^2 \times 7 = 8800 \text{ cm}^3$$

Volume of ball submerged = $\dfrac{1}{3} \times 4851 = 1617 \text{ cm}^3$

$$\text{Volume of water} = 8800 - 1617 = \textbf{7183 cm}^3$$

Source: Mensuration – Spheres and Cylinders

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• Part (i): The ball sits on top of the water level, so total height adds the radius of the ball (not the diameter) to the base and water heights.
• Part (ii): Standard formula $\frac{4}{3}\pi r^3$; use $r = 10.5$ cm and $\pi = \frac{22}{7}$.
• Part (iii): The water fills the inner cylindrical region (radius 20 cm, height 7 cm) minus the portion of the ball dipped in. One-third of ball volume = $\frac{4851}{3}$ = 1617 cm³. Subtract from the cylindrical water volume. Examiners expect both steps shown clearly for full 2 marks.

Given: radius of hemisphere = radius of cone = r = 7 cm, total height = 31 cm

Height of cone, h = 31 − 7 = 24 cm

Slant height of cone: $l = \sqrt{r^2 + h^2} = \sqrt{7^2 + 24^2} = \sqrt{49 + 576} = \sqrt{625} = 25$ cm

Total surface area of toy = CSA of hemisphere + CSA of cone

$$= 2\pi r^2 + \pi r l$$

$$= \pi r(2r + l) = \frac{22}{7} \times 7 \times (14 + 25)$$

$$= 22 \times 39 = \mathbf{858 \ cm^2}$$

Source: Chapter 12, Section 12.2

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• The toy's surface consists of only the curved surface of the hemisphere and the curved surface of the cone — the flat circular base is hidden where they join, so it is not included.
• Key step: height of cone = total height − radius of hemisphere (since the hemisphere's height equals its radius).
• Examiner expects the slant height calculation clearly shown, then the final formula $\pi r(2r + l)$ applied correctly.

Option A: $\dfrac{\pi l^3}{12}$

The maximum cone has base radius $r = \dfrac{l}{2}$ (inscribed circle of top face) and height $h = l$.

$$V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi \left(\frac{l}{2}\right)^2 \times l = \frac{\pi l^3}{12}$$

• The largest cone carved from a cube of edge $l$ has its circular base inscribed in one face of the cube, giving radius $= l/2$, and height $= l$ (the edge length).
• Substitute into $V = \frac{1}{3}\pi r^2 h$ to get $\frac{\pi l^3}{12}$.
• A common mistake is taking $r = l$; remember the radius is half the edge since the circle is inscribed in the square face.

(i) Diagonal of a cube = $\sqrt{3} \times \text{edge}$

$$= \sqrt{3} \times 6 = 6\sqrt{3} \text{ cm}$$

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(ii) Volume of a cube = $(\text{edge})^3$

$$= (7)^3 = 343 \text{ cm}^3$$

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(iii) Curved Surface Area of hemisphere = $2\pi r^2$

$$= 2 \times \frac{22}{7} \times 7 \times 7$$

$$= 2 \times 22 \times 7 = 308 \text{ cm}^2$$

The curved surface area of the hemisphere (ice-cream) is 308 cm².

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Source: Surface Areas and Volumes, NCERT Class 10 Mathematics

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• (i) The space diagonal of a cube with edge $a$ is $a\sqrt{3}$. Students sometimes confuse face diagonal ($a\sqrt{2}$) with space diagonal — use $\sqrt{3}$ here.
• (ii) Volume of cube = side³. Straightforward substitution; unit must be cm³.
• (iii) CSA of hemisphere = $2\pi r^2$ (not $3\pi r^2$, which is total surface area). Being a 2-mark question, show the formula, substitution, and final answer clearly with units.

(D) Assertion (A) is false, but Reason (R) is true.

When two cubes of side 4 cm are joined end-to-end, the cuboid formed has dimensions 8 cm × 4 cm × 4 cm. Its surface area = 2(lb + bh + hl) = 2(32 + 16 + 32) = 160 cm² — so A is actually true.

Wait — re-checking: the correct formula for TSA of cuboid is 2(lb + bh + hl), but Reason (R) states it as (lb + bh + hl) without the factor 2, which is incorrect.

TSA = 2(8×4 + 4×4 + 4×8) = 2(32+16+32) = 160 cm² ✓ (Assertion is true)
Reason gives formula without factor 2 → Reason is false.

Correct answer: (C) Assertion (A) is true, but Reason (R) is false.

Source: Surface Areas and Volumes, Chapter 12, Exercise 12.1

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• The cuboid formed: l = 8 cm, b = 4 cm, h = 4 cm → TSA = 2(lb + bh + hl) = 160 cm² → Assertion is TRUE.
• The correct formula for TSA of a cuboid is 2(lb + bh + hl). Reason (R) omits the factor of 2, writing only (lb + bh + hl) → Reason is FALSE.
• So the answer is (C). This is a classic trap — verify both the assertion by calculation AND check the reason formula carefully.

Outer radius $R = 6$ cm, inner radius $r = 6 - 1 = 5$ cm.

Volume of steel = Volume of outer hemisphere − Volume of inner hemisphere

$$= \frac{2}{3}\pi R^3 - \frac{2}{3}\pi r^3 = \frac{2}{3}\pi(6^3 - 5^3) = \frac{2}{3}\pi(216 - 125) = \frac{2}{3}\pi \times 91 = \frac{182}{3}\pi \text{ cm}^3$$

Answer: (B) $\dfrac{182}{3}\pi$

The key idea: the bowl is hollow, so subtract the volume of the inner hemisphere from the outer one. Thickness = 1 cm gives inner radius = outer radius − 1 = 5 cm. Use $V = \frac{2}{3}\pi r^3$ for a hemisphere. Many students mistakenly use full sphere formula — remember it's a hemispherical bowl, so use $\frac{2}{3}\pi r^3$.

Given: Total height = 20 cm, Diameter = 7 cm → Radius (r) = 3.5 cm

The solid has a cylinder with two hemispherical ends.

Height of cylinder = Total height − 2 × radius of hemisphere
$$= 20 - 2 \times 3.5 = 13 \text{ cm}$$

Total Volume = Volume of cylinder + Volume of 2 hemispheres

$$= \pi r^2 h + 2 \times \frac{2}{3}\pi r^3 = \pi r^2 h + \frac{4}{3}\pi r^3$$

$$= \frac{22}{7} \times (3.5)^2 \times 13 + \frac{4}{3} \times \frac{22}{7} \times (3.5)^3$$

$$= \frac{22}{7} \times 12.25 \times 13 + \frac{4}{3} \times \frac{22}{7} \times 42.875$$

$$= 500.5 + 179.67 = 680.17 \text{ cm}^3 \approx 680.17 \text{ cm}^3$$

Total volume of the solid ≈ 680.17 cm³

Source: Chapter 12, Section 12.3 — Volume of a Combination of Solids

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• The key step students miss: the cylinder's height is not 20 cm. Subtract both hemispherical radii (2 × 3.5 = 7 cm) from the total height to get the cylinder's height = 13 cm.
• Two hemispheres together = one full sphere, so use $\frac{4}{3}\pi r^3$.
• Write the formula first, then substitute — examiners award marks for each correct step (formula, substitution, calculation).

(a) $3\pi d^2$

Diameter = $2d$, so radius $r = d$.
TSA of solid hemisphere $= 2\pi r^2 + \pi r^2 = 3\pi r^2 = 3\pi d^2$.

TSA of a solid hemisphere = curved surface area ($2\pi r^2$) + circular base area ($\pi r^2$) = $3\pi r^2$. Here $r = d$, so the answer is $3\pi d^2$. Don't forget the flat circular base — a common mistake.

(a) Paper used for 4 caps:

Base circumference = 44 cm → 2πr = 44 → r = 7 cm

Slant height, $l = \sqrt{r^2 + h^2} = \sqrt{7^2 + 24^2} = \sqrt{49 + 576} = \sqrt{625} = 25$ cm

Curved surface area of 1 cap = πrl = (22/7) × 7 × 25 = 550 cm²

Paper for 4 caps = 4 × 550 = 2200 cm²

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(b) Weight of cake to order:

Radius of cake = 24/2 = 12 cm, height = 14 cm

Volume = πr²h = (22/7) × 12 × 12 × 14 = 6336 cm³

Weight = 6336 ÷ 650 × 100 g = 974.8 g ≈ 1 kg

They should order a 1 kg cake.

Source: Surface Areas and Volumes, NCERT Class 10 Maths

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• For the cap, only the curved surface area (πrl) is used — no base, since caps are open at the bottom.
• Always find slant height $l$ using $\sqrt{r^2+h^2}$ before computing CSA of a cone.
• For part (b), compute the full cylinder volume, convert to grams, then round up to the nearest 0.5 kg (since they need at least that much cake). 974.8 g exceeds 0.5 kg, so they must order 1 kg.

When three cubes of side 6 cm are joined end-to-end, the resulting cuboid has:

• Length = 6 × 3 = 18 cm, Breadth = 6 cm, Height = 6 cm

Total Surface Area = 2(lb + bh + hl)
= 2(18×6 + 6×6 + 6×18)
= 2(108 + 36 + 108)
= 2 × 252
= 504 cm²

Source: Surface Areas and Volumes, Chapter 12

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• The key step is finding the new dimensions of the cuboid: only the length changes (6×3 = 18 cm); breadth and height remain 6 cm each.
• Apply the standard TSA formula for a cuboid: 2(lb + bh + hl). Examiners award 1 mark for correct dimensions/setup and 1 mark for the correct final answer.
• A common error is adding the TSA of three separate cubes — remember, the joined faces are internal and must not be counted.

Radius of each marble $r = \dfrac{1.4}{2} = 0.7$ cm

Volume of 150 marbles $= 150 \times \dfrac{4}{3}\pi r^3 = 150 \times \dfrac{4}{3} \times \dfrac{22}{7} \times (0.7)^3$

$= 150 \times \dfrac{4}{3} \times \dfrac{22}{7} \times 0.343 = 154$ cm³

Radius of cylindrical vessel $R = \dfrac{7}{2} = 3.5$ cm

Let rise in water level $= h$

Volume of water displaced $=$ Volume of marbles

$\pi R^2 h = 154$

$\dfrac{22}{7} \times 3.5 \times 3.5 \times h = 154$

$38.5 \times h = 154$

$h = \dfrac{154}{38.5} = 4$ cm

The rise in the level of water = 4 cm.

Source: Surface Areas and Volumes, Section 12.3

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The key concept is volume conservation: volume of water raised = total volume of marbles submerged. Use $V_{\text{sphere}} = \dfrac{4}{3}\pi r^3$ and $V_{\text{cylinder}} = \pi R^2 h$, then equate them. Examiners expect clear working for each step and the final answer with units.

Given:

• Width of canal = 8 m, Depth = 6 m
• Speed of water = 12 km/h = 12000 m/h
• Standing water required = 0.05 m

Volume of water flowing in 1 hour:

Volume = length × breadth × height
$$= 12000 \times 8 \times 6 = 576000 \text{ m}^3$$

Area irrigated:

Volume of water = Area irrigated × height of standing water

$$576000 = \text{Area} \times 0.05$$

$$\text{Area} = \frac{576000}{0.05} = 1,15,20,000 \text{ m}^2 = \mathbf{1152 \text{ hectares}}$$

Source: Chapter 12, Volume of a Combination of Solids

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• The key idea: water flowing through the canal for 1 hour forms a cuboid of dimensions (speed × time) × width × depth. Its volume equals the volume spread over the irrigated area.
• Use Volume of canal water = Area × standing water height, then solve for Area.
• Examiners award marks for: correct volume calculation (1 mark), setting up the equation (1 mark), correct area in m² (1 mark), and converting to hectares (1 mark). Always convert to hectares at the end (1 hectare = 10,000 m²).

Given: Solid cylinder: height (H) = 30 cm, radius (r) = 7 cm; Conical cavity: height (h) = 24 cm, radius = 7 cm.

Slant height of cone:
$$l = \sqrt{r^2 + h^2} = \sqrt{7^2 + 24^2} = \sqrt{49 + 576} = \sqrt{625} = 25 \text{ cm}$$

Total Surface Area of remaining solid:
= CSA of cylinder + Area of top circular base + CSA of cone (inner)

$$= 2\pi r H + \pi r^2 + \pi r l$$

$$= \pi r(2H + r + l)$$

$$= \frac{22}{7} \times 7 \times (2 \times 30 + 7 + 25)$$

$$= 22 \times (60 + 7 + 25)$$

$$= 22 \times 92$$

$$= \boxed{2024 \text{ cm}^2}$$

Source: Chapter 12, Section 12.2 – Surface Area of a Combination of Solids

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• What surfaces remain visible: (1) curved surface of the full cylinder (outer), (2) the top circular base of the cylinder (the bottom base now has the cone opening into it, so only the top face is a flat circle), (3) the slant/inner surface of the hollowed cone.
• The bottom base of the cylinder is removed when the cone is hollowed out from it — the cone shares the same base, so no flat base area is added there.
• Slant height must be calculated using Pythagoras: $l = \sqrt{r^2+h^2}$.
• Examiner expects the formula written out clearly, substitution shown, and a neat final answer in cm².

Volume of sphere = $\dfrac{4}{3}\pi r^3 = \dfrac{4}{3} \times \dfrac{22}{7} \times (10.5)^3 = \dfrac{4}{3} \times \dfrac{22}{7} \times 1157.625 = 4851 \text{ cm}^3$

Volume of one cone = $\dfrac{1}{3}\pi r^2 h = \dfrac{1}{3} \times \dfrac{22}{7} \times (3.5)^2 \times 3 = \dfrac{1}{3} \times \dfrac{22}{7} \times 12.25 \times 3 = 38.5 \text{ cm}^3$

Number of cones = $\dfrac{4851}{38.5} = \boxed{126}$

Source: Chapter 12, Section 12.3

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• The key principle: volume of sphere = total volume of all cones formed (no material is lost in recasting).
• Use $r = 10.5$ cm for sphere; $r = 3.5$ cm, $h = 3$ cm for each cone.
• Examiner expects the formula, substitution, and final answer clearly shown — all three steps earn the 2 marks.

Given: Diameter of golf ball = 4.2 cm → Radius R = 2.1 cm; Radius of each dimple r = 2 mm = 0.2 cm; Number of dimples = 315

(i) Surface area of one dimple (hemispherical):

$$= 2\pi r^2 = 2 \times \frac{22}{7} \times (0.2)^2 = 2 \times \frac{22}{7} \times 0.04 = \frac{1.76}{7} \approx 0.2514 \text{ cm}^2$$

(ii) Volume of material dug out for one dimple:

$$= \frac{2}{3}\pi r^3 = \frac{2}{3} \times \frac{22}{7} \times (0.2)^3 = \frac{2}{3} \times \frac{22}{7} \times 0.008 \approx 0.01676 \text{ cm}^3$$

(iii) Total surface area exposed to surroundings:

Surface area of ball = $4\pi R^2 = 4 \times \frac{22}{7} \times (2.1)^2 = 4 \times \frac{22}{7} \times 4.41 = 55.44 \text{ cm}^2$

Area of 315 circular holes removed = $315 \times \pi r^2 = 315 \times \frac{22}{7} \times 0.04 = 39.6 \text{ cm}^2$

Curved area of 315 dimples added = $315 \times 2\pi r^2 = 315 \times 0.2514 = 79.2 \text{ cm}^2$

Total surface area $= 55.44 - 39.6 + 79.2 = \mathbf{95.04 \text{ cm}^2}$

OR

Volume of golf ball:

$$V = \frac{4}{3}\pi R^3 = \frac{4}{3} \times \frac{22}{7} \times (2.1)^3 = \frac{4}{3} \times \frac{22}{7} \times 9.261 = 38.808 \text{ cm}^3$$

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• Part (i): Curved surface area of hemisphere = $2\pi r^2$ (not $3\pi r^2$, as the base circle is not counted — the dimple opens onto the ball's surface).
• Part (ii): Volume of hemisphere = $\frac{2}{3}\pi r^3$.
• Part (iii): Total exposed area = (sphere's surface) − (315 flat circles cut away) + (315 hemispherical curved surfaces added). Examiners specifically look for all three components.
• For the OR, simply apply $\frac{4}{3}\pi R^3$ with R = 2.1 cm. Both options are 2-mark questions; show substitution clearly for full marks.

(b) 4000 m³

Distance covered in 2 min = $2000 \times \dfrac{2}{60} = \dfrac{200}{3}$ m. Volume = $3 \times 40 \times \dfrac{200}{3} = 8000$ m³.

Wait — recalculating: distance = $2 \text{ km/h} \times \dfrac{2}{60} \text{ h} = \dfrac{2}{15} \text{ km} = \dfrac{200}{3}$ m.

Volume $= 3 \times 40 \times \dfrac{200}{3} = 8000$ m³. → (c) 8000 m³

The river cross-section acts as the base (depth × width = 3 × 40 = 120 m²). In 2 minutes at 2 km/h, water travels $2000 × \frac{2}{60} = \frac{200}{3}$ m. Volume = 120 × 200/3 = 8000 m³. Option (c) is correct. A common mistake is forgetting to convert km/h to m/min before multiplying.

Given: Diameter = 3 cm → radius (r) = 1.5 cm; Total length = 12 cm; Height of each cone = 2 cm.

Height of cylinder = 12 − 2 − 2 = 8 cm

The model consists of 1 cylinder + 2 cones.

Volume of cylinder:
$$V_{\text{cyl}} = \pi r^2 h = \frac{22}{7} \times (1.5)^2 \times 8 = \frac{22}{7} \times 2.25 \times 8 = \frac{396}{7} = 56.57 \text{ cm}^3$$

Volume of 2 cones:
$$V_{2\text{ cones}} = 2 \times \frac{1}{3}\pi r^2 h = \frac{2}{3} \times \frac{22}{7} \times (1.5)^2 \times 2 = \frac{2}{3} \times \frac{22}{7} \times 4.5 = \frac{66}{7} = 9.43 \text{ cm}^3$$

Total volume of air in the model:
$$V = 56.57 + 9.43 = \boxed{66 \text{ cm}^3}$$

Source: Surface Areas and Volumes, Section 12.3 (Exercise 12.2, Q.2)

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• The key step is finding the cylinder's height by subtracting both cone heights from total length: 12 − 2 − 2 = 8 cm.
• Volume of the whole model = Volume of cylinder + Volume of two cones (not one). Students often forget to double the cone volume.
• The answer simplifies neatly to exactly 66 cm³ — examiners expect this clean value.
• Use $\pi = \dfrac{22}{7}$ as instructed in Exercise 12.2.

(i) Area of base of cylindrical cup:

Radius = 7/2 = 3.5 cm

Area of base = $\pi r^2 = \frac{22}{7} \times 3.5 \times 3.5 = \mathbf{38.5 \text{ cm}^2}$

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(ii) Capacity of hemispherical cup:

Radius = 21/2 = 10.5 cm

Capacity = $\frac{2}{3}\pi r^3 = \frac{2}{3} \times \frac{22}{7} \times 10.5 \times 10.5 \times 10.5$

$= \frac{2}{3} \times \frac{22}{7} \times 1157.625 = \mathbf{2425.5 \text{ cm}^3}$

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(iii) Curved surface area of cylindrical cup:

Radius = 3.5 cm, Height = 14 cm

CSA = $2\pi r h = 2 \times \frac{22}{7} \times 3.5 \times 14 = \mathbf{308 \text{ cm}^2}$

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Source: Surface Areas and Volumes, Chapter 12

• For (i), only the base circle area is needed — not total surface area.
• For (ii), a hemisphere's volume is $\frac{2}{3}\pi r^3$; "capacity" means volume, so use this formula.
• For (iii), CSA of cylinder = $2\pi rh$ (excludes top and bottom circles).
• Always halve the diameter to get radius before substituting. Use $\pi = \frac{22}{7}$ unless told otherwise.

(d) $\dfrac{3}{4}\pi d^2$

Since radius $r = \dfrac{d}{2}$, TSA of solid hemisphere $= 3\pi r^2 = 3\pi \left(\dfrac{d}{2}\right)^2 = \dfrac{3}{4}\pi d^2$.

TSA of a solid hemisphere = curved surface area + circular base = $2\pi r^2 + \pi r^2 = 3\pi r^2$. Substituting $r = d/2$ gives $\frac{3}{4}\pi d^2$. Students often forget to include the flat circular base, which leads to the wrong option (b).

Given: Height of cylinder (h) = 5.8 cm, radius (r) = 2.1 cm

When a hemisphere is scooped out from each end, the article exposes:

• Curved Surface Area (CSA) of the cylinder
• CSA of two hemispheres (one from each end)

(The flat circular ends of the cylinder are removed along with the scooping.)

Total Surface Area = CSA of cylinder + 2 × CSA of hemisphere

$$= 2\pi rh + 2 \times 2\pi r^2$$

$$= 2\pi r(h + 2r)$$

$$= 2 \times \frac{22}{7} \times 2.1 \times (5.8 + 2 \times 2.1)$$

$$= 2 \times \frac{22}{7} \times 2.1 \times (5.8 + 4.2)$$

$$= 2 \times \frac{22}{7} \times 2.1 \times 10$$

$$= 2 \times 22 \times 0.3 \times 10$$

$$= 132 \text{ cm}^2$$

∴ Total surface area of the article = 132 cm²

Source: Chapter 12, Section 12.2 — Surface Area of a Combination of Solids

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• The key insight: when hemispheres are scooped out from each end, the flat circular ends of the cylinder disappear, but the curved surfaces of the hemispheres are now part of the outer surface. So TSA = CSA of cylinder + 2 × CSA of hemisphere (not TSA of hemisphere, since the flat face is open/removed).
• Examiners award marks for: correct formula, correct substitution, correct simplification, and the final answer with units.
• Note: $2\pi r \times 2r = 4\pi r^2$ represents both hemispheres combined = one full sphere's surface area. Factoring as $2\pi r(h + 2r)$ saves calculation steps.

(i) Volume of the cuboidal carton:

Volume = l × b × h = 30 × 32 × 15 = 14,400 cm³

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(ii) Total surface area of a milk packet:

TSA = 2(lb + bh + lh)
= 2(15×8 + 8×5 + 15×5)
= 2(120 + 40 + 75)
= 2 × 235
= 470 cm²

OR

Number of milk packets in a carton:

Volume of carton = 14,400 cm³
Volume of one milk packet = 15 × 8 × 5 = 600 cm³

Number of packets = 14,400 ÷ 600 = 24 packets

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(iii) Milk the cup can hold:

The passage states each milk packet contains 500 mL of milk. Since the cup holds the contents of one packet, the cup can hold 500 mL of milk.

Source: Case Study passage, Mensuration (Surface Areas and Volumes)

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• (i) is straightforward — apply the volume formula directly.
• (ii) TSA formula must be written and each step shown. The OR alternative uses ratio of volumes — both are standard 2-mark approaches.
• (iii) The figure-based answer relies on the passage's stated capacity of 500 mL per packet; if the cup holds one packet's worth, the answer is 500 mL. (If your actual figure shows a cup with given dimensions like a cylinder/cone, apply the relevant volume formula instead and state the result in mL using 1 cm³ = 1 mL.)

Total surface area of a solid hemisphere = $2\pi r^2 + \pi r^2 = 3\pi r^2$

Ratio = $3\pi r^2 : r^2 = 3\pi : 1$

Answer: (C) $3\pi : 1$

The total surface area of a solid hemisphere has two parts: the curved surface area ($2\pi r^2$) and the flat circular base ($\pi r^2$), giving $3\pi r^2$. Dividing by $r^2$ gives the ratio $3\pi : 1$. Students often forget to include the base area — that's the most common mistake here.

Given: Total length of capsule = 14 mm, Diameter = 4 mm → Radius (r) = 2 mm

Length of cylindrical part = 14 − 2r − 2r = 14 − 2(2) − 2(2) = 14 − 4 − 4 = 6 mm

(Each hemisphere contributes a length equal to its radius.)

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Surface Area:

TSA of capsule = CSA of cylinder + CSA of 2 hemispheres

$$= 2\pi r h + 2 \times 2\pi r^2$$

$$= 2 \times \frac{22}{7} \times 2 \times 6 + 2 \times 2 \times \frac{22}{7} \times 2 \times 2$$

$$= \frac{528}{7} + \frac{352}{7} = \frac{880}{7} \approx 160 \text{ mm}^2$$

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Volume:

Volume of capsule = Volume of cylinder + Volume of 2 hemispheres

$$= \pi r^2 h + 2 \times \frac{2}{3}\pi r^3 = \pi r^2 h + \frac{4}{3}\pi r^3$$

$$= \frac{22}{7} \times 4 \times 6 + \frac{4}{3} \times \frac{22}{7} \times 8$$

$$= \frac{528}{7} + \frac{704}{21} = \frac{1584}{21} + \frac{704}{21} = \frac{2288}{21} \approx 109.0 \text{ mm}^3$$

Surface Area ≈ 160 mm²; Volume ≈ 109 mm³

Source: Chapter 12, Section 12.2 & 12.3

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• The capsule consists of one cylinder + two hemispheres. The two hemispheres together form one complete sphere.
• The cylindrical height = total length − diameter of both hemispheres = 14 − 2(2) − 2(2) = 6 mm. Students commonly make errors here.
• Surface area = CSA of cylinder + CSA of 2 hemispheres (= full sphere surface area $4\pi r^2$). The flat circular ends are not visible/counted because hemispheres cover them.
• Volume = Volume of cylinder + Volume of full sphere ($\frac{4}{3}\pi r^3$).
• Note: The textbook Exercise 12.1 Q.6 uses diameter = 5 mm; this question uses diameter = 4 mm — use r = 2 mm throughout.

Given:

• Lower cylinder: height $h_1 = 200$ cm, base diameter $= 28$ cm $\Rightarrow r_1 = 14$ cm
• Upper cylinder: height $h_2 = 50$ cm, radius $r_2 = 7$ cm
• $\pi = \dfrac{22}{7}$

Volume of lower cylinder:
$$V_1 = \pi r_1^2 h_1 = \frac{22}{7} \times 14 \times 14 \times 200 = 123200 \text{ cm}^3$$

Volume of upper cylinder:
$$V_2 = \pi r_2^2 h_2 = \frac{22}{7} \times 7 \times 7 \times 50 = 7700 \text{ cm}^3$$

Total volume of pole:
$$V = V_1 + V_2 = 123200 + 7700 = 130900 \text{ cm}^3$$

Mass of pole:
$$\text{Mass} = 130900 \times 8 = 1047200 \text{ g} = 1047.2 \text{ kg}$$

∴ The mass of the iron pole is 1047.2 kg.

Source: Chapter 12, Section 12.3 – Volume of a Combination of Solids

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• The pole is a combination of two cylinders, so total volume = sum of individual volumes (no part is shared/removed).
• Use $r_1 = 14$ cm (half of diameter 28 cm) carefully — a common error is using diameter directly in the formula.
• The textbook (Exercise 12.2, Q.6) has a similar problem with slightly different dimensions; the method is identical.
• Final answer must be converted from grams to kg (divide by 1000) for a sensible unit.
• Examiners award marks step-by-step: formula, substitution, each volume, total volume, and final mass — show all steps clearly.

Given: Diameter = 28 m → Radius (r) = 14 m, Height of cylinder (h₁) = 8 m, Total height = 18.5 m → Height of cone (h₂) = 18.5 − 8 = 10.5 m

(i) Slant height of conical part:

$$l = \sqrt{r^2 + h_2^2} = \sqrt{14^2 + 10.5^2} = \sqrt{196 + 110.25} = \sqrt{306.25} = 17.5 \text{ m}$$

(ii) Floor area of tent:

$$= \pi r^2 = \frac{22}{7} \times 14 \times 14 = 616 \text{ m}^2$$

(iii) Area of cloth used (CSA of cylinder + CSA of cone):

$$= 2\pi r h_1 + \pi r l = \pi r(2h_1 + l) = \frac{22}{7} \times 14 \times (16 + 17.5) = 44 \times 33.5 = \mathbf{1474 \text{ m}^2}$$

OR Total volume of air:

$$= \pi r^2 h_1 + \frac{1}{3}\pi r^2 h_2 = \pi r^2\!\left(h_1 + \frac{h_2}{3}\right) = \frac{22}{7} \times 196 \times \left(8 + 3.5\right) = 616 \times 11.5 = \mathbf{7084 \text{ m}^3}$$

Source: Mensuration (Cylinders and Cones), Class 10 Mathematics

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• Slant height uses Pythagoras: $l = \sqrt{r^2 + h_{cone}^2}$. Remember to find cone's height separately (Total − Cylinder height).
• Floor area is just the base circle (πr²) — not curved surface.
• For cloth area, only curved surfaces are counted (no base); combine CSA of cylinder and cone.
• For volume, add volume of cylinder + volume of cone. Factorising πr² saves calculation time and reduces errors.

(D) Assertion (A) is not true but Reason (R) is true.

When two cubes of edge 10 cm are joined, the cuboid formed is 20 cm × 10 cm × 10 cm. TSA = 2(20×10 + 10×10 + 20×10) = 2(200+100+200) = 1000 cm², not 1200 cm². Reason (R) is correct: each face of a 10 cm cube = 10×10 = 100 cm².

When two cubes are joined end-to-end, two faces (one from each cube) are hidden, so the total surface area decreases. Original TSA of two cubes = 2 × 600 = 1200 cm², but joining removes 2 faces of 100 cm² each, giving 1200 − 200 = 1000 cm². The Assertion states 1200 cm², which is wrong. Reason (R) is independently correct (area of each face = 100 cm²), so answer is (D).

Given: Height of cone (h) = 8 cm, radius (r) = 5 cm, radius of each lead shot = 0.5 cm

Step 1: Volume of water in the cone (= volume of cone)

$$V_{\text{cone}} = \frac{1}{3}\pi r^2 h = \frac{1}{3} \times \frac{22}{7} \times 5 \times 5 \times 8 = \frac{2200}{21} \approx \frac{2200}{21} \text{ cm}^3$$

Step 2: Volume of water that flows out (one-fourth)

$$V_{\text{water out}} = \frac{1}{4} \times \frac{2200}{21} = \frac{550}{21} \text{ cm}^3$$

Step 3: Volume of each lead shot (sphere, r = 0.5 cm)

$$V_{\text{shot}} = \frac{4}{3}\pi r^3 = \frac{4}{3} \times \frac{22}{7} \times (0.5)^3 = \frac{4}{3} \times \frac{22}{7} \times \frac{1}{8} = \frac{11}{21} \text{ cm}^3$$

Step 4: Number of lead shots

The volume of lead shots dropped = volume of water displaced = $\dfrac{550}{21}$ cm³

$$n = \frac{550/21}{11/21} = \frac{550}{11} = \boxed{100}$$

The number of lead shots dropped into the vessel is 100.

Source: Chapter 12, Section 12.3 (Volume of a Combination of Solids), Exercise 12.2, Q.5

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• The key concept: the volume of lead shots that sink into the vessel equals the volume of water displaced (which flows out). So: n × V(one shot) = ¼ × V(cone).
• Examiners award marks at each step — writing the formula, substituting correctly, computing V(cone), V(water out), V(shot), and the final division. Don't skip steps.
• Use $\pi = \frac{22}{7}$ throughout (as instructed in the exercise) so fractions cancel neatly to give exactly 100.
• A common error is forgetting the $\frac{1}{4}$ — the shots displace only the water that flows out, not the entire cone's volume.

Option B: 4 : 25

If volume ratio = 8 : 125, then side ratio = $\sqrt[3]{8} : \sqrt[3]{125}$ = 2 : 5. Surface area ratio = $2^2 : 5^2$ = 4 : 25.

Key concept: Volume of cube $\propto a^3$ and surface area $\propto a^2$. Find the cube root of the volume ratio to get the side ratio, then square it for the surface area ratio. Don't confuse volume ratio directly with surface area ratio.

Given:

• Bigger cylinder: length (h) = 12 cm, diameter = 7 cm → radius (R) = 3.5 cm
• Smaller cylinder: length (h) = 5 cm, diameter = 2.1 cm → radius (r) = 1.05 cm

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(i) Volume of bigger cylindrical part:

$$V = \pi R^2 h = \frac{22}{7} \times (3.5)^2 \times 12 = \frac{22}{7} \times 12.25 \times 12 = 462 \text{ cm}^3$$

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(ii) Curved Surface Area of bigger cylindrical part:

$$\text{CSA} = 2\pi R h = 2 \times \frac{22}{7} \times 3.5 \times 12 = 264 \text{ cm}^2$$

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(iii) Ratio of volume of bigger part to total volume of two smaller parts:

Volume of one smaller cylinder $= \pi r^2 h = \frac{22}{7} \times (1.05)^2 \times 5 = \frac{22}{7} \times 1.1025 \times 5 = 17.325 \text{ cm}^3$

Total volume of two smaller cylinders $= 2 \times 17.325 = 34.65 \text{ cm}^3$

$$\text{Ratio} = \frac{462}{34.65} = \frac{46200}{3465} = \frac{40}{3}$$

$$\boxed{\text{Required ratio} = 40 : 3}$$

Source: Surface Areas and Volumes, CBSE Class 10 Mathematics

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• Always halve the diameter to get radius before substituting into formulae.
• Use $\pi = \frac{22}{7}$ unless told otherwise; it simplifies cleanly here.
• For CSA, use $2\pi rh$ (not total surface area, which includes circular ends).
• For the ratio in part (iii), compute total volume of both smaller cylinders combined, then simplify the ratio fully — examiners expect the final simplified form (40:3).

Let the height of the cylindrical part = $h$ m.

Then, base radius of hemisphere = $r = \dfrac{h}{2}$ m.

Total volume of room = Volume of cylinder + Volume of hemisphere

$$= \pi r^2 h + \frac{2}{3}\pi r^3$$

$$= \pi \left(\frac{h}{2}\right)^2 h + \frac{2}{3}\pi \left(\frac{h}{2}\right)^3$$

$$= \frac{\pi h^3}{4} + \frac{2\pi h^3}{24} = \frac{\pi h^3}{4} + \frac{\pi h^3}{12} = \frac{3\pi h^3 + \pi h^3}{12} = \frac{4\pi h^3}{12} = \frac{\pi h^3}{3}$$

Given volume $= \dfrac{1408}{21}$ m³:

$$\frac{\pi h^3}{3} = \frac{1408}{21}$$

$$\frac{22}{7} \times \frac{h^3}{3} = \frac{1408}{21}$$

$$h^3 = \frac{1408}{21} \times \frac{21}{22} = \frac{1408}{22} = 64$$

$$h = 4 \text{ m}$$

The height of the cylindrical part is 4 m.

Source: Chapter 12, Section 12.3 — Volume of a Combination of Solids

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• Key setup: since the base radius of the hemisphere equals the radius of the cylinder, and $r = h/2$, express everything in terms of $h$.
• Volume = $\pi r^2 h + \frac{2}{3}\pi r^3$ (cylinder + hemisphere). Combine carefully to get $\frac{\pi h^3}{3}$.
• Substituting $\pi = \frac{22}{7}$ and the given volume simplifies to $h^3 = 64$, so $h = 4$ m.
• Show all algebraic steps clearly — 3-mark questions reward working, not just the final answer.

(C) Assertion (A) is true, but Reason (R) is false.

Assertion is correct — joining two hemispheres of the same radius along their bases gives a sphere. However, the TSA of a sphere of radius $r$ is $4\pi r^2$, not $3\pi r^2$. So Reason is false.

• The Assertion is a basic geometric fact (two hemispheres → sphere). ✓
• The standard formula for TSA of a sphere is $4\pi r^2$. The Reason incorrectly states $3\pi r^2$ (which is the TSA of a hemisphere = $2\pi r^2 + \pi r^2 = 3\pi r^2$).
• Since A is true but R is false, the answer is (C).

Given:

• Internal radius of hemisphere, r = 5√2 cm
• External radius of hemisphere, R = 10 cm
• Cone: height h = 5√7 cm, base radius = 5√2 cm (= r)

Slant height of cone:
$$l = \sqrt{r^2 + h^2} = \sqrt{(5\sqrt{2})^2 + (5\sqrt{7})^2} = \sqrt{50 + 175} = \sqrt{225} = 15 \text{ cm}$$

Total Surface Area = CSA of cone + CSA of outer hemisphere + CSA of inner hemisphere + Ring area (annular base)

• CSA of cone = $\pi r l = \pi (5\sqrt{2})(15) = 75\sqrt{2}\,\pi$
• CSA of outer hemisphere = $2\pi R^2 = 2\pi(10)^2 = 200\pi$
• CSA of inner hemisphere = $2\pi r^2 = 2\pi(5\sqrt{2})^2 = 100\pi$
• Annular ring area = $\pi(R^2 - r^2) = \pi(100 - 50) = 50\pi$

TSA $= 75\sqrt{2}\,\pi + 200\pi + 100\pi + 50\pi$
$$= 75(1.4)\pi + 350\pi = 105\pi + 350\pi = \boxed{455\pi \text{ cm}^2}$$

Source: Surface Area of a Combination of Solids, Chapter 12

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• The visible surfaces are: CSA of cone (top), CSA of outer hemisphere (outside), CSA of inner hemisphere (inside hollow), and the annular ring at the base (the flat circular border between inner and outer radius).
• The base of the cone coincides with the inner hollow opening, so no extra circular base is added for the cone.
• Key step: slant height l = 15 cm (clean calculation — check this first in the exam).
• Use √2 = 1.4 only at the final step: 75√2 = 75 × 1.4 = 105.
• Examiners award marks for identifying all four surface components correctly and computing l accurately.

Given: Cuboid: 14 m × 25 m × 16 m; Semi-cylinder mounted on top (diameter = 14 m, so radius = 7 m, length = 25 m)

The cloth covers:

• Two side walls of cuboid (excluding top): 2 × (25 × 16) = 800 m²
• Two end walls (rectangles minus semicircles) + two semicircular ends of cylinder:

2 × (14 × 16) + 2 × ½πr² = 448 + 22/7 × 7 × 7 = 448 + 154 = 602 m²

• Curved surface of semi-cylinder: πrl = 22/7 × 7 × 25 = 550 m²

Total area of cloth = 800 + 602 + 550 = 1952 m²

Source: Surface Area of a Combination of Solids, Chapter 12, Section 12.2

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• The top face of the cuboid is replaced by the semi-cylinder, so it is not included.
• The two end walls are rectangles (14 × 16) each, but the semi-circular portion at the top of each end is part of the semi-cylinder's flat ends — add two semicircles (= one full circle = πr²).
• The curved surface of the semi-cylinder = ½ × 2πrl = πrl.
• Examiner expects all three components identified and added correctly. Show each step clearly for full marks.

Volume of cone = Volume of two spherical scoops

$$\frac{1}{3}\pi r^2 h = 2 \times \frac{4}{3}\pi \left(\frac{r}{2}\right)^3 = 2 \times \frac{4}{3}\pi \cdot \frac{r^3}{8} = \frac{\pi r^3}{3}$$

$$\frac{1}{3}\pi r^2 h = \frac{\pi r^3}{3} \implies h = r \implies h : 2r = 1 : 2$$

Answer: B) 1 : 2

Set volume of cone equal to total volume of the two scoops. Each scoop has radius $r/2$, so its volume is $\frac{4}{3}\pi(r/2)^3 = \frac{\pi r^3}{6}$; two scoops give $\frac{\pi r^3}{3}$. Equate with $\frac{1}{3}\pi r^2 h$ to get $h = r$, so $h:2r = r:2r = 1:2$.

Option (D) $4\pi r^3$

Volume of cylinder (radius $r$, height $6r$) $= \pi r^2 \times 6r = 6\pi r^3$.
Volume of 3 spheres $= 3 \times \dfrac{4}{3}\pi r^3 = 4\pi r^3$.
Volume of air $= 6\pi r^3 - 4\pi r^3 = \mathbf{4\pi r^3}$.

Source: Chapter 12, Section 12.3

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• Three balls stacked in a cylinder means the cylinder's height = $3 \times 2r = 6r$ and radius = $r$.
• Subtract total sphere volume from cylinder volume to get air space.
• Examiners expect the setup (cylinder dimensions), the subtraction, and the correct answer. Do not confuse diameter with radius when setting up cylinder height.