CBSE Class 10 Mathematics Surface Areas and Volumes Assignment Set 21

Exercise-1 Objective Questions: Surface Area and Volume of Solid Figures

Teacher's Note:
1. Multiply the three edges and equate to the volume.

Question A-2. The volume of a cube (in cubic cm) plus three times the total length of its edges (in cms) is equal to twice its surface area (in sq. cm). The length of its diagonal is:
(A) 6   (B) 6√3   (C) 3√6   (D) 6√6
Answer:
Ans. (B) 6√3
a³ + 3(12a) = 2(6a²) ⇒ a² − 12a + 36 = 0 ⇒ a = 6.
Diagonal = a√3 = 6√3.

Teacher's Note:
1. A cube has 12 equal edges.

Question A-3. An aquarium has a rectangular base that measures 100 cm by 40 cm and has a height of 50 cm. It is filled with water to a height of 40 cm. A brick with a rectangular base measures 40 cm by 20 cm and a height of 10 cm is placed in the aquarium. The water level (in cm) rises by
(A) 0.5   (B) 1   (C) 1.5   (D) 2
Answer:
Ans. (D) 2
Volume of the brick = 40 × 20 × 10 = 8000 cm³ (fully under water).
Rise = \( \dfrac{8000}{100 \times 40} \) = 2 cm.

Teacher's Note:
1. Rise in level = volume of the brick ÷ area of the base of the aquarium.

Question A-4. The curved surface area of a cylinder is 264 m² and its volume is 924 m³. The ratio of its diameter to its height is :
(A) 7 : 6   (B) 6 : 7   (C) 3 : 7   (D) 7 : 3
Answer:
Ans. (D) 7 : 3
\( \dfrac{V}{\text{C.S.A.}} = \dfrac{\pi r^2 h}{2\pi r h} = \dfrac{r}{2} = \dfrac{924}{264} = 3.5 \Rightarrow r = 7 \).
2π(7)h = 264 ⇒ h = 6. Diameter : height = 14 : 6 = 7 : 3.

Teacher's Note:
1. Dividing volume by curved area gives r/2.

Question A-5. A cylindrical tank has base radius 7 m and height 9 m. If 2/3 of the tank is filled with water, find the volume of water in the tank. [Use π = 22/7]
(A) 1386 cu m   (B) 462 cu m   (C) 824 cu m   (D) 924 cu m
Answer:
Ans. (D) 924 cu m
Volume of tank = \( \dfrac{22}{7} \times 49 \times 9 \) = 1386 m³.
Water = \( \dfrac{2}{3} \times 1386 \) = 924 m³.

Teacher's Note:
1. Option (A) is the whole tank, not 2/3 of it.

Question A-6. Two steel sheets each of length a₁ and breadth a₂ are used to prepare the surface of two right circular cylinders: one having volume V₁ and height a₂ and the other having volume V₂ and height a₁. Then
(A) V₁ = V₂   (B) a₁V₁ = a₂V₂   (C) a₂V₁ = a₁V₂   (D) V₁²/a₁ = V₂²/a₂
Answer:
Ans. (C) a₂V₁ = a₁V₂
First cylinder: height a₂, circumference a₁, so r₁ = \( \dfrac{a_1}{2\pi} \) and V₁ = \( \dfrac{a_1^2 a_2}{4\pi} \).
Second cylinder: V₂ = \( \dfrac{a_2^2 a_1}{4\pi} \). So \( \dfrac{V_1}{V_2} = \dfrac{a_1}{a_2} \), that is a₂V₁ = a₁V₂.

Teacher's Note:
1. The side that is not the height becomes the circumference.

Question A-7. The height of a conical tent at the centre is 5 m. The distance of any point on its circular base from the top of the tent is 13 m. The area of the slant surface is :
(A) 144π sq m   (B) 130π sq m   (C) 156π sq m   (D) 169π sq m
Answer:
Ans. (C) 156π sq m
Slant height ℓ = 13 m, so r = \( \sqrt{13^2 - 5^2} \) = 12 m.
Slant area = πrℓ = π(12)(13) = 156π m².

Teacher's Note:
1. The distance from the top to the rim is the slant height.

Question A-8. The slant height of a cone is increased by P%. If the radius remains the same, the curved surface area is increased by :
(A) P %   (B) P² %   (C) 2P %   (D) None of these
Answer:
Ans. (A) P %
C.S.A. = πrℓ is proportional to ℓ when r is fixed, so the area increases by the same P%.

Teacher's Note:
1. With r fixed, C.S.A. changes in the same ratio as ℓ.

Question A-9. It is required to construct a conical circus tent of radius 21 m and 35 m slant height. The width of the canvas cloth is 3 meters. What will be the length of the cloth which shall do the needful?
(A) 700 m   (B) 1250 m   (C) 776.5 m   (D) 770 m
Answer:
Ans. (D) 770 m
C.S.A. = \( \dfrac{22}{7} \times 21 \times 35 \) = 2310 m².
Length = \( \dfrac{2310}{3} \) = 770 m.

Teacher's Note:
1. Length of cloth = area ÷ width.

Question A-10. If a hemi-spherical dome has an inner diameter of 28 m, then its volume (in m³) is :
(A) 6186.60   (B) 5749.33   (C) 7099.33   (D) 7459.33
Answer:
Ans. (B) 5749.33
r = 14 m. Volume = \( \dfrac{2}{3} \times \dfrac{22}{7} \times 14^3 \) = 5749.33 m³.

Teacher's Note:
1. Use the radius, not the diameter.

Question A-11. The radius of a sphere is increased by P%. Its surface area increases by :
(A) P %   (B) P² %   (C) (2P + P²/100) %   (D) P²/2 %
Answer:
Ans. (C) (2P + P²/100) %
New area = 4π (r (1 + P/100))², so the ratio is \( (1 + P/100)^2 \).
Increase = \( \left( 2P + \dfrac{P^2}{100} \right) \)%.

Teacher's Note:
1. Area depends on the square of the radius.

Conversion of Solids

Question B-1. The number of cubes of volume 4 cubic units which can be cut from a cube with a surface area of 96 square units is
(A) 4   (B) 8   (C) 12   (D) 16
Answer:
Ans. (D) 16
6a² = 96 ⇒ a = 4, volume = 64.
Number of small cubes = 64 ÷ 4 = 16.

Teacher's Note:
1. Find the edge from the surface area first.

Question B-2. 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. Then the number of marbles that the cube can accommodate is
(A) 142296   (B) 142396   (C) 142496   (D) 142596
Answer:
Ans. (A) 142296
Number = \( \dfrac{\frac{7}{8} \times 22^3}{\frac{4}{3} \times \frac{22}{7} \times (0.25)^3} \) = 142296.

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

Question B-3. A lead ball of radius 24 cm is melted down and recast into smaller balls of radius 6 cm. Assuming that no metal is lost in this process, number of complete smaller balls that can be made, is :
(A) 4   (B) 16   (C) 36   (D) 64
Answer:
Ans. (D) 64
Number = \( \left( \dfrac{24}{6} \right)^3 \) = 64.

Teacher's Note:
1. Ratio of volumes = cube of ratio of radii.

Question B-4. A sphere of radius 3 cms is dropped into a cylindrical vessel of radius 4 cms. If the sphere is submerged completely, then the height (in cm) to which the water rises, is :
(A) 2.35   (B) 2.30   (C) 2.25   (D) 2.15
Answer:
Ans. (C) 2.25
π(4)² h = \( \dfrac{4}{3}\pi (3)^3 \) ⇒ 16h = 36 ⇒ h = 2.25.

Teacher's Note:
1. Rise in level = volume of the sphere ÷ base area.

Question B-5. Twelve solid spheres of the same size are made by melting a solid metallic cylinder of base diameter 2 cm and height 16 cm. The diameter of each sphere is :
(A) 4 cm   (B) 3 cm   (C) 2 cm   (D) 6 cm
Answer:
Ans. (C) 2 cm
Volume of cylinder = π(1)²(16) = 16π. Each sphere = \( \dfrac{16\pi}{12} = \dfrac{4\pi}{3} \).
\( \dfrac{4}{3}\pi r^3 = \dfrac{4\pi}{3} \Rightarrow r = 1 \), diameter = 2 cm.

Teacher's Note:
1. Divide the cylinder volume equally among 12 spheres.

Question B-6. A sphere of radius 6 cm is dropped into a cylindrical vessel partly filled with water. The radius of the vessel is 8 cm. If the sphere is submerged completely, then the surface of the water rises by :
(A) 4.5 cm   (B) 4 cm   (C) 3 cm   (D) 2 cm
Answer:
Ans. (A) 4.5 cm
π(8)² h = \( \dfrac{4}{3}\pi (6)^3 \) ⇒ 64h = 288 ⇒ h = 4.5 cm.

Teacher's Note:
1. Displaced water equals the volume of the sphere.

Question B-7. A metallic spherical shell of internal and external diameters 8 cm and 12 cm, respectively is melted and recast into the form of a cone of base diameter 8 cm. The height of the cone is :
(A) 114 cm   (B) 76 cm   (C) 38 cm   (D) 19 cm
Answer:
Ans. (C) 38 cm
Volume of shell = \( \dfrac{4}{3}\pi (6^3 - 4^3) = \dfrac{608\pi}{3} \).
\( \dfrac{1}{3}\pi (4)^2 h = \dfrac{608\pi}{3} \Rightarrow h = \) 38 cm.

Teacher's Note:
1. Convert diameters to radii (4 cm and 6 cm) first.

Question B-8. A solid piece of iron of dimensions 66 cm × 49 cm × 12 cm is moulded into a sphere. The radius of the sphere is :
(A) 7 cm   (B) 3 cm   (C) 21 cm   (D) 14 cm
Answer:
Ans. (C) 21 cm
Volume = 66 × 49 × 12 = 38808 cm³. \( \dfrac{4}{3} \times \dfrac{22}{7} r^3 = 38808 \Rightarrow r^3 = 9261 \Rightarrow r = \) 21 cm.

Teacher's Note:
1. Note 38808 = 4/3 × 22/7 × 21³.

Question B-9. The volume of the greatest sphere that can be cut off from a cylindrical log of wood of base radius 3 cm and height 7 cm is :
(A) 108π cm³   (B) 36π cm³   (C) 12π cm³   (D) (4/3)π cm³
Answer:
Ans. (B) 36π cm³
The largest sphere has radius 3 cm (diameter 6 cm < height 7 cm).
Volume = \( \dfrac{4}{3}\pi (3)^3 \) = 36π cm³.

Teacher's Note:
1. The sphere's diameter is limited by the smaller of base diameter and height.

Question B-10. The number of solid spheres, each of diameter 6 cm, that could be moulded to form a solid metal cylinder of height 45 cm and diameter 4 cm is :
(A) 3   (B) 4   (C) 5   (D) 6
Answer:
Ans. (C) 5
Volume of cylinder = π(2)²(45) = 180π. Volume of a sphere = \( \dfrac{4}{3}\pi (3)^3 \) = 36π.
Number = \( \dfrac{180}{36} \) = 5.

Teacher's Note:
1. Divide the cylinder volume by the volume of one sphere.

Combination of Solids

Question C-1. There is a cylinder circumscribing a hemisphere such that their bases are common. The ratio of their volumes (hemisphere : cylinder) is :
(A) 1 : 3   (B) 1 : 2   (C) 2 : 3   (D) 3 : 4
Answer:
Ans. (C) 2 : 3
Cylinder of radius r and height r has volume πr³. The hemisphere has volume \( \dfrac{2}{3}\pi r^3 \).
Ratio = \( \dfrac{2}{3} : 1 \) = 2 : 3.

Teacher's Note:
1. A circumscribing cylinder has the same radius, and its height equals the radius.

Question C-2. A medicine-capsule is in the shape of a cylinder of diameter 0.5 cm with two hemispheres stuck to each of its ends. The length of entire capsule is 2 cm. The capacity of the capsule is :
(A) 0.36 cm³   (B) 0.40 cm³   (C) 0.45 cm³   (D) 0.30 cm³
Answer:
Ans. (A) 0.36 cm³
r = 0.25 cm, cylinder length = 1.5 cm.
V = π(0.25)²(1.5) + \( \dfrac{4}{3}\pi (0.25)^3 \) ≈ 0.2946 + 0.0655 = 0.36 cm³.

Teacher's Note:
1. Cylinder plus one full sphere (two hemispheres).

Question C-3. A solid consists of a circular cylinder with an exact fitting right circular cone placed at the top. If the height of the cone is h and the total volume of the solid is 3 times the volume of the cone, then the height of the circular cylinder is :
(A) 2h   (B) 2h/3   (C) 3h/2   (D) 4h
Answer:
Ans. (B) 2h/3
Cone: \( \dfrac{1}{3}\pi r^2 h \). Cylinder: πr²H. Total = 3 × cone ⇒ πr²H = 2 × \( \dfrac{1}{3}\pi r^2 h \).
So H = 2h/3.

Teacher's Note:
1. Cylinder volume = 2 × cone volume.

Question C-4. A solid cylinder of diameter 12 cm and height 15 cm is melted and recast into 12 toys in the shape of a right circular cone mounted on a hemisphere. Find the total height of the toy if the height of the conical part is 3 times its radius.
(A) 9 cm   (B) 12 cm   (C) 15 cm   (D) 18 cm
Answer:
Ans. (B) 12 cm
Volume of each toy = \( \dfrac{\pi (6)^2 (15)}{12} \) = 45π cm³.
Toy: \( \dfrac{1}{3}\pi r^2 (3r) + \dfrac{2}{3}\pi r^3 = \dfrac{5}{3}\pi r^3 = 45\pi \Rightarrow r = 3 \).
Total height = 3r + r = 12 cm.

Teacher's Note:
1. Express the toy volume in terms of r alone.

Question C-5. A tent is in the form of a cylinder of diameter 4.2 m and height 4 m, surmounted by a cone of equal base and height 2.8 m. Find the cost of canvas for making the tent at Rs. 100 per sq. m.
(A) Rs. 7590   (B) Rs. 7950   (C) Rs. 7580   (D) Rs. 7850
Answer:
Ans. (A) Rs. 7590
r = 2.1 m, slant height = \( \sqrt{2.1^2 + 2.8^2} \) = 3.5 m.
Canvas = 2π(2.1)(4) + π(2.1)(3.5) = \( \dfrac{22}{7} \times 2.1 \times 11.5 \) = 75.9 m².
Cost = 75.9 × 100 = Rs. 7590.

Teacher's Note:
1. Only curved surfaces of the cylinder and cone need canvas.