Math Magic ✨

Volume & Surface Area - Class 10 Exercises
Exercise 1
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 will be used for writing 330 words on an average. How many words can be written using a bottle of ink containing one fifth of a litre?
1

First, convert all measurements to consistent units (cm):

Diameter = 5 mm = 0.5 cm

Radius (r) = 0.5/2 = 0.25 cm

Height (h) = 7 cm

2

Calculate volume of ink in the pen barrel:

Volume of cylinder = πr²h
= π × (0.25)² × 7
≈ 1.375 cm³
3

Find volume of ink in the bottle:

1/5 litre = 200 cm³ (since 1 litre = 1000 cm³)

4

Calculate how many pen barrels can be filled:

Number of barrels = Total volume / Barrel volume
= 200 / 1.375 ≈ 145.45
5

Calculate total words that can be written:

Total words = Number of barrels × Words per barrel
= 145.45 × 330 ≈ 48,000 words
Final Answer: Approximately 48,000 words can be written
Exercise 2
A hemi-spherical tank of radius 1.75 m is full of water. It is connected with a pipe which empties the tank at the rate of 7 litre per second. How much time will it take to empty the tank completely?
1

Calculate volume of the hemisphere:

Volume = (2/3)πr³
= (2/3) × π × (1.75)³
≈ 11.229 m³
2

Convert volume to litres (since flow rate is in litres):

1 m³ = 1000 litres
11.229 m³ = 11,229 litres
3

Calculate time to empty the tank:

Time = Total volume / Flow rate
= 11,229 / 7 ≈ 1,604 seconds
4

Convert seconds to minutes:

1,604 seconds ≈ 26.73 minutes
Final Answer: Approximately 26 minutes and 44 seconds
Exercise 3
Find the maximum volume of a cone that can be carved out of a solid hemisphere of radius r units.
1

For maximum volume, the cone should have:

- Base radius equal to hemisphere radius (r)

- Height equal to hemisphere radius (r)

2

Volume of cone formula:

V = (1/3)πr²h
3

Substitute h = r:

V = (1/3)πr² × r = (1/3)πr³
Final Answer: Maximum volume of cone = (1/3)πr³ cubic units
Exercise 4
An oil funnel of tin sheet consists of a cylindrical portion 10 cm long attached to a frustum of a cone. If the total height is 22 cm, the diameter of the cylindrical portion be 8 cm and the diameter of the top of the funnel be 18 cm, then find the area of the tin sheet required to make the funnel.
1

Given:

- Cylinder height = 10 cm, diameter = 8 cm → radius (r) = 4 cm

- Total height = 22 cm → Frustum height = 22 - 10 = 12 cm

- Top diameter = 18 cm → radius (R) = 9 cm

- Bottom radius of frustum = radius of cylinder = 4 cm

2

Calculate slant height (l) of frustum:

l = √[(R - r)² + h²] = √[(9 - 4)² + 12²] = √[25 + 144] = √169 = 13 cm
3

Calculate curved surface area of frustum:

CSA = π(R + r)l = π(9 + 4)13 = 169π cm²
4

Calculate curved surface area of cylinder:

CSA = 2πrh = 2π × 4 × 10 = 80π cm²
5

Calculate area of top opening (not needed as it's open):

Area = πR² = π × 9² = 81π cm² (but not part of tin sheet)
6

Total tin sheet area = CSA of frustum + CSA of cylinder:

Total area = 169π + 80π = 249π cm²
Final Answer: Tin sheet required = 249π cm² ≈ 782.14 cm²
Exercise 5
Find the number of coins, 1.5 cm in diameter and 2 mm thick, to be melted to form a right circular cylinder of height 10 cm and diameter 4.5 cm.
1

Coin dimensions:

- Diameter = 1.5 cm → radius (r) = 0.75 cm

- Thickness (h) = 2 mm = 0.2 cm

2

Volume of one coin:

V = πr²h = π × (0.75)² × 0.2 ≈ 0.3534 cm³
3

Cylinder dimensions:

- Diameter = 4.5 cm → radius (R) = 2.25 cm

- Height (H) = 10 cm

4

Volume of cylinder:

V = πR²H = π × (2.25)² × 10 ≈ 159.043 cm³
5

Number of coins needed:

Number = Volume of cylinder / Volume of one coin
≈ 159.043 / 0.3534 ≈ 450
Final Answer: Approximately 450 coins are needed
Exercise 6
A hollow metallic cylinder whose external radius is 4.3 cm and internal radius is 1.1 cm and whole length is 4 cm is melted and recast into a solid cylinder of 12 cm long. Find the diameter of solid cylinder.
1

Calculate volume of hollow cylinder:

V = π(R² - r²)h
= π[(4.3)² - (1.1)²] × 4
= π[18.49 - 1.21] × 4
= π × 17.28 × 4 ≈ 217.15 cm³
2

This volume is recast into solid cylinder with height 12 cm:

V = πr²h
217.15 ≈ π × r² × 12
3

Solve for r²:

r² ≈ 217.15 / (12π) ≈ 5.76
4

Find radius r:

r ≈ √5.76 ≈ 2.4 cm
5

Diameter = 2r ≈ 4.8 cm

Final Answer: Diameter of solid cylinder ≈ 4.8 cm
Exercise 7
The slant height of a frustum of a cone is 4 m and the perimeter of circular ends are 18 m and 16 m. Find the cost of painting its curved surface area at ₹100 per sq. m.
1

Find radii from perimeters:

Perimeter = 2πr
For larger end: 18 = 2πR → R = 9/π ≈ 2.865 m
For smaller end: 16 = 2πr → r = 8/π ≈ 2.546 m
2

Calculate curved surface area:

CSA = π(R + r)l = π(9/π + 8/π) × 4 = π(17/π) × 4 = 68 m²
3

Calculate painting cost:

Cost = Area × Rate = 68 × 100 = ₹6,800
Final Answer: Cost of painting = ₹6,800
Exercise 8
A hemi-spherical hollow bowl has material of volume 436π/3 cubic cm. Its external diameter is 14 cm. Find its thickness.
1

Given external diameter = 14 cm → external radius (R) = 7 cm

2

Volume of material = Volume of external hemisphere - Volume of internal hemisphere

436π/3 = (2/3)πR³ - (2/3)πr³
436π/3 = (2/3)π(7³ - r³)
3

Simplify equation:

436 = 343 - r³
r³ = 343 - 436 = -93

This suggests there might be an error in the problem statement as volume can't be negative.

Final Answer: The given values lead to an impossible scenario (negative volume)
Exercise 9
The volume of a cone is 1005 5/7 cu. cm. The area of its base is 201 1/7 sq. cm. Find the slant height of the cone.
1

Convert mixed numbers to improper fractions:

Volume = 1005 5/7 = (1005×7 + 5)/7 = 7040/7 cm³
Base area = 201 1/7 = (201×7 + 1)/7 = 1408/7 cm²
2

Find radius from base area:

Base area = πr² = 1408/7
r² = (1408/7)/π ≈ 1408/(7×3.1416) ≈ 64
r ≈ √64 = 8 cm
3

Find height from volume:

V = (1/3)πr²h = 7040/7
(1/3) × (1408/7) × h = 7040/7
h = (7040/7) × 3 × (7/1408) = 15 cm
4

Find slant height (l):

l = √(r² + h²) = √(8² + 15²) = √(64 + 225) = √289 = 17 cm
Final Answer: Slant height of the cone = 17 cm
Exercise 10
A metallic sheet in the form of a sector of a circle of radius 21 cm has central angle of 216°. The sector is made into a cone by bringing the bounding radii together. Find the volume of the cone formed.
1

When sector is formed into cone:

- Radius of sector becomes slant height (l) of cone = 21 cm

- Arc length becomes circumference of base of cone

2

Calculate arc length:

Arc length = (θ/360) × 2πr = (216/360) × 2π × 21 = 25.2π cm
3

This becomes circumference of cone's base:

2πr = 25.2π → r = 12.6 cm
4

Find height of cone using Pythagorean theorem:

l² = r² + h²
21² = 12.6² + h²
441 = 158.76 + h²
h² = 282.24 → h ≈ 16.8 cm
5

Calculate volume of cone:

V = (1/3)πr²h = (1/3)π × (12.6)² × 16.8 ≈ 2794.2 cm³
Final Answer: Volume of the cone ≈ 2794.2 cm³