Extra Practice Questions
Surface Areas and Volumes (R.D. Sharma Level)
1. A metallic sphere of radius 10.5 cm is melted and then recast into smaller cones, each of radius 3.5 cm and height 3 cm. Find the number of cones so formed.
Step 1: Find the volume of the sphere.
Volumesphere = (4/3)πR³ = (4/3)π(10.5)³ = 1543.5π cm³.
Step 2: Find the volume of one small cone.
Volumecone = (1/3)πr²h = (1/3)π(3.5)²(3) = 12.25π cm³.
Step 3: Calculate the number of cones.
Number of cones = Volumesphere / Volumecone
= (1543.5π) / (12.25π) = 126.
Answer: 126 cones can be formed.
2. Water in a canal, 6 m wide and 1.5 m deep, is flowing with a speed of 10 km/hr. How much area will it irrigate in 30 minutes, if 8 cm of standing water is needed?
Step 1: Calculate the volume of water flowing in 30 minutes.
Speed = 10 km/hr = 10000 m/hr.
Length of water column in 30 mins (0.5 hr) = 10000 × 0.5 = 5000 m.
Volume of water = Length × Width × Depth = 5000 × 6 × 1.5 = 45000 m³.
Step 2: Calculate the area it can irrigate.
Volume of water = Area of field × Height of standing water.
Height = 8 cm = 0.08 m.
Area = Volume / Height = 45000 / 0.08 = 562500 m².
Answer: It will irrigate 562,500 m² (or 56.25 hectares).
3. A rocket is in the form of a closed cylinder with a cone of the same radius attached to one end. The cylinder is of radius 2.5 m and height 21 m and the cone has a slant height of 8 m. Calculate the total surface area of the rocket.
Step 1: Identify dimensions.
Radius (r) = 2.5 m, Cylinder height (h) = 21 m, Cone slant height (l) = 8 m.
Step 2: Determine the required surface area.
TSA = (CSA of Cone) + (CSA of Cylinder) + (Area of circular base of cylinder)
= πrl + 2πrh + πr² = πr(l + 2h + r)
Step 3: Calculate the area.
= (22/7) × 2.5 × (8 + 2(21) + 2.5) = (22/7) × 2.5 × (8 + 42 + 2.5)
= (22/7) × 2.5 × 52.5 = 22 × 2.5 × 7.5 = 412.5 m².
Answer: The total surface area of the rocket is 412.5 m².
4. 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.
Step 1: Calculate the volume of earth dug out (volume of the well).
Well radius (r) = 1.5 m, Depth (h) = 14 m.
Volumeearth = πr²h = (22/7) × (1.5)² × 14 = 22 × 2.25 × 2 = 99 m³.
Step 2: Calculate the area of the embankment.
Inner radius (r) = 1.5 m. Outer radius (R) = 1.5 + 4 = 5.5 m.
Area = π(R² - r²) = π( (5.5)² – (1.5)² ) = π(30.25 – 2.25) = 28π m².
Step 3: Find the height of the embankment.
Height = Volumeearth / Areaembankment
= 99 / (28 × 22/7) = 99 / 88 = 9/8 = 1.125 m.
Answer: The height of the embankment is 1.125 m.
… (16 more questions follow)
5. A bucket is in the form of a frustum of a cone of height 30 cm with radii of its lower and upper ends as 10 cm and 20 cm, respectively. Find the capacity of the bucket. (Use π = 3.14)
Step 1: Identify dimensions of the frustum.
Height (h) = 30 cm, Lower radius (r₁) = 10 cm, Upper radius (r₂) = 20 cm.
Step 2: Use the formula for the volume of a frustum.
Volume = (1/3)πh(r₁² + r₂² + r₁r₂)
Step 3: Calculate the capacity (volume).
= (1/3) × 3.14 × 30 × (10² + 20² + 10×20)
= 10 × 3.14 × (100 + 400 + 200) = 31.4 × 700 = 21980 cm³.
Answer: The capacity of the bucket is 21980 cm³ (or 21.98 litres).
6. The volume of a right circular cone is 9856 cm³. If the diameter of the base is 28 cm, find the curved surface area of the cone.
Step 1: Find the height of the cone.
Diameter = 28 cm, Radius (r) = 14 cm. Volume = 9856 cm³.
(1/3)πr²h = 9856 => (1/3) × (22/7) × 14² × h = 9856
(1/3) × 22 × 2 × 14 × h = 9856 => h = (9856 × 3) / 616 = 48 cm.
Step 2: Find the slant height (l).
l = √(r² + h²) = √(14² + 48²) = √(196 + 2304) = √2500 = 50 cm.
Step 3: Calculate the Curved Surface Area (CSA).
CSA = πrl = (22/7) × 14 × 50 = 22 × 2 × 50 = 2200 cm².
Answer: The CSA of the cone is 2200 cm².
7. A solid wooden toy is in the shape of a hemisphere surmounted by a cone. If the radius of the hemisphere is 4.2 cm and the total height of the toy is 10.2 cm, find the volume of the wooden toy.
Step 1: Identify dimensions.
Radius (r) = 4.2 cm. Total height = 10.2 cm.
Height of cone (h) = Total height – radius = 10.2 – 4.2 = 6 cm.
Step 2: Calculate the total volume.
Volume = (Vol. of Cone) + (Vol. of Hemisphere)
= (1/3)πr²h + (2/3)πr³ = (1/3)πr²(h + 2r)
= (1/3) × (22/7) × (4.2)² × (6 + 2×4.2)
= (1/3) × (22/7) × 17.64 × (6 + 8.4) = (1/3) × 22 × 2.52 × 14.4 ≈ 266.11 cm³.
Answer: The volume of the toy is approximately 266.11 cm³.
8. A solid sphere of radius 6 cm is melted into a hollow cylinder of uniform thickness. If the external radius of the base of the cylinder is 5 cm and its height is 32 cm, find the uniform thickness of the cylinder.
Step 1: Calculate the volume of the sphere.
Volumesphere = (4/3)πR³ = (4/3)π(6)³ = 288π cm³.
Step 2: Set up the volume for the hollow cylinder.
Let internal radius be r. External radius Rext = 5 cm. Height h = 32 cm.
Volumecylinder = π(Rext² - r²)h = π(5² – r²) × 32.
Step 3: Equate volumes and solve for r.
288π = 32π(25 – r²) => 288/32 = 25 – r² => 9 = 25 – r² => r² = 16 => r = 4 cm.
Step 4: Find the thickness.
Thickness = Rext – r = 5 – 4 = 1 cm.
Answer: The thickness of the cylinder is 1 cm.
9. From a solid cube of side 7 cm, a conical cavity of height 7 cm and radius 3 cm is hollowed out. Find the volume of the remaining solid.
Step 1: Calculate the volume of the cube.
Volumecube = a³ = 7³ = 343 cm³.
Step 2: Calculate the volume of the conical cavity.
Volumecone = (1/3)πr²h = (1/3) × (22/7) × 3² × 7 = 66 cm³.
Step 3: Calculate the volume of the remaining solid.
Volumeremaining = Volumecube – Volumecone = 343 – 66 = 277 cm³.
Answer: The volume of the remaining solid is 277 cm³.
10. A hemispherical bowl of internal radius 9 cm is full of water. Its contents are emptied into a cylindrical vessel of internal radius 6 cm. Find the height of water in the cylindrical vessel.
Step 1: Find the volume of water in the bowl.
Volumewater = Volumehemisphere = (2/3)πr³ = (2/3)π(9)³ = 486π cm³.
Step 2: Find the height in the cylindrical vessel.
Volume in cylinder = πR²h. Here R = 6 cm.
486π = π(6)²h => 486 = 36h => h = 486/36 = 13.5 cm.
Answer: The height of the water in the cylindrical vessel is 13.5 cm.
11. A solid is composed of a cylinder with hemispherical ends. If the whole length of the solid is 104 cm and the radius of each hemispherical end is 7 cm, find the cost of polishing its surface at the rate of ₹10 per dm².
Step 1: Find dimensions.
Radius (r) = 7 cm. Total length = 104 cm.
Cylinder height (h) = 104 – 2(7) = 90 cm.
Step 2: Calculate the total surface area in cm².
TSA = (CSA of cylinder) + 2 × (CSA of hemisphere) = 2πrh + 2(2πr²)
= 2πr(h+2r) = 2 × (22/7) × 7 × (90 + 14) = 44 × 104 = 4576 cm².
Step 3: Convert area to dm² and find the cost.
1 dm = 10 cm, so 1 dm² = 100 cm².
Area in dm² = 4576 / 100 = 45.76 dm².
Cost = Area × Rate = 45.76 × 10 = ₹457.60.
Answer: The cost of polishing is ₹457.60.
12. How many spherical bullets can be made out of a solid cube of lead whose edge measures 44 cm, each bullet being 4 cm in diameter?
Step 1: Find the volume of the cube.
Volumecube = a³ = 44³ = 85184 cm³.
Step 2: Find the volume of one bullet.
Diameter = 4 cm, Radius (r) = 2 cm.
Volumebullet = (4/3)πr³ = (4/3) × (22/7) × 2³ ≈ 33.52 cm³.
Step 3: Calculate the number of bullets.
Number = Volumecube / Volumebullet = 85184 / ((4/3)*(22/7)*8) = (85184 * 21) / 704 = 2541.
Answer: 2541 spherical bullets can be made.
13. The largest sphere is carved out of a cube of side 7 cm. Find the volume of the sphere.
The diameter of the largest possible sphere is equal to the side of the cube.
Diameter = 7 cm, so Radius (r) = 3.5 cm.
Volume = (4/3)πr³ = (4/3) × (22/7) × (3.5)³
= (4/3) × (22/7) × 42.875 ≈ 179.67 cm³.
Answer: The volume of the sphere is approximately 179.67 cm³.
14. Water flows at the rate of 10 m/minute through a cylindrical pipe 5 mm in diameter. How long would it take to fill a conical vessel whose diameter at the base is 40 cm and depth is 24 cm?
Step 1: Calculate the volume of the conical vessel. (all in cm)
Radius R = 20 cm, Height H = 24 cm.
Volumevessel = (1/3)πR²H = (1/3)π(20)²(24) = 3200π cm³.
Step 2: Calculate the volume of water flowing per minute.
Pipe diameter = 5 mm = 0.5 cm, Radius r = 0.25 cm.
Flow rate = 10 m/min = 1000 cm/min.
Volumepipe/min = πr² × rate = π(0.25)²(1000) = 62.5π cm³/min.
Step 3: Calculate the time required.
Time = Volumevessel / Volumepipe/min = (3200π) / (62.5π) = 51.2 minutes.
Answer: It would take 51.2 minutes (or 51 minutes and 12 seconds).
15. A building is in the form a cylinder surmounted by a hemispherical dome. The base diameter of the dome is equal to 2/3 of the total height of the building. Find the height of the building, if it contains 67 1/21 m³ of air. (Take π = 22/7).
Let total height = H. Diameter = Radius(r) × 2 = (2/3)H, so r = H/3.
Height of cylinder (h) = H – r = H – H/3 = 2H/3.
Volume = 67 1/21 = 1408/21 m³.
Volume = (Vol. cylinder) + (Vol. hemisphere) = πr²h + (2/3)πr³
1408/21 = πr²(h + 2r/3) = (22/7)(H/3)²(2H/3 + 2(H/3)/3) = (22/7)(H²/9)(2H/3 + 2H/9)
1408/21 = (22/63)H²(8H/9) => 1408/21 = (176/567)H³
H³ = (1408/21) × (567/176) = 8 × 27 = 216.
H = ∛216 = 6 m.
Answer: The height of the building is 6 m.
16. The radii of the ends of a frustum of a cone 45 cm high are 28 cm and 7 cm. Find its volume and total surface area.
Step 1: Calculate Volume.
h=45, R=28, r=7.
Volume = (1/3)πh(R²+r²+Rr) = (1/3)(22/7)(45)(28²+7²+28*7) = (15*22/7)(784+49+196) = (330/7)(1029) = 48510 cm³.
Step 2: Calculate Slant Height (l).
l = √(h² + (R-r)²) = √(45² + (28-7)²) = √(2025 + 21²) = √(2025+441) = √2466 ≈ 49.66 cm.
Step 3: Calculate Total Surface Area.
TSA = CSA + Area of bases = πl(R+r) + πR² + πr²
= (22/7)[49.66(35) + 28² + 7²] = (22/7)[1738.1 + 784 + 49] = (22/7)(2571.1) ≈ 8079.5 cm².
Answer: Volume ≈ 48510 cm³, TSA ≈ 8079.5 cm².
17. A solid is in the form of a right circular cylinder with a hemisphere at one end and a cone at the other end. The radius of the common base is 8 cm and the heights of the cylindrical and conical portions are 10 cm and 6 cm respectively. Find the total surface area of the solid.
r=8, hcyl=10, hcone=6.
Slant height of cone l = √(r² + hcone²) = √(8²+6²) = √100 = 10 cm.
TSA = (CSA of Hemisphere) + (CSA of Cylinder) + (CSA of Cone)
= 2πr² + 2πrhcyl + πrl = πr(2r + 2hcyl + l)
= (22/7) × 8 × (2*8 + 2*10 + 10) = (176/7)(16+20+10) = (176/7)(46) ≈ 1156.57 cm².
Answer: The total surface area is approximately 1156.57 cm².
18. A shuttlecock used for playing badminton has the shape of a frustum of a cone surmounted on a hemisphere. The diameters of the frustum are 5 cm and 2 cm. The height of the entire shuttlecock is 7 cm. Find its external surface area.
Hemisphere radius r₁ = 1 cm (diameter 2 cm).
Height of frustum h = Total height – hemisphere radius = 7 – 1 = 6 cm.
Frustum radii: r₁=1 cm, r₂=2.5 cm (diameter 5 cm).
Slant height l = √(h² + (r₂-r₁)²) = √(6² + (2.5-1)²) = √(36 + 1.5²) = √38.25 ≈ 6.18 cm.
External Area = (CSA of frustum) + (CSA of hemisphere)
= πl(r₁+r₂) + 2πr₁²
= (22/7) [6.18(1+2.5) + 2(1)²] = (22/7)[21.63 + 2] = (22/7)(23.63) ≈ 74.26 cm².
Answer: The external surface area is approximately 74.26 cm².
19. From a solid cylinder of height 14 cm and base diameter 7 cm, two equal conical holes of radius 2.1 cm and height 4 cm are cut off. Find the volume of the remaining solid.
Step 1: Calculate volume of the cylinder.
h=14, r=3.5. Volumecyl = πr²h = (22/7)(3.5)²(14) = 539 cm³.
Step 2: Calculate volume of the two conical holes.
rcone=2.1, hcone=4.
Volume2cones = 2 × (1/3)πr²h = 2 × (1/3)(22/7)(2.1)²(4) = 36.96 cm³.
Step 3: Find the volume of the remaining solid.
Volume = 539 – 36.96 = 502.04 cm³.
Answer: The volume of the remaining solid is 502.04 cm³.
20. A right circular cone is divided into two parts by a plane parallel to the base at a distance of h/3 from the base. Find the ratio of the volumes of the smaller cone (top part) to the frustum (bottom part).
Let the original cone have height H and radius R. Its volume is V = (1/3)πR²H.
The smaller cone is formed by the cut. Its height H’ = H – H/3 = 2H/3.
By similar triangles, R’/R = H’/H. So, R’ = R(H’/H) = R((2H/3)/H) = 2R/3.
Volume of smaller cone V’ = (1/3)π(R')²(H') = (1/3)π(2R/3)²(2H/3) = (1/3)π(4R²/9)(2H/3) = (8/27) * (1/3)πR²H = (8/27)V.
Volume of frustum = V – V’ = V – (8/27)V = (19/27)V.
Ratio = (Volume of smaller cone) / (Volume of frustum) = ((8/27)V) / ((19/27)V) = 8/19.
Answer: The ratio of the volumes is 8:19.
