How to Use This Page
Solve every exercise question on your own first, then compare your answers below. Focus on understanding why each step works — that builds exam confidence. Show all working, include units at every step, and verify your final answer by estimation. If a question says "use pi = 22/7", always use that value; otherwise the default is pi = 22/7.
Key Concepts (Quick Recap)
- Cylinder:
- CSA = 2 pi r * h
- TSA = 2 pi r * (r + h)
- Volume = pi r^2 h
- Hollow Cylinder:
- CSA = 2 pi h * (R + r)
- TSA = 2 pi h (R + r) + 2 pi * (R^2 - r^2)
- Cone:
- CSA = pi r l (l = slant height)
- TSA = pi r (l + r)
- Volume = (1/3) pi r^2 * h
- l = sqrt(r^2 + h^2)
- Sphere:
- Surface area = 4 pi r^2
- Volume = (4/3) pi r^3
- Hemisphere:
- CSA = 2 pi r^2
- TSA = 3 pi r^2 (CSA + base circle)
- Volume = (2/3) pi r^3
- Frustum of a Cone:
- CSA = pi (R + r) l
- TSA = pi (R + r) l + pi R^2 + pi r^2
- Volume = (1/3) pi h * (R^2 + Rr + r^2)
- l = sqrt(h^2 + (R - r)^2)
- Combined Solids: Split into parts, calculate each, add volumes; for surface area, add CSAs but subtract hidden/shared surfaces.
- Conversion of Solids: Volume before melting = Volume after recasting.
- Unit conversions: 1 litre = 1000 cm^3; 1 m^3 = 1000 litres; 1 m^2 = 10,000 cm^2; 1 m^3 = 1,000,000 cm^3.
Exercise 7.1 — Surface Area of Solids
Q1. A cylindrical vessel has radius 7 cm and height 15 cm. Find its curved surface area.
Given: r = 7 cm, h = 15 cm
CSA of cylinder = 2 * pi * r * h
= 2 * (22/7) * 7 * 15
= 2 * 22 * 15
= 660 cm^2
Q2. A cylinder has radius 14 cm and height 10 cm. Find its total surface area.
Given: r = 14 cm, h = 10 cm
TSA = 2 * pi * r * (r + h)
= 2 * (22/7) * 14 * (14 + 10)
= 2 * (22/7) * 14 * 24
= 2 * 22 * 2 * 24
= 2112 cm^2
Q3. A conical tent has a base radius of 7 m and slant height 24 m. Find the curved surface area of the tent. Also find the cost of painting the canvas at Rs 15 per m^2.
Given: r = 7 m, l = 24 m
CSA of cone = pi * r * l
= (22/7) * 7 * 24
= 22 * 24 = 528 m^2
Cost of painting = CSA * Rate
= 528 * 15
= 7920
Q4. Find the total surface area of a hemisphere of radius 10.5 cm.
Given: r = 10.5 cm = 21/2 cm
TSA of hemisphere = 3 * pi * r^2
= 3 * (22/7) * (21/2)^2
= 3 * (22/7) * (441/4)
= 3 * 22 * 63 / 4
= 4158 / 4
= 1039.5 cm^2
Q5. The curved surface area of a cone is 308 cm^2 and its slant height is 14 cm. Find the radius of the base.
Given: CSA = 308 cm^2, l = 14 cm
CSA = pi * r * l
308 = (22/7) * r * 14
308 = 22 * 2 * r
308 = 44r
r = 308/44 = 7 cm
Q6. A hollow cylinder has outer radius 8 cm, inner radius 6 cm, and height 10 cm. Find its curved surface area.
Given: R = 8 cm, r = 6 cm, h = 10 cm
CSA of hollow cylinder = 2 * pi * h * (R + r)
= 2 * (22/7) * 10 * (8 + 6)
= 2 * (22/7) * 10 * 14
= 2 * 22 * 10 * 2
= 880 cm^2
Q7. A hemispherical dome of a building needs to be painted. If the circumference of the base is 44 m, find the cost of painting it at Rs 20 per m^2.
Given: Circumference of base = 44 m
Circumference = 2 * pi * r
44 = 2 * (22/7) * r
r = 44 * 7 / 44 = 7 m
CSA of hemisphere = 2 * pi * r^2
= 2 * (22/7) * 7^2
= 2 * 22 * 7
= 308 m^2
Cost = 308 * 20 = 6160
Exercise 7.2 — Volume of Solids
Q1. A cylindrical water tank has radius 1.4 m and height 2 m. Find the capacity of the tank in litres. (1 m^3 = 1000 litres)
Given: r = 1.4 m = 14/10 = 7/5 m, h = 2 m
Volume = pi * r^2 * h
= (22/7) * (7/5)^2 * 2
= (22/7) * (49/25) * 2
= (22 * 49 * 2) / (7 * 25)
= 2156 / 175
= 12.32 m^3
Capacity in litres = 12.32 * 1000 = 12320 litres
Q2. A cone has base radius 6 cm and height 8 cm. Find its volume.
Given: r = 6 cm, h = 8 cm
Volume of cone = (1/3) * pi * r^2 * h
= (1/3) * (22/7) * 36 * 8
= (1/3) * (22/7) * 288
= (22 * 288) / (3 * 7)
= 6336 / 21
= 301.71 cm^3 (approx)
Exact: = (1/3) * 3.14 * 36 * 8 = 301.44 cm^3
Q3. Find the volume of a sphere of radius 5.6 cm.
Given: r = 5.6 cm = 56/10 = 28/5 cm
Volume = (4/3) * pi * r^3
= (4/3) * (22/7) * (28/5)^3
= (4/3) * (22/7) * (21952/125)
= (4 * 22 * 21952) / (3 * 7 * 125)
= 1931776 / 2625
= 735.92 cm^3 (approx)
Q4. A well of diameter 3 m is dug to a depth of 14 m. The earth taken out is spread evenly around the well to form an embankment of width 4 m. Find the height of the embankment.
Given: Diameter of well = 3 m, so r = 1.5 m
Depth of well (h) = 14 m
Width of embankment = 4 m
Volume of earth dug out = pi * r^2 * h
= (22/7) * (1.5)^2 * 14
= (22/7) * 2.25 * 14
= 22 * 2.25 * 2
= 99 m^3
The embankment is a hollow cylinder:
Inner radius R1 = 1.5 m (radius of well)
Outer radius R2 = 1.5 + 4 = 5.5 m
Volume of embankment = pi * (R2^2 - R1^2) * H
99 = (22/7) * (5.5^2 - 1.5^2) * H
99 = (22/7) * (30.25 - 2.25) * H
99 = (22/7) * 28 * H
99 = 88H
H = 99/88 = 9/8 = 1.125 m
Q5. A metallic sphere of radius 10.5 cm is melted and recast into a cylinder of radius 3.5 cm. Find the height of the cylinder.
Given: r_sphere = 10.5 cm, r_cylinder = 3.5 cm
Volume of sphere = Volume of cylinder
(4/3) * pi * r_sphere^3 = pi * r_cylinder^2 * h
(4/3) * (10.5)^3 = (3.5)^2 * h
(4/3) * 1157.625 = 12.25 * h
1543.5 = 12.25h
h = 1543.5 / 12.25 = 126 cm
Q6. How many litres of water can a hemispherical tank of radius 2.1 m hold?
Given: r = 2.1 m = 21/10 m
Volume of hemisphere = (2/3) * pi * r^3
= (2/3) * (22/7) * (21/10)^3
= (2/3) * (22/7) * (9261/1000)
= (2 * 22 * 9261) / (3 * 7 * 1000)
= 407484 / 21000
= 19.404 m^3
Capacity in litres = 19.404 * 1000 = 19404 litres
Q7. The volume of a cone is 1232 cm^3 and its height is 24 cm. Find its radius.
Given: V = 1232 cm^3, h = 24 cm
Volume = (1/3) * pi * r^2 * h
1232 = (1/3) * (22/7) * r^2 * 24
1232 = (22/7) * r^2 * 8
1232 = (176/7) * r^2
r^2 = 1232 * 7 / 176
r^2 = 8624 / 176 = 49
r = 7 cm
Exercise 7.3 — Combined Solids
Q1. A vessel is in the form of a hemisphere mounted on a cylinder. The radius of the hemisphere is 7 cm and the height of the cylindrical part is 13 cm. Find the total surface area of the vessel.
Given: r = 7 cm, h_cylinder = 13 cm
The vessel has:
- A hemisphere on top (open at bottom, attached to cylinder)
- A cylinder below (open at top, attached to hemisphere; closed at bottom)
TSA = CSA of hemisphere + CSA of cylinder + Area of base
= 2 * pi * r^2 + 2 * pi * r * h + pi * r^2
= 3 * pi * r^2 + 2 * pi * r * h
= pi * r * (3r + 2h)
= (22/7) * 7 * (3 * 7 + 2 * 13)
= 22 * (21 + 26)
= 22 * 47
= 1034 cm^2
Q2. A solid is composed of a cylinder with hemispheres on both ends. The radius of each hemisphere is 3.5 cm and the total length of the solid is 21 cm. Find the volume of the solid.
Given: r = 3.5 cm = 7/2 cm
Total length = 21 cm
Length of each hemisphere = r = 3.5 cm
Length of two hemispheres = 2 * 3.5 = 7 cm
Height of cylindrical part = 21 - 7 = 14 cm
Volume = Volume of cylinder + 2 * Volume of hemisphere
= pi * r^2 * h + 2 * (2/3) * pi * r^3
= pi * r^2 * h + (4/3) * pi * r^3
= pi * r^2 * [h + (4/3)r]
= (22/7) * (7/2)^2 * [14 + (4/3)(7/2)]
= (22/7) * (49/4) * [14 + 14/3]
= (22 * 7/4) * [(42 + 14)/3]
= (154/4) * (56/3)
= (77/2) * (56/3)
= 4312/3
= 1437.33 cm^3 (approx)
Q3. A medicine capsule is in the shape 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.
Given: Total length = 14 mm, diameter = 5 mm, so r = 2.5 mm = 5/2 mm
Length of two hemispheres = 2 * 2.5 = 5 mm
Height of cylindrical part = 14 - 5 = 9 mm
Surface area = CSA of cylinder + 2 * CSA of hemisphere
= 2 * pi * r * h + 2 * (2 * pi * r^2)
= 2 * pi * r * h + 4 * pi * r^2
= 2 * pi * r * (h + 2r)
= 2 * (22/7) * (5/2) * (9 + 5)
= 2 * (22/7) * (5/2) * 14
= 2 * 22 * 5
= 220 mm^2
Q4. A toy is in the shape of a cone mounted on a hemisphere of radius 3.5 cm. The total height of the toy is 15.5 cm. Find the volume of the toy.
Given: r = 3.5 cm = 7/2 cm
Total height = 15.5 cm
Height of hemisphere = r = 3.5 cm
Height of cone = 15.5 - 3.5 = 12 cm
Volume of hemisphere = (2/3) * pi * r^3
= (2/3) * (22/7) * (7/2)^3
= (2/3) * (22/7) * (343/8)
= (2 * 22 * 343) / (3 * 7 * 8)
= 15092 / 168
= 89.83 cm^3
Volume of cone = (1/3) * pi * r^2 * h
= (1/3) * (22/7) * (7/2)^2 * 12
= (1/3) * (22/7) * (49/4) * 12
= (22 * 49 * 12) / (3 * 7 * 4)
= 12936 / 84
= 154 cm^3
Total volume = 89.83 + 154 = 243.83 cm^3
Q5. A solid consisting of a right circular cone of height 12 cm and radius 5 cm standing on a hemisphere of radius 5 cm is placed upright in a right circular cylinder full of water such that it touches the bottom. Find the volume of water displaced by the solid, if the radius of the cylinder is 5 cm and height is 18 cm.
Volume of water displaced = Volume of the solid
= Volume of cone + Volume of hemisphere
Volume of cone = (1/3) * pi * r^2 * h
= (1/3) * (22/7) * 25 * 12
= (1/3) * (22/7) * 300
= 2200/7
= 314.29 cm^3
Volume of hemisphere = (2/3) * pi * r^3
= (2/3) * (22/7) * 125
= 5500/21
= 261.90 cm^3
Total = 314.29 + 261.90 = 576.19 cm^3
Exercise 7.4 — Conversion of Solids
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.
Volume of sphere = Volume of cylinder
(4/3) * pi * R^3 = pi * r^2 * h
(4/3) * (4.2)^3 = (6)^2 * h
(4/3) * 74.088 = 36 * h
98.784 = 36h
h = 98.784 / 36 = 2.744 cm
Using pi = 22/7 method:
(4/3) * pi * (42/10)^3 = pi * 6^2 * h
(4/3) * (42/10)^3 = 36 * h
(4/3) * (74088/1000) = 36h
98.784 = 36h
h = 2.744 cm
Q2. A solid metallic hemisphere of radius 8 cm is melted and recast into a cone of base radius 8 cm. Find the height of the cone.
Volume of hemisphere = Volume of cone
(2/3) * pi * R^3 = (1/3) * pi * r^2 * h
(2/3) * 8^3 = (1/3) * 8^2 * h
(2/3) * 512 = (1/3) * 64 * h
1024/3 = 64h/3
1024 = 64h
h = 1024/64 = 16 cm
Q3. A cone of height 24 cm and radius 6 cm is melted and recast into a sphere. Find the radius of the sphere.
Volume of cone = Volume of sphere
(1/3) * pi * r^2 * h = (4/3) * pi * R^3
(1/3) * 6^2 * 24 = (4/3) * R^3
(1/3) * 36 * 24 = (4/3) * R^3
288 = (4/3) * R^3
R^3 = 288 * 3 / 4 = 216
R = cube_root(216) = 6 cm
Q4. A solid metallic cylinder of radius 14 cm and height 21 cm is melted and recast into 288 solid spheres of equal size. Find the radius of each sphere.
Volume of cylinder = 288 * Volume of each sphere
pi * R^2 * H = 288 * (4/3) * pi * r^3
R^2 * H = 288 * (4/3) * r^3
14^2 * 21 = 288 * (4/3) * r^3
196 * 21 = 384 * r^3
4116 = 384 * r^3
r^3 = 4116 / 384 = 10.71875
r = cube_root(10.71875) = 2.2 cm (approx)
Let me verify: 2.2^3 = 10.648... close enough.
Actually let me try with exact values using pi = 22/7:
14^2 = 196, 196 * 21 = 4116
4116 / 384 = 10.71875
r = cube_root(10.71875) ≈ 2.2 cm
Q5. Water flows through a cylindrical pipe of radius 1 cm at 3 m/s into a conical tank of base radius 4 m and height 6 m. How long will it take to fill the tank?
Radius of pipe = 1 cm = 0.01 m
Speed of water = 3 m/s
Volume of water flowing per second = pi * r^2 * speed
= pi * (0.01)^2 * 3 = 3 * 0.0001 * pi = 0.0003 * pi m^3/s
Volume of conical tank = (1/3) * pi * R^2 * H
= (1/3) * pi * 16 * 6 = 32 * pi m^3
Time = Volume of tank / Volume per second
= 32 * pi / (0.0003 * pi)
= 32 / 0.0003
= 106666.67 seconds
In hours: 106666.67 / 3600 = 29.63 hours (approx)
Q6. A spherical cannon ball of diameter 28 cm is melted and recast into a right circular cone of base diameter 28 cm. Find the height of the cone.
Given: Diameter of sphere = 28 cm, so r_sphere = 14 cm
Diameter of cone base = 28 cm, so r_cone = 14 cm
Volume of sphere = Volume of cone
(4/3) * pi * r^3 = (1/3) * pi * r^2 * h
(4/3) * 14^3 = (1/3) * 14^2 * h
4 * 14 = h * (14/3) ... let me redo:
(4/3) * (14)^3 = (1/3) * (14)^2 * h
4 * 14 = h
h = 56 cm
Exercise 7.5 — Multiple Choice Questions
Q1. The curved surface area of a right circular cone of height 15 cm and base radius 8 cm is
(a) 440 cm^2 (b) 427.7 cm^2 (c) 352 cm^2 (d) 308 cm^2
h = 15 cm, r = 8 cm
l = sqrt(r^2 + h^2) = sqrt(64 + 225) = sqrt(289) = 17 cm
CSA = pi * r * l = (22/7) * 8 * 17 = 2992/7 = 427.43 cm^2 ≈ 427.7 cm^2
Q2. The total surface area of a hemisphere of radius r is
(a) 4 pi r^2 (b) 2 pi r^2 (c) 3 pi r^2 (d) pi r^2
Q3. The volume of a frustum of a cone is
(a) (1/3) pi h (R^2 + Rr + r^2) (b) (1/3) pi h (R^2 - r^2) (c) pi h (R^2 + r^2) (d) (1/3) pi h (R + r)
Q4. If a sphere of radius r is melted and recast into two smaller spheres of radii r1 and r2, then
(a) r = r1 + r2 (b) r^3 = r1^3 + r2^3 (c) r^2 = r1^2 + r2^2 (d) r = r1 * r2
Volume is conserved: (4/3) * pi * r^3 = (4/3) * pi * r1^3 + (4/3) * pi * r2^3
r^3 = r1^3 + r2^3
Q5. A shuttlecock used in badminton is a combination of
(a) Cylinder and hemisphere (b) Frustum and hemisphere (c) Cone and hemisphere (d) Cylinder and cone
Q6. The ratio of the volumes of two spheres is 8:27. The ratio of their surface areas is
(a) 2:3 (b) 4:9 (c) 8:27 (d) 16:81
V1/V2 = r1^3/r2^3 = 8/27
r1/r2 = 2/3
SA1/SA2 = (r1/r2)^2 = (2/3)^2 = 4/9
Q7. If the radius of a cylinder is doubled and the height is halved, the volume becomes
(a) 2 times (b) 4 times (c) Same (d) Half
Original: V = pi * r^2 * h
New: V' = pi * (2r)^2 * (h/2) = pi * 4r^2 * h/2 = 2 * pi * r^2 * h = 2V
Q8. The slant height of a frustum of a cone is
(a) sqrt(h^2 + (R + r)^2) (b) sqrt(h^2 + (R - r)^2) (c) sqrt(h^2 - (R - r)^2) (d) h^2 + (R - r)^2
Unit Exercise 7 — Advanced Mixed Problems
Q1. A metallic sphere of radius 12 cm is melted and recast into a cone of base radius 6 cm. Find the height of the cone.
Volume of sphere = Volume of cone
(4/3) * pi * 12^3 = (1/3) * pi * 6^2 * h
4 * 1728 = 36h
6912 = 36h
h = 192 cm
Q2. A hollow cylindrical pipe is 21 dm long. Its outer and inner diameters are 10 cm and 6 cm respectively. Find the volume of the copper used in making the pipe.
Given: Length h = 21 dm = 210 cm
Outer diameter = 10 cm, R = 5 cm
Inner diameter = 6 cm, r = 3 cm
Volume of copper = pi * (R^2 - r^2) * h
= (22/7) * (25 - 9) * 210
= (22/7) * 16 * 210
= 22 * 16 * 30
= 10560 cm^3
Q3. A hemispherical tank full of water is emptied by a pipe at the rate of 3.5 litres per second. How much time will it take to empty half the tank if the radius of the tank is 1.5 m?
Given: r = 1.5 m = 150 cm
Volume of hemisphere = (2/3) * pi * r^3
= (2/3) * (22/7) * (150)^3
= (2/3) * (22/7) * 3375000
= 7425000000 / 21
= 353571.43 cm^3 (Hmm, that's not right)
Let me recalculate in metres:
Volume = (2/3) * pi * (1.5)^3
= (2/3) * (22/7) * 3.375
= (2 * 22 * 3.375) / (3 * 7)
= 148.5 / 21
= 7.0714 m^3
Half tank volume = 7.0714 / 2 = 3.5357 m^3 = 3535.7 litres
Time = 3535.7 / 3.5 = 1010.2 seconds = 16 min 50 sec (approx)
Q4. 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 is 8 cm, and the diameter of the top of the funnel is 18 cm, find the area of the tin sheet required to make the funnel.
Cylindrical part: r = 4 cm, h = 10 cm
CSA of cylinder = 2 * pi * r * h = 2 * (22/7) * 4 * 10 = 1760/7 = 251.43 cm^2
Frustum part:
R = 9 cm (top radius), r = 4 cm (bottom radius = same as cylinder)
Height of frustum = 22 - 10 = 12 cm
l = sqrt(h^2 + (R - r)^2) = sqrt(144 + 25) = sqrt(169) = 13 cm
CSA of frustum = pi * (R + r) * l = (22/7) * (9 + 4) * 13 = (22/7) * 13 * 13 = 3718/7 = 531.14 cm^2
Total area = 251.43 + 531.14 = 782.57 cm^2
Q5. The radius of a sphere is increased by 10%. Find the percentage increase in its surface area.
Original surface area = 4 * pi * r^2
New radius = 1.1r
New surface area = 4 * pi * (1.1r)^2 = 4 * pi * 1.21r^2 = 1.21 * (4 * pi * r^2)
Percentage increase = (1.21 - 1) * 100 = 21%
Q6. A right circular cone of height 8 cm and base radius 6 cm is carved out from a solid wooden hemisphere of radius 6 cm. Find the volume of the remaining wood.
Volume of hemisphere = (2/3) * pi * r^3 = (2/3) * (22/7) * 216 = 9504/21 = 452.57 cm^3
Volume of cone = (1/3) * pi * r^2 * h = (1/3) * (22/7) * 36 * 8 = 6336/21 = 301.71 cm^3
Remaining volume = 452.57 - 301.71 = 150.86 cm^3
Common Mistakes to Avoid
1. Mixing up CSA and TSA — CSA excludes base areas. TSA includes all surfaces. Read the question carefully for keywords like "open top" or "painted on the outside only".
2. Using height instead of slant height in cone CSA — Cone CSA uses slant height l, not vertical height h. Always compute l = sqrt(r^2 + h^2) first.
3. Forgetting unit conversions — If dimensions are in different units (cm and m), convert everything to one unit before substituting. 1 m = 100 cm.
4. Not equating volumes in conversion problems — When a solid is melted and recast, the total volume is conserved. Never use surface area for these problems.
5. Double-counting hidden surfaces in combined solids — When a hemisphere sits on a cylinder, the shared circular face is not visible. Do not include it in the surface area.
6. Wrong hemisphere TSA formula — TSA of a hemisphere = 3 pi r^2 (CSA + flat base). CSA alone = 2 pi r^2. Students often confuse these.
7. Ignoring "open" or "closed" in cylinder/cone problems — An open cylinder has no top circle. Subtract that circle area from TSA.
8. Not writing units — Surface area needs square units (cm^2, m^2), volume needs cubic units (cm^3, m^3). Missing units costs marks in the exam.
Exam Checklist
- [ ] All key formulae memorised (cylinder, cone, sphere, hemisphere, frustum)
- [ ] Can distinguish CSA vs TSA vs Volume from question keywords
- [ ] Know to compute slant height l = sqrt(r^2 + h^2) before cone CSA
- [ ] Unit conversion skills solid (cm to m, cm^3 to litres)
- [ ] Combined solid strategy: split, calculate each part, add/subtract correctly
- [ ] Conversion of solids: always equate volumes, never surface areas
- [ ] Frustum formulae practised (different from full cone)
- [ ] Write units with every answer — sq units for area, cubic units for volume
- [ ] Verify answer by estimation (e.g., volume of a 10 cm sphere should be thousands, not single digits)
- [ ] Draw a rough diagram for every word problem before starting calculations
Frequently Asked Questions
If the question mentions painting, covering, plastering, or wrapping, use surface area. If it mentions filling, capacity, water, melting, or how much material, use volume.