RowQ
The Vault
RowQ
The Vault
CBSE Class 10 Maths · 11 questions · 26 marks
Here you glue basic solids together — a cone on a hemisphere, a cylinder topped by a dome — and work out how much surface they show and how much space they fill. The one rule that decides most answers is that surfaces which are joined together disappear from the total, while volumes simply add. When a solid is melted and recast into a new shape, nothing is lost, so equating the two volumes is the whole solution.
A solid metallic sphere of radius 6 cm is melted and recast into smaller solid spheres each of radius 2 cm. The number of small spheres formed is:
Answer
The number of small spheres is 27. Melting keeps the total volume unchanged, so: Number = volume of big sphere / volume of one small sphere. Volume of big sphere = (4/3)π(6)³ = (4/3)π × 216 = 288π cm³. Volume of one small sphere = (4/3)π(2)³ = (4/3)π × 8 = (32/3)π cm³. Number = 288π ÷ (32π/3) = 288 × 3/32 = 27. Shortcut: the ratio of radii is 3 : 1, so the ratio of volumes is 3³ : 1 = 27 : 1.
The total surface area of a solid hemisphere of radius 7 cm is (take π = 22/7):
Answer
The total surface area is 462 cm². A solid hemisphere shows a curved dome of area 2πr² plus a flat circular face of area πr². Total surface area = 2πr² + πr² = 3πr². = 3 × (22/7) × 7 × 7 = 3 × 22 × 7 = 462 cm². The answer 308 cm² is only the curved part, and it misses the flat circular base.
A right circular cone has a vertical height of 24 cm and a base radius of 7 cm. Its slant height is:
Answer
The slant height is 25 cm. For a cone, l = √(r² + h²). l = √(7² + 24²) = √(49 + 576) = √625 = 25 cm. The wrong option 31 cm comes from simply adding 7 + 24 instead of using Pythagoras.
A solid cylinder and a solid cone have equal radii and equal heights. The ratio of the volume of the cylinder to the volume of the cone is:
Answer
The ratio is 3 : 1. Volume of cylinder = πr²h. Volume of cone = (1/3)πr²h. Ratio = πr²h : (1/3)πr²h. Cancelling the common factor πr²h gives 1 : 1/3, which is the same as 3 : 1. In words, three such cones exactly fill one such cylinder.
Assertion (A): If a solid metallic cylinder is melted and recast into a solid sphere, the volume of the sphere is equal to the volume of the cylinder. Reason (R): Melting and recasting a solid changes its shape but not the quantity of material in it, so the volume is conserved.
Answer
Both A and R are true and R is the correct explanation of A. Reason R states the conservation principle for recasting: the same metal is simply given a new shape, so the space it occupies stays the same. Applying R to the situation in A, the metal of the cylinder becomes the metal of the sphere, so πr²h = (4/3)πR³ where R is the radius of the sphere. That is exactly the statement in A, so A is true and R is the reason behind it. Note that surface areas are not conserved in such a change — only volume is.
The surface area of a solid sphere is 616 cm². Find its radius. (Take π = 22/7)
Answer
For a sphere, surface area = 4πr². 4 × (22/7) × r² = 616 (88/7)r² = 616 r² = 616 × 7/88 = 4312/88 = 49. Therefore r = 7 cm.
A solid is made by fixing a hemisphere on the flat top of a solid cylinder, both having radius 3.5 cm. The cylindrical part has height 10 cm. Find the total surface area of the solid. (Take π = 22/7)
Answer
The exposed surfaces are: the curved surface of the cylinder, the flat circular base of the cylinder, and the curved surface of the hemisphere. The top face of the cylinder is covered by the hemisphere and is not counted. Curved surface of cylinder = 2πrh = 2 × (22/7) × 3.5 × 10 = 220 cm². Flat base of cylinder = πr² = (22/7) × 3.5 × 3.5 = 38.5 cm². Curved surface of hemisphere = 2πr² = 2 × 38.5 = 77 cm². Total surface area = 220 + 38.5 + 77 = 335.5 cm².
A conical paper hat has base radius 6 cm and slant height 10 cm. Find the area of paper used in its curved surface. (Take π = 3.14)
Answer
The paper forms only the curved surface of the cone, since a hat has no base. Curved surface area = πrl. = 3.14 × 6 × 10 = 188.4 cm². So 188.4 cm² of paper is used.
A solid metallic cylinder of diameter 12 cm and height 15 cm is melted down and recast into small solid cones, each of diameter 6 cm and height 10 cm. Find how many complete cones are obtained.
Answer
Step 1 — Find the volume of the cylinder. The diameter is 12 cm, so its radius R = 6 cm and height H = 15 cm. Volume of cylinder = πR²H = π × 6² × 15 = π × 36 × 15 = 540π cm³. Step 2 — Find the volume of one cone. The diameter is 6 cm, so its radius r = 3 cm and height h = 10 cm. Volume of one cone = (1/3)πr²h = (1/3) × π × 3² × 10 = (1/3) × π × 90 = 30π cm³. Step 3 — Since melting does not change the total volume: Number of cones = volume of cylinder / volume of one cone = 540π / 30π = 18. Step 4 — The division is exact, so no metal is left over. Hence 18 cones are obtained.
A circus tent is cylindrical up to a height of 4 m and conical above it. The diameter of its circular base is 42 m and the slant height of the conical top is 25 m. Find the total area of canvas required for the tent, and the cost of the canvas at ₹60 per square metre. (Take π = 22/7)
Answer
Step 1 — Find the radius. Diameter = 42 m, so r = 21 m. Step 2 — Canvas for the cylindrical part (its curved surface only; the tent has no canvas floor and no lid at the join). Curved surface of cylinder = 2πrh = 2 × (22/7) × 21 × 4. (22/7) × 21 = 66, so this is 2 × 66 × 4 = 528 m². Step 3 — Canvas for the conical top. Curved surface of cone = πrl = (22/7) × 21 × 25 = 66 × 25 = 1650 m². Step 4 — Total canvas required. = 528 + 1650 = 2178 m². Step 5 — Cost of canvas. Cost = 2178 × 60 = ₹1,30,680. So 2178 m² of canvas is needed, costing ₹1,30,680.
Read the following and answer the questions that follow: A workshop turns solid wooden blocks into decorative toys. Each toy is made of a right circular cone standing on the flat face of a hemisphere, and both parts have the same radius 3.5 cm. The conical part has a vertical height of 12 cm. (Take π = 22/7) (a) Find the volume of the hemispherical part. (b) Find the volume of the conical part. (c) Find the total volume of wood in one toy. (d) Find the slant height of the cone and hence the total surface area of the toy.
Answer
Here r = 3.5 cm for both parts, and the cone has h = 12 cm. (a) Volume of hemisphere = (2/3)πr³. r³ = 3.5 × 3.5 × 3.5 = 42.875 cm³. (22/7) × 42.875 = 134.75. Volume = (2/3) × 134.75 = 89.83 cm³ (correct to two decimal places). (b) Volume of cone = (1/3)πr²h. πr² = (22/7) × 12.25 = 38.5. Volume = (1/3) × 38.5 × 12 = (1/3) × 462 = 154 cm³. (c) Total volume of wood = 154 + 89.83 = 243.83 cm³. (d) Slant height l = √(r² + h²) = √(12.25 + 144) = √156.25 = 12.5 cm. The exposed surfaces are the curved surface of the cone and the curved surface of the hemisphere; the flat circular join is hidden. Curved surface of cone = πrl = (22/7) × 3.5 × 12.5 = 11 × 12.5 = 137.5 cm². Curved surface of hemisphere = 2πr² = 2 × 38.5 = 77 cm². Total surface area = 137.5 + 77 = 214.5 cm².
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