SURFACE AREAS AND VOLUMES

EXERCISE 12.1

Q1. 2 cubes each of volume 64 cm³ are joined end to end. Find the surface area of the resulting cuboid.

Solution:

1. Volume of each cube = 64 cm³ ⇒ side = cube root of 64 = 4 cm.

2. Joining end to end makes a cuboid of dimensions:
Length = 4 + 4 = 8 cm, Breadth = 4 cm, Height = 4 cm.

3. Surface area of a cuboid = 2 × (lb + lh + bh)
= 2 × (8×4 + 8×4 + 4×4)
= 2 × (32 + 32 + 16)
= 2 × 80
= 160 cm².

Answer: 160 cm².

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Q2: A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel.

Solution:

1. Diameter of hemisphere = 14 cm ⇒ radius r = 14/2 = 7 cm.

2. Height of hemisphere = radius = 7 cm.
Total height = 13 cm ⇒ height of cylinder h = 13 − 7 = 6 cm.

3. Inner curved surface area of hemisphere = 2πr²
= 2 × π × 7²
= 2 × π × 49
= 98π cm².

4. Inner curved surface area of cylinder = 2πrh
= 2 × π × 7 × 6
= 84π cm².

5. Total inner surface area = 98π + 84π = 182π cm².

6. Taking π = 22/7 (to get a simple exact numerical value):
Total = 182 × 22/7 = 26 × 22 = 572 cm².

Answer: 572 cm².

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Q3: 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:

1. Radius of hemisphere and cone = 3.5 cm.

2. Total height of the toy = 15.5 cm.
Height of cone = 15.5 − 3.5 = 12 cm.

3. Slant height of cone = √(r² + h²)
= √(3.5² + 12²)
= √(12.25 + 144)
= √156.25
= 12.5 cm.

4. Curved surface area of cone = πrl
= (22/7) × 3.5 × 12.5
= 22 × 0.5 × 12.5
= 137.5 cm².

5. Curved surface area of hemisphere = 2πr²
= 2 × (22/7) × (3.5)²
= 2 × (22/7) × 12.25
= 77 cm².

6. Total surface area = 137.5 + 77
= 214.5 cm².

Answer: 214.5 cm²

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Q4: A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have? Find the surface area of the solid.

Solution:

1. Side of cube = 7 cm.
Greatest diameter of hemisphere = side of cube = 7 cm.
So radius r = 7/2 = 3.5 cm.

2. Surface area of cube = 6 × side² = 6 × 7² = 6 × 49 = 294 cm².
Top face area (covered by hemisphere) = 7 × 7 = 49 cm².
So exposed surface area of cube = 294 − 49 = 245 cm².

3. Curved surface area of hemisphere = 2πr²
= 2 × π × (3.5)²
= 2 × π × 12.25
= 24.5π cm².
Taking π = 22/7 to get a simple value:
24.5π = (49/2) × 22/7 = 77 cm².

4. Total surface area of the solid = exposed cube area + curved area of hemisphere
= 245 + 77 = 322 cm².

Answer: Greatest diameter = 7 cm.
Total surface area = 322 cm².

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Q5: A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter l of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid.

Solution:

Step 1: Dimensions

• Edge of cube = l.

• Diameter of hemisphere = l ⇒ radius r = l/2.

Step 2: Surface area of cube before cutting
Total surface area of cube = 6l².

Step 3: Effect of cutting the hemisphere
On one face of the cube, we cut out a hemisphere:

• That face had area = l².

• The circular part (πr²) is removed.

• But instead of leaving it blank, the inner curved surface of the hemisphere is now exposed.

So the new surface area is:
= (5 other faces) + (remaining part of cut face) + (curved surface of hemisphere).

Step 4: Remaining flat area of cut face
Remaining area = l² − πr².
= l² − π(l/2)²
= l² − (πl²)/4.

Step 5: Curved surface area of hemisphere
CSA = 2πr²
= 2π (l/2)²
= 2π (l²/4)
= (πl²)/2.

Step 6: Total surface area
TSA = 5l² + (l² − (πl²)/4) + (πl²)/2
= 6l² + (−πl²/4 + πl²/2)
= 6l² + (πl²/4).

Factorize:
= (1/4) l² (24 + π).

Final Answer: Surface area = (1/4) l² (π + 24).

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Q6: A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends (see Fig. 12.10 (refer book)). The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area.

Solution:

1. Diameter = 5 mm ⇒ radius r = 5/2 = 2.5 mm.

2. Total length = 14 mm. The two hemispheres together contribute 2r = 5 mm to the length.
So height (length) of the cylindrical part h = 14 − 5 = 9 mm.

3. Curved surface area of the cylinder = 2πrh
= 2 × π × 2.5 × 9
= 45π mm².

4. Curved surface area of the two hemispheres = curved surface area of a full sphere = 4πr²
= 4 × π × (2.5)²
= 4 × π × 6.25
= 25π mm².

5. Total surface area = 45π + 25π = 70π mm².

6. Taking π = 22/7 for a numerical value:
Total = 70 × 22/7 = 10 × 22 = 220 mm².

Answer: Total surface area = 70π mm² = 220 mm² (using π = 22/7).

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Q7: A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 2.1 m and 4 m respectively, and the slant height of the top is 2.8 m, find the area of the canvas used for making the tent. Also, find the cost of the canvas of the tent at the rate of ₹ 500 per m². (Note that the base of the tent will not be covered with canvas.)

Solution:

1. Curved surface area of the cylinder = 2πrh
= 2 × π × 2 × 2.1
= 8.4π m².

2. Curved surface area of the conical top = πrl
= π × 2 × 2.8
= 5.6π m².

3. Total area of canvas required = 8.4π + 5.6π = 14π m².

4. Taking π = 22/7 to get a simple numerical value:
Total area = 14 × 22/7 = 2 × 22 = 44 m².

5. Cost of canvas at the rate of ₹ 500 per m²:
Cost = 44 × 500 = ₹ 22,000.

Answer:
Total area of canvas = 14π m² = 44 m².
Cost of the canvas = ₹ 22,000.

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Q8: 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:

1. Diameter = 1.4 cm ⇒ radius r = 1.4/2 = 0.7 cm.
Height h = 2.4 cm.

2. Slant height of the conical cavity l = √(r² + h²)
= √(0.7² + 2.4²)
= √(0.49 + 5.76)
= √6.25 = 2.5 cm.

3. Curved surface area of the cylinder = 2πrh
= 2 × π × 0.7 × 2.4
= 3.36π cm².

4. Area of the bottom base of the cylinder (the base that remains) = πr²
= π × (0.7)²
= 0.49π cm².

5. Curved surface area of the conical cavity (inner exposed surface) = πrl
= π × 0.7 × 2.5
= 1.75π cm².

6. Total surface area of the remaining solid = 3.36π + 0.49π + 1.75π = 5.6π cm².

7. Numerical value: 5.6 × π ≈ 5.6 × 3.14159 = 17.5929 ≈ 18 cm² (to the nearest cm²).

Answer: 18 cm².

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Q9: A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in Fig. 12.11( refer book). 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:

1. Radius r = 3.5 cm. Height of cylinder h = 10 cm.

2. Curved surface area of the cylinder = 2πrh
= 2 × π × 3.5 × 10
= 70π cm².

3. Curved surface area of one hemisphere = 2πr²
So curved surface area of two hemispheres = 2 × 2πr² = 4πr²
= 4 × π × (3.5)²
= 4 × π × 12.25
= 49π cm².

4. Total surface area of the article = curved area of cylinder + curved areas of two hemispheres
= 70π + 49π
= 119π cm².

5. Taking π = 22/7 for a simple numerical value:
Total = 119 × 22/7 = 17 × 22 = 374 cm².

Answer: Total surface area = 119π cm² = 374 cm² (using π = 22/7).

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EXERCISE 12.2

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.

Solution:

1. Radius of hemisphere = radius of cone = r = 1 cm.
Height of cone h = r = 1 cm.

2. Volume of hemisphere = (2/3)πr³
= (2/3)π(1)³
= (2/3)π cm³.

3. Volume of cone = (1/3)πr²h
= (1/3)π(1)²(1)
= (1/3)π cm³.

4. Total volume of the solid = volume of hemisphere + volume of cone
= (2/3)π + (1/3)π
= π cm³.

Answer: Volume of the solid = π cm³.

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Q2: Rachel, an engineering student, was asked to make a model shaped like a cylinder with two cones attached at its two ends by using a thin aluminium sheet. 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 that Rachel made. (Assume the outer and inner dimensions of the model to be nearly the same.)

Solution:

1. Diameter = 3 cm ⇒ radius r = 3/2 = 1.5 cm.

2. Each cone has height 2 cm, so total length contributed by two cones = 2 × 2 = 4 cm.
Total length of model = 12 cm ⇒ height (length) of cylindrical part = 12 − 4 = 8 cm.

3. Volume of cylinder = π r² h
= π × (1.5)² × 8
= π × 2.25 × 8
= 18π cm³.

4. Volume of one cone = (1/3) π r² h
= (1/3) × π × (1.5)² × 2
= (1/3) × π × 2.25 × 2
= 1.5π cm³.
Volume of two cones = 2 × 1.5π = 3π cm³.

5. Total volume of air inside the model = volume of cylinder + volume of two cones
= 18π + 3π
= 21π cm³.

6. Numerical value (using π ≈ 3.1416):
21π ≈ 21 × 3.1416 = 65.97 ≈ 66.0 cm³ (to one decimal place).

Answer: Volume of air = 21π cm³ ≈ 66.0 cm³.

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Q3: 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. (see Fig. 12.15(refer b0ok)).

Solution:

1. Diameter = 2.8 cm ⇒ radius r = 2.8/2 = 1.4 cm.
Total length = 5 cm. The two hemispheres together have length = 2r = 2.8 cm.
So length of cylindrical part h = 5 − 2.8 = 2.2 cm.

2. Volume of the cylindrical part = π r² h
= π × (1.4)² × 2.2
= π × 1.96 × 2.2
= 4.312π cm³.

3. Volume of the two hemispheres = volume of a sphere = (4/3) π r³
= (4/3) π × (1.4)³
= (4/3) π × 2.744
= 3.658666...π cm³.

4. Volume of one gulab jamun = 4.312π + 3.658666...π = 7.970666...π cm³
= approximately 25.04 cm³ (using π ≈ 3.1416).

5. Total volume of 45 gulab jamuns = 45 × 25.0406 ≈ 1126.83 cm³.

6. Syrup is 30% of this volume: Syrup volume = 0.30 × 1126.83 ≈ 338.05 cm³.

7. Rounding to the nearest cubic centimetre: approximately 338 cm³ of syrup.

Answer: approximately 338 cm³ of syrup in 45 gulab jamuns.

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Q4: 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. (Refer Fig. 12.16. (refer book))

Solution:

1. Volume of the cuboid = length × breadth × height
= 15 × 10 × 3.5 = 525 cm³.

2. Each depression is a cone of radius r = 0.5 cm and height h = 1.4 cm.
Volume of one cone = (1/3) × π × r² × h
= (1/3) × π × (0.5)² × 1.4
= (7/60) π cm³. (exact form)

3. Volume of four cones = 4 × (7/60) π = (7/15) π cm³.
Numerical value = (7/15) × π ≈ 1.4661 cm³.

4. Volume of wood in the stand = Volume of cuboid − Volume of four conical depressions
= 525 − (7/15) π cm³.
Numerical value ≈ 525 − 1.4661 = 523.5339 cm³.

5. Rounding: approximately 523.53 cm³ (or 524 cm³ to the nearest cm³).

Answer: Exact form = 525 − (7/15) π cm³.
Approximate value = 523.53 cm³ (≈ 524 cm³).

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Q5: A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of its top, which is
open, is 5 cm. It is filled with water up to the brim. When lead shots, each of which is a sphere of radius
0.5 cm are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots
dropped in the vessel.

Solution:

1. Volume of water initially (volume of the cone)
= (1/3) × π × R² × h
= (1/3) × π × 5² × 8
= (1/3) × π × 25 × 8
= (200/3) π cm³.

2. Water that flows out is one-fourth of the initial water:
Overflow volume = (1/4) × (200/3) π = (50/3) π cm³.

3. Volume of one lead shot (sphere of radius 0.5 cm)
= (4/3) × π × (0.5)³
= (4/3) × π × 0.125
= (1/6) π cm³.

4. Let n be the number of shots. Displaced volume equals overflow:
n × (1/6) π = (50/3) π
⇒ n/6 = 50/3
⇒ n = (50/3) × 6 = 100.

Answer: 100 lead shots.

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Q6: 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:

1. Cylinder (lower part)
Base diameter = 24 cm ⇒ radius = 12 cm
Height = 220 cm
Volume = π × r² × h
= 3.14 × (12)² × 220
= 3.14 × 144 × 220
= 3.14 × 31,680
= 99,475.2 cm³

2. Cylinder (upper part)
Radius = 8 cm
Height = 60 cm
Volume = π × r² × h
= 3.14 × (8)² × 60
= 3.14 × 64 × 60
= 3.14 × 3,840
= 12,057.6 cm³

3. Total volume of the pole
= 99,475.2 + 12,057.6
= 111,532.8 cm³

4. Mass
1 cm³ of iron ≈ 8 g
Total mass = 111,532.8 × 8 g
= 892,262.4 g
= 892.2624 kg

Answer: The mass of the pole is 892,262.4 g ≈ 892.26 kg (approximately).

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Q7: A solid consisting of a right circular cone of height 120 cm and radius 60 cm standing on a hemisphere of radius 60 cm is placed upright in a right circular cylinder full of water such that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is 60 cm and its height is 180 cm.

Solution:

1) Volume of the cylinder (container with water)
Radius = 60 cm, Height = 180 cm
Volume = π × r² × h
= 3.14 × (60)² × 180
= 3.14 × 3600 × 180
= 3.14 × 648000
= 2,034,720 cm³

2) Volume of the cone (upper part of the solid)
Radius = 60 cm, Height = 120 cm
Volume = (1/3) × π × r² × h
= (1/3) × 3.14 × (60)² × 120
= (1/3) × 3.14 × 3600 × 120
= (1/3) × 3.14 × 432000
= 3.14 × 144000
= 452,160 cm³

3) Volume of the hemisphere (lower part of the solid)
Radius = 60 cm
Volume = (2/3) × π × r³
= (2/3) × 3.14 × (60)³
= (2/3) × 3.14 × 216000
= 3.14 × 144000
= 452,160 cm³

4) Total volume of the solid placed inside the cylinder
= Volume of cone + Volume of hemisphere
= 452,160 + 452,160
= 904,320 cm³

5) Volume of water left in the cylinder
= Volume of cylinder − Volume of solid
= 2,034,720 − 904,320
= 1,130,400 cm³

6) Convert into cubic metres
1 m³ = 1,000,000 cm³
So, 1,130,400 cm³ = 1,130,400 ÷ 1,000,000
= 1.1304 m³ ≈ 1.131 m³

Final Answer:
The volume of water left in the cylinder is 1.131 m³ (approx.)

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Q8: A spherical glass vessel has a cylindrical neck 8 cm long, 2 cm in diameter; the diameter of the spherical part is 8.5 cm. By measuring the amount of water it holds, a child finds its volume to be 345 cm3. Check whether she is correct, taking the above as the inside measurements, and π = 3.14.

Solution:

1. Volume of the spherical part
Diameter = 8.5 cm ⇒ radius = 4.25 cm
Volume of sphere = (4/3) × π × r³
= (4/3) × 3.14 × (4.25)³
Calculate (4.25)³ = 4.25 × 4.25 × 4.25 = 76.765625
So volume = (4/3) × 3.14 × 76.765625
= 1.3333333333 × 3.14 × 76.765625
= 4.1866666667 × 76.765625
= 321.3920833 cm³ (approximately)

2. Volume of the cylindrical neck
Diameter = 2 cm ⇒ radius = 1 cm, Height = 8 cm
Volume of cylinder = π × r² × h
= 3.14 × (1)² × 8
= 3.14 × 8
= 25.12 cm³

3. Total (theoretical) internal volume of the vessel
= Volume of sphere + Volume of cylinder
= 321.3920833 + 25.12
= 346.5120833 cm³ (approximately)

4. Compare with the child's measurement
Child’s measured volume = 345 cm³
Difference = 346.5120833 − 345
= 1.5120833 cm³ (≈ 1.51 cm³)
Percentage error = (1.5120833 / 346.5120833) × 100 ≈ 0.44%

Conclusion:
The theoretical volume (using π = 3.14) is approximately 346.51 cm³. The child’s measured value 345 cm³ differs by only about 1.51 cm³ (≈ 0.44%), so her measurement is essentially correct (very close to the expected value).

Answer: The correct volume ≈ 346.51 cm³, so the child’s 345 cm³ is acceptably accurate.

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