CBSE Class 10 Mathematics Surface Areas and Volumes Assignment Set 18

Board Level Exercise: Very Short Answer Questions (1 Mark Each)

Teacher's Note:
1. Split a solid into simple solids before measuring it.

Question 3. A solid piece of iron in the form of a cuboid of dimensions 49 cm × 33 cm × 24 cm is melted to form a solid sphere. Find the radius of the sphere.
Answer:
Volume of cuboid = 49 × 33 × 24 = 38808 cm³.
\( \dfrac{4}{3}\pi r^3 = 38808 \Rightarrow r^3 = \dfrac{38808 \times 3 \times 7}{4 \times 22} = 9261 \Rightarrow r = 21 \)
Radius of the sphere = 21 cm.

Teacher's Note:
1. Volume stays the same when a solid is melted and recast.

Question 5. A right circular cylinder of radius r cm and height h cm (h > 2r) just encloses a sphere. Find the diameter of the sphere.
Answer:
The sphere just fits inside the cylinder, so it touches the curved surface and the two bases.
Its diameter equals the diameter of the cylinder base.
Diameter of the sphere = 2r cm.

Teacher's Note:
1. A sphere that just fits in a cylinder has the same radius as the cylinder.

Question 6. During conversion of a solid from one shape to another, how will the volume of the new shape change?
Answer:
Melting and recasting does not add or remove any material.
The volume of the new shape remains unaltered.

Teacher's Note:
1. Surface area may change on recasting, but volume does not.

Question 7. Volumes of two spheres are in the ratio 64 : 27. Find the ratio of their surface areas.
Answer:
\( \dfrac{r_1^3}{r_2^3} = \dfrac{64}{27} \Rightarrow \dfrac{r_1}{r_2} = \dfrac{4}{3} \)
Ratio of surface areas = \( \dfrac{4\pi r_1^2}{4\pi r_2^2} = \left( \dfrac{4}{3} \right)^2 \) = 16 : 9.

Teacher's Note:
1. Volumes go as the cube of the radius, areas as the square.

Short Answer Questions (2 Marks Each)

Question 8. A hollow cube of internal edge 22 cm is filled with spherical marbles of diameter 0.5 cm and it is assumed that \( \dfrac{1}{8} \) space of the cube remains unfilled. Find the number of marbles that the cube can accommodate.
Answer:
Volume occupied by marbles = \( \dfrac{7}{8} \times 22^3 \) = 9317 cm³.
Volume of one marble = \( \dfrac{4}{3} \times \dfrac{22}{7} \times \left( \dfrac{1}{4} \right)^3 \) cm³ = \( \dfrac{11}{168} \) cm³ (radius 0.25 cm).
Number of marbles = \( 9317 \times \dfrac{168}{11} \) = 142296.

Teacher's Note:
1. Only 7/8 of the cube is filled by the marbles.

Question 9. A metallic spherical shell of internal and external diameters 4 cm and 8 cm, respectively, is melted and recast into the form of a cone of base diameter 8 cm. Find the height of the cone.
Answer:
Volume of shell = \( \dfrac{4}{3}\pi (4^3 - 2^3) = \dfrac{224\pi}{3} \) cm³.
Volume of cone = \( \dfrac{1}{3}\pi (4)^2 h = \dfrac{16\pi h}{3} \).
\( \dfrac{16\pi h}{3} = \dfrac{224\pi}{3} \Rightarrow h = 14 \). Height of the cone = 14 cm.

Teacher's Note:
1. Use radii (2 cm and 4 cm), not diameters, in the formulas.

Question 10. A mason constructs a wall of dimensions 270 cm × 300 cm × 350 cm with bricks each of size 22.5 cm × 11.25 cm × 8.75 cm and it is assumed that \( \dfrac{1}{8} \) space is covered by the mortar. Find the number of bricks used to construct the wall.
Answer:
Volume of wall = 270 × 300 × 350 = 28350000 cm³. Volume of bricks = \( \dfrac{7}{8} \) of this.
Volume of one brick = 22.5 × 11.25 × 8.75 = 2214.84375 cm³.
Number of bricks = \( \dfrac{\frac{7}{8} \times 28350000}{2214.84375} \) = 11200.

Teacher's Note:
1. Bricks fill only 7/8 of the wall volume; the rest is mortar.

Question 11. A medicine capsule is in the form of a cylinder of diameter 0.5 cm with two hemispheres stuck to each of its ends. The length of the entire capsule is 2 cm. Find the capacity of the capsule.
Answer:
r = 0.25 cm. Length of the cylindrical part = 2 − 2 × 0.25 = 1.5 cm.
Volume = \( \pi r^2 h + \dfrac{4}{3}\pi r^3 = \dfrac{22}{7} \times 0.0625 \times 1.5 + \dfrac{4}{3} \times \dfrac{22}{7} \times 0.015625 \) ≈ 0.2946 + 0.0655
Capacity ≈ 0.36 cm³.

Teacher's Note:
1. The two hemispheres together make one full sphere.

Question 13. Three metallic solid cubes whose edges are 3 cm, 4 cm and 5 cm are melted and formed into a single cube. Find the edge of the cube so formed.
Answer:
Volume = 3³ + 4³ + 5³ = 27 + 64 + 125 = 216 cm³.
a³ = 216 ⇒ a = 6.
Edge of the new cube = 6 cm.

Teacher's Note:
1. Add the volumes first, then take the cube root.

Question 15. Two identical cubes each of volume 64 cm³ are joined together end to end. What is the surface area of the resulting cuboid?
Answer:
Edge of each cube = 4 cm, so the cuboid is 8 cm × 4 cm × 4 cm.
T.S.A. = 2 (8×4 + 4×4 + 4×8) = 2 (32 + 16 + 32) = 160 cm².

Teacher's Note:
1. The joined faces are no longer on the surface.

Question 16. How many spherical lead shots of diameter 4 cm can be made out of a solid cube of lead whose edge measures 44 cm?
Answer:
Volume of cube = 44³ = 85184 cm³. Volume of one shot = \( \dfrac{4}{3} \times \dfrac{22}{7} \times 2^3 \) = \( \dfrac{704}{21} \) cm³.
Number of shots = \( \dfrac{85184 \times 21}{704} \) = 2541.

Teacher's Note:
1. Divide the volume of the big solid by the volume of one small solid.

Question 17. Find the number of metallic circular discs with 1.5 cm base diameter and of height 0.2 cm to be melted to form a right circular cylinder of height 10 cm and diameter 4.5 cm.
Answer:
Volume of one disc = \( \pi (0.75)^2 (0.2) \) = 0.1125π cm³.
Volume of cylinder = \( \pi (2.25)^2 (10) \) = 50.625π cm³.
Number of discs = \( \dfrac{50.625}{0.1125} \) = 450.

Teacher's Note:
1. A disc is a very short cylinder.

Long Answer Questions (3 Marks Each)

Teacher's Note:
1. Cones on the same base have volumes in the ratio of their heights.

Teacher's Note:
1. Find the full volume first, then take 5/6 of it.

Question 20. Marbles of diameter 1.4 cm are dropped into a cylindrical beaker of diameter 7 cm containing some water. Find the number of marbles that should be dropped into the beaker so that the water level rises by 5.6 cm.
Answer:
Volume of water displaced = \( \pi (3.5)^2 (5.6) \) = 68.6π cm³.
Volume of one marble = \( \dfrac{4}{3}\pi (0.7)^3 \) = 0.45733π cm³.
Number of marbles = \( \dfrac{68.6}{0.45733} \) = 150.

Teacher's Note:
1. Marbles displace water equal to their own volume.

Question 21. A wall 24 m long, 0.4 m thick and 6 m high is constructed with bricks each of dimensions 25 cm × 16 cm × 10 cm. If the mortar occupies \( \dfrac{1}{10} \)th of the volume of the wall, find the number of bricks used in constructing the wall.
Answer:
Volume of wall = 24 × 0.4 × 6 = 57.6 m³. Volume of bricks = \( \dfrac{9}{10} \) × 57.6 = 51.84 m³.
Volume of one brick = 0.25 × 0.16 × 0.10 = 0.004 m³.
Number of bricks = \( \dfrac{51.84}{0.004} \) = 12960.

Teacher's Note:
1. Convert all lengths to the same unit before multiplying.

Question 22. A solid metallic hemisphere of radius 8 cm is melted and recast into a right circular cone of base radius 6 cm. Determine the height of the cone.
Answer:
\( \dfrac{2}{3}\pi (8)^3 = \dfrac{1}{3}\pi (6)^2 h \Rightarrow 1024 = 36 h \)
h = 28.44. Height of the cone = 28.44 cm (exactly \( \dfrac{256}{9} \) cm).

Teacher's Note:
1. Equate the volume of the hemisphere with the volume of the cone.

Question 23. A rectangular water tank of base 11 m × 6 m contains water up to a height of 5 m. If the water in the tank is transferred to a cylindrical tank of radius 3.5 m, find the height of the water level in the tank.
Answer:
Volume of water = 11 × 6 × 5 = 330 m³.
\( \dfrac{22}{7} \times 3.5^2 \times h = 330 \Rightarrow 38.5\, h = 330 \Rightarrow h = 8.57 \)
Height of water level = 8.6 m (approx).

Teacher's Note:
1. The volume of water is unchanged when it is poured into another tank.

Question 24. The barrel of a fountain pen, cylindrical in shape, is 7 cm long and 5 mm in diameter. A full barrel of ink in the pen is used up on writing 3300 words on an average. How many words can be written in a bottle of ink containing one fifth of a litre?
Answer:
Radius = 0.25 cm. Volume of barrel = \( \dfrac{22}{7} \times 0.25^2 \times 7 \) = 1.375 cm³.
One fifth of a litre = 200 cm³, which is \( \dfrac{200}{1.375} \) barrels of ink.
Words = \( \dfrac{200}{1.375} \times 3300 \) = 480000.

Teacher's Note:
1. Find how many barrels the bottle holds, then multiply by 3300.

Question 26. A heap of rice is in the form of a cone of diameter 9 m and height 3.5 m. Find the volume of the rice. How much canvas cloth is required to just cover the heap?
Answer:
r = 4.5 m, h = 3.5 m. Volume = \( \dfrac{1}{3}\pi r^2 h = \dfrac{1}{3} \times \dfrac{22}{7} \times 20.25 \times 3.5 \) = 74.25 m³.
Slant height ℓ = \( \sqrt{4.5^2 + 3.5^2} = \sqrt{32.5} \) = 5.70 m.
Canvas required = C.S.A. = πrℓ = \( \dfrac{22}{7} \times 4.5 \times 5.70 \) ≈ 80.6 m² (80.63 m² with the unrounded ℓ).

Teacher's Note:
1. Covering the heap needs only the curved surface, not the base.

Question 27. A factory manufactures 120000 pencils daily. The pencils are cylindrical in shape each of length 25 cm and circumference of base 1.5 cm. Determine the cost of colouring the curved surfaces of the pencils manufactured in one day at Rs. 0.05 per dm².
Answer:
Curved surface of one pencil = circumference × length = 1.5 × 25 = 37.5 cm².
For 120000 pencils = 4500000 cm² = 45000 dm² (1 dm² = 100 cm²).
Cost = 45000 × 0.05 = Rs. 2250.

Teacher's Note:
1. C.S.A. of a cylinder = circumference × height.

Question 28. 500 persons are taking a dip into a cuboidal pond which is 80 m long and 50 m broad. What is the rise of water level in the pond, if the average displacement of the water by a person is 0.04 m³?
Answer:
Volume of water displaced = 500 × 0.04 = 20 m³.
Rise in level = \( \dfrac{20}{80 \times 50} \) = 0.005 m = 0.5 cm.

Teacher's Note:
1. Rise in level = volume displaced ÷ area of the water surface.

Question 29. Water is flowing at the rate of 15 km/h through a pipe of diameter 14 cm into a cuboidal pond which is 50 m long and 44 m wide. In what time will the level of water in pond rise by 21 cm?
Answer:
Volume needed = 50 × 44 × 0.21 = 462 m³.
Area of pipe cross-section = \( \dfrac{22}{7} \times 0.07^2 \) = 0.0154 m². Volume flowing per hour = 0.0154 × 15000 = 231 m³.
Time = \( \dfrac{462}{231} \) = 2 hours.

Teacher's Note:
1. Flow per hour = cross-section area × speed.

Question 31. A cylindrical bucket of height 32 cm and base radius 18 cm is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm, find the radius and slant height of the heap.
Answer:
Volume of sand = π × 18² × 32 = \( \dfrac{1}{3}\pi r^2 (24) \Rightarrow 8r^2 = 10368 \Rightarrow r^2 = 1296 \Rightarrow r = 36 \)
Slant height = \( \sqrt{36^2 + 24^2} = \sqrt{1872} \) = 43.27 cm.
Radius = 36 cm, slant height = 43.27 cm.

Teacher's Note:
1. Equal volumes: cylinder of sand equals conical heap.

Question 32. A rocket is in the form of a right circular cylinder closed at the lower end and surmounted by a cone with the same radius as that of the cylinder. The diameter and height of the cylinder are 6 cm and 12 cm, respectively. If the slant height of the conical portion is 5 cm, find the total surface area and volume of the rocket. [Use π = 3.14]
Answer:
r = 3 cm, cone height = \( \sqrt{5^2 - 3^2} \) = 4 cm.
T.S.A. = C.S.A. of cylinder + base + C.S.A. of cone = 2π(3)(12) + π(3)² + π(3)(5) = 96π = 301.44 cm².
Volume = π(3)²(12) + \( \dfrac{1}{3}\pi (3)^2 (4) \) = 120π = 376.8 cm³.

Teacher's Note:
1. Only the curved parts of the cone and cylinder plus one circular base are visible.

Question 33. A building is in the form of a cylinder surmounted by a hemispherical vaulted dome and contains \( 41\dfrac{19}{21} \) m³ of air. If the internal diameter of the dome is equal to its total height above the floor, find the height of the building.
Answer:
Let r be the radius of the dome. Total height = diameter of dome = 2r, so the cylinder has height 2r − r = r.
Volume = \( \pi r^2 \cdot r + \dfrac{2}{3}\pi r^3 = \dfrac{5}{3}\pi r^3 = \dfrac{880}{21} \Rightarrow r^3 = 8 \Rightarrow r = 2 \)
Height of the building = 2r = 4 m.

Teacher's Note:
1. Express the cylinder height in terms of the radius first.

Question 34. A solid right circular cone of height 120 cm and radius 60 cm is placed in a right circular cylinder full of water of height 180 cm such that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is equal to the radius of the cone.
Answer:
Volume of cylinder = π(60)²(180) = 648000π cm³. Volume of cone = \( \dfrac{1}{3}\pi (60)^2 (120) \) = 144000π cm³.
Water left = 504000π = 504000 × \( \dfrac{22}{7} \) = 1584000 cm³ = 1.584 m³.

Teacher's Note:
1. Water left = volume of cylinder − volume of the immersed cone.