CHAPTER 10 TELANGANA CLASS X MATHEMATICS - CHAPTER 10: MENSURATION (EXERCISE 10.1 TO 10.4 FULL AUDITED SOLUTIONS WITH SVG FIGURES)

BOARD OF SECONDARY EDUCATION, TELANGANA

CLASS 10 MATHEMATICS - CHAPTER 10: MENSURATION

Complete Exercise-Wise Questions & Step-by-Step Solutions with 3D Figures (Ex 10.1 - 10.4) | Muqeem Tutorial

SUBJECT: MATHEMATICS CHAPTER 10 (FULL EXERCISES WITH FIGURES) WEIGHTAGE: 8 - 10 MARKS

EXERCISE 10.1 — Surface Areas of Combination of Solids

Q1. A joker's cap is in the form of a right circular cone whose base radius is 7 cm and height is 24 cm. Find the area of the sheet required to make 10 such caps.

Figure 10.1: Conical Cap with Radius 7 cm and Height 24 cm

Solution:
Radius r = 7 cm, height h = 24 cm.
Slant height l = √(r² + h²) = √(7² + 24²) = √(49 + 576) = √625 = 25 cm.
CSA of 1 cap = π r l = (22/7) × 7 × 25 = 550 cm².
Area for 10 caps = 10 × 550 = 5500 cm².
👉 Total Area of Sheet Required = 5500 cm²

Q2. A sports cap is in the form of a hemisphere surmounted by a cylinder. The total height of the cap is 14 cm and the diameter of the cylinder is 14 cm. Find its outer surface area.

Solution:
Diameter d = 14 cm ⇒ Radius r = 7 cm.
Total height = 14 cm ⇒ Height of cylinder h = 14 - 7 = 7 cm.
• CSA of Cylinder = 2πrh = 2 × (22/7) × 7 × 7 = 308 cm².
• CSA of Hemisphere = 2πr² = 2 × (22/7) × 7² = 308 cm².
Total Outer Surface Area = 308 + 308 = 616 cm².
👉 Total Surface Area = 616 cm²

Q3. A medicine capsule is in the form of a cylinder with two hemispheres stuck to each of its ends. The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area.

Figure 10.2: Medicine Capsule with Two Hemispherical Ends

Solution:
Diameter d = 5 mm ⇒ Radius r = 2.5 mm.
Height of cylindrical part h = 14 - 2(2.5) = 9 mm.
• CSA of Cylinder = 2πrh = 2 × (22/7) × 2.5 × 9 = 990 / 7 mm².
• CSA of 2 Hemispheres = 4πr² = 4 × (22/7) × (2.5)² = 550 / 7 mm².
Total Surface Area = (990 + 550) / 7 = 1540 / 7 = 220 mm².
👉 Surface Area of Capsule = 220 mm²

Q4. A wooden article was made by scooping out a hemisphere from each end of a solid cylinder. If the height of the cylinder is 10 cm and its base is of radius 3.5 cm, find the total surface area of the article.

Solution:
Radius r = 3.5 cm = 7/2 cm, height h = 10 cm.
Total Surface Area = CSA of Cylinder + 2 × (CSA of Hemisphere)
TSA = 2πrh + 2 × (2πr²) = 2πr (h + 2r)
TSA = 2 × (22/7) × (7/2) × [10 + 2(3.5)] = 22 × [10 + 7] = 22 × 17 = 374 cm².
👉 Total Surface Area = 374 cm²

Q5. A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy.

Solution:
Radius r = 3.5 cm, total height = 15.5 cm.
Height of cone h = 15.5 - 3.5 = 12 cm.
Slant height l = √(3.5² + 12²) = √(12.25 + 144) = √156.25 = 12.5 cm.
TSA of Toy = CSA of Cone + CSA of Hemisphere = π r l + 2 π r² = π r (l + 2r)
TSA = (22/7) × 3.5 × (12.5 + 2(3.5)) = 11 × (12.5 + 7) = 11 × 19.5 = 214.5 cm².
👉 Total Surface Area of Toy = 214.5 cm²

Q6. From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest cm².

Solution:
Height h = 2.4 cm, diameter d = 1.4 cm ⇒ Radius r = 0.7 cm.
Slant height of conical cavity l = √(r² + h²) = √(0.7² + 2.4²) = √(0.49 + 5.76) = √6.25 = 2.5 cm.
TSA of Remaining Solid = CSA of Cylinder + CSA of Cone + Area of Top Base
TSA = 2πrh + πrl + πr² = πr (2h + l + r)
TSA = (22/7) × 0.7 × [2(2.4) + 2.5 + 0.7] = 2.2 × [4.8 + 2.5 + 0.7] = 2.2 × 8.0 = 17.6 cm².
👉 Total Surface Area = 17.6 cm² ≈ 18 cm²

Q7. A hemispherical depression is cut out from one face of a cubical wooden block of side 7 cm such that the diameter of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid.

Solution:
Edge of cube l = 7 cm ⇒ Radius of hemisphere r = 3.5 cm = 7/2 cm.
TSA = Surface Area of Cube - Base Area of Hemisphere + CSA of Hemisphere
TSA = 6 l² - π r² + 2 π r² = 6 l² + π r²
TSA = 6(7)² + (22/7)(7/2)² = 294 + 38.5 = 332.5 cm².
👉 Surface Area of Remaining Solid = 332.5 cm²

EXERCISE 10.2 — Volumes of Combined Solids

Q1. A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1 cm and the height of the cone is equal to its radius. Find the volume of the solid in terms of π.

Figure 10.3: Cone Mounted on Hemisphere (r = h = 1 cm)

Solution:
Radius r = 1 cm, height of cone h = 1 cm.
Total Volume = Volume of Cone + Volume of Hemisphere
Volume = (1/3) π r² h + (2/3) π r³ = (1/3) π (1) + (2/3) π (1) = π cm³.
👉 Volume of Solid = π cm³

Q2. Rachel, an engineering student, was asked to make a model shaped like a cylinder with two cones attached at its two ends. The diameter of the model is 3 cm and its length is 12 cm. If each cone has a height of 2 cm, find the volume of air contained in the model.

Solution:
Diameter = 3 cm ⇒ Radius r = 1.5 cm.
Total length = 12 cm, height of each cone h_cone = 2 cm.
Height of cylindrical part h_cyl = 12 - 2(2) = 8 cm.
Total Volume = Volume of Cylinder + 2 × (Volume of Cone)
Volume = π r² h_cyl + 2 × [(1/3) π r² h_cone]
Volume = π (1.5)² (8) + (2/3) π (1.5)² (2) = 18 π + 3 π = 21 π = 21 × (22/7) = 66 cm³.
👉 Volume of Air Contained = 66 cm³

Q3. A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm. The radius of each of the depressions is 0.5 cm and the depth is 1.4 cm. Find the volume of wood in the entire stand.

Solution:
1. Volume of Cuboid = l × b × h = 15 × 10 × 3.5 = 525 cm³.
2. Volume of 1 Conical Depression = (1/3) π r² h = (1/3) × (22/7) × (0.5)² × 1.4 = 0.367 cm³.
3. Volume of 4 Depressions = 4 × 0.367 = 1.47 cm³.
Volume of Wood = 525 - 1.47 = 523.53 cm³.
👉 Volume of Wood = 523.53 cm³

Q4. A gulab jamun, contains sugar syrup up to about 30% of its volume. Find approximately how much syrup would be found in 45 gulab jamuns, each shaped like a cylinder with two hemispherical ends with length 5 cm and diameter 2.8 cm.

Solution:
Diameter d = 2.8 cm ⇒ Radius r = 1.4 cm.
Cylinder length h = 5 - 2(1.4) = 2.2 cm.
• Volume of 1 Gulab Jamun = π r² h + (4/3) π r³ = 25.05 cm³.
• Volume of 45 Gulab Jamuns = 45 × 25.05 = 1127.25 cm³.
Quantity of Syrup = 30% of 1127.25 = 0.30 × 1127.25 = 338.17 cm³.
👉 Syrup Volume = 338 cm³ (approx)

Q5. A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, which is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the pole, given that 1 cm³ of iron has approximately 8g mass. (Use π = 3.14)

Solution:
Lower cylinder: R = 12 cm, H = 220 cm ⇒ Vol₁ = π (12)² (220) = 31680 π cm³.
Upper cylinder: r = 8 cm, h = 60 cm ⇒ Vol₂ = π (8)² (60) = 3840 π cm³.
Total Volume = 31680 π + 3840 π = 35520 π = 35520 × 3.14 = 111,532.8 cm³.
Total Mass = 111,532.8 × 8 g = 892,262.4 g = 892.26 kg.
👉 Mass of Iron Pole = 892.26 kg

EXERCISE 10.3 — Conversion of Solid from One Shape to Another

Q1. A metallic sphere of radius 4.2 cm is melted and recast into the shape of a cylinder of radius 6 cm. Find the height of the cylinder.

Solution:
Volume of Cylinder = Volume of Sphere
π r₁² h = (4/3) π r₂³ ⇒ 6² × h = (4/3) × (4.2)³ ⇒ 36 h = 98.784 ⇒ h = 2.74 cm.
👉 Height of Cylinder = 2.74 cm

Q2. Metallic spheres of radii 6 cm, 8 cm and 10 cm respectively are melted to form a single solid sphere. Find the radius of the resulting sphere.

Solution:
R³ = 6³ + 8³ + 10³ = 216 + 512 + 1000 = 1728 ⇒ R = ∛1728 = 12 cm.
👉 Radius of Resulting Sphere = 12 cm

Q3. A 20 m deep well with diameter 7 m is dug and the earth from digging is evenly spread out to form a platform 22 m by 14 m. Find the height of the platform.

Solution:
Volume of Earth Dug = π r² h = (22/7) × (3.5)² × 20 = 770 m³.
Volume of Platform = 22 × 14 × H = 308 H = 770 ⇒ H = 2.5 m.
👉 Height of Platform = 2.5 m

Q4. A well of diameter 3 m is dug 14 m deep. The earth taken out of it has been spread evenly all around it in the shape of a circular ring of width 4 m to form an embankment. Find the height of the embankment.

Solution:
Well radius r = 1.5 m, depth h = 14 m ⇒ Vol = π (1.5)² (14) = 31.5 π m³.
Embankment inner radius r = 1.5 m, outer radius R = 1.5 + 4 = 5.5 m.
Base Area of Embankment = π (R² - r²) = π (5.5² - 1.5²) = 28 π m².
Height H = Volume / Area = 31.5 π / 28 π = 1.125 m.
👉 Height of Embankment = 1.125 m

Q5. A container shaped like a right circular cylinder having diameter 12 cm and height 15 cm is full of ice cream. The ice cream is to be filled into cones of height 12 cm and diameter 6 cm, having a hemispherical shape on the top. Find the number of such cones which can be filled with ice cream.

Solution:
• Cylinder Volume = π (6)² (15) = 540 π cm³.
• 1 Cone Volume = (1/3) π (3)² (12) + (2/3) π (3)³ = 36 π + 18 π = 54 π cm³.
Number of Cones = 540 π / 54 π = 10.
👉 Number of Cones = 10

Q6. How many silver coins, 1.75 cm in diameter and of thickness 2 mm, must be melted to form a cuboid of dimensions 5.5 cm × 10 cm × 3.5 cm?

Solution:
Coin radius r = 0.875 cm = 7/8 cm, thickness h = 2 mm = 0.2 cm = 1/5 cm.
Volume of 1 Coin = π r² h = (22/7) × (7/8)² × (1/5) = 77 / 160 cm³.
Volume of Cuboid = 5.5 × 10 × 3.5 = 192.5 cm³.
Number of Coins = 192.5 / (77/160) = (192.5 × 160) / 77 = 400.
👉 Number of Silver Coins = 400

Q7. A cylindrical bucket, 32 cm high and with radius of base 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.

Solution:
(1/3) π r² (24) = π (18)² (32) ⇒ 8 r² = 10368 ⇒ r² = 1296 ⇒ r = 36 cm.
Slant height l = √(36² + 24²) = √(1296 + 576) = √1872 = 12√13 cm ≈ 43.27 cm.
👉 Radius = 36 cm, Slant Height = 12√13 cm

EXERCISE 10.4 — Water Flow Applications & Frustum of a Cone

Q1. Water in a canal, 6 m wide and 1.5 m deep, is flowing with a speed of 10 km/h. How much area will it irrigate in 30 minutes, if 8 cm of standing water is needed?

Solution:
In 30 min (0.5 h), length L = 10,000 × 0.5 = 5000 m.
Volume of Water = 5000 × 6 × 1.5 = 45,000 m³.
Standing water height H = 8 cm = 0.08 m.
Area = 45,000 / 0.08 = 562,500 m² = 56.25 hectares.
👉 Area Irrigated = 562,500 m² (56.25 hectares)

Q2. The radii of the ends of a frustum of a cone 45 cm high are 28 cm and 7 cm. Find its volume, curved surface area and total surface area. (Use π = 22/7)

Figure 10.4: Frustum of a Cone (R = 28 cm, r = 7 cm, h = 45 cm)

Solution:
h = 45 cm, R = 28 cm, r = 7 cm.
l = √[45² + (28 - 7)²] = √[2025 + 441] = √2466 ≈ 49.65 cm.
• Volume = (1/3) π h [R² + r² + R·r] = 15 × (22/7) × [784 + 49 + 196] = 48,510 cm³.
• CSA = π l (R + r) = (22/7) × 49.65 × 35 = 5461.5 cm².
• TSA = CSA + πR² + πr² = 5461.5 + 2464 + 154 = 8079.5 cm².
👉 Volume = 48,510 cm³  |  CSA = 5461.5 cm²  |  TSA = 8079.5 cm²

Q3. A drinking glass is in the shape of a frustum of a cone of height 14 cm. The diameters of its two circular ends are 4 cm and 2 cm. Find the capacity of the glass.

Solution:
Height h = 14 cm, R = 2 cm, r = 1 cm.
Capacity (Volume) = (1/3) π h [R² + r² + R·r]
Volume = (1/3) × (22/7) × 14 × [2² + 1² + (2 × 1)] = (1/3) × 44 × [4 + 1 + 2] = (1/3) × 44 × 7 = 308 / 3 = 102.67 cm³.
👉 Capacity of Glass = 102.67 cm³ (102 2/3 cm³)