1
The curved surface of a cone uses —
💡 Hint:
Slant height.
Explanation:
πrl.
Class 10 · Maths · Chapter 12 · बिहार बोर्ड (BSEB)CBSE · NCERT 2026-27
Surface Areas and Volumes
Surface Areas and Volumes
How to use this page:
1. Read — Combined surface · Exercise 12.1 · combined volume · Exercise 12.2, diagram, worked example, board tip
2. Check — each lesson has its own questions; the number follows the lesson
3. Mastery ★ — all of that lesson correct. Redo the wrong ones
4. The memory figure shows a hemisphere on a cylinder and keeps the joint circle out of the outer surface.
In NCERT 2026-27 this is Chapter 12; in Bihar’s older book Surface Areas and Volumes is Chapter 13. Progress stays in this browser.
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- 1 Five solids, two kinds of surface
- 2 Do not count the joint circle
- 3 A hemisphere on a cone, and a hollow vessel
- 4 Capsule, tent and cavity
- 5 A hemisphere on a cube, and both ends
- 6 Add the volumes
- 7 Holes, syrup and water left
- Chapter winner — every lesson at mastery ★
🎬 Shorts
This chapter has no Short on the channel yet, so ten empty cards are not shown. The notes, diagrams and checks are complete.
📄 PDF download — notes + answer key
Five solids, two kinds of surface
पाँच ठोस, दो तरह के पृष्ठ · NCERT 12.1 · the Class 9 formulas
Curved surface and total surface are differentNotes
The solids in this chapter are the cube, cuboid, cylinder, cone, sphere and hemisphere. Curved surface is only the bent part. Total surface also includes the bases.
A cone has curved surface πrl, a hemisphere 2πr², a cylinder 2πrh. A sphere has total surface 4πr². In Exercises 12.1 and 12.2, π = 22/7 when nothing else is written.
Keep l and h apartActivity
For a cone the slant height is l = √(r² + h²). Surface uses l. Volume (1/3)πr²h uses the straight height h. Do not push l into the volume. The unit is cm² on a surface and cm³ on a volume.
Worked exampleExample
Question: A cone has radius 3.5 cm and height 12 cm. Find the slant height and the curved surface. π = 22/7.
Formula: l = √(r² + h²), curved surface = πrl.
Substitution: l = √(12.25 + 144) = √156.25 = 12.5. Curved surface = (22/7) × 3.5 × 12.5 = 137.5.
Answer: l = 12.5 cm, curved surface 137.5 cm².
| Solid | Curved / total | Volume |
|---|---|---|
| Cylinder | 2πrh | πr²h |
| Cone | πrl | (1/3)πr²h |
| Hemisphere | 2πr² | (2/3)πr³ |
| Sphere | 4πr² | (4/3)πr³ |
10-second revision
- l in surface, h in volume
- Hemisphere curved surface 2πr²
- π = 22/7 unless written otherwise
Board tip · BSEBBoard tip
In a BSEB answer write the formula line with the unit.
Board tip · CBSEBoard tip
On CBSE do not add the base πr² into a curved surface on your own.
Check your understandingall correct = mastery ★
2
The volume of a cone uses the slant height l.
Explanation:
False — volume uses h.
3
If r = 3.5 cm and h = 12 cm, then l = ______ cm.
💡 Hint:
√156.25.
Explanation:
12.5 cm.
4
A hemisphere has radius 7 cm. Write the curved surface and the volume. π = 22/7.
Model answer (board format)
Curved surface = 2×(22/7)×49 = 308 cm². Volume = (2/3)×(22/7)×343 = 2156/3 cm³.
Check your answer:
Do not count the joint circle
जोड़ का वृत्त मत गिनो · NCERT 12.2 · Exercise 12.1 question 1
Only the surface that shows outsideNotes
When two solids stick, the circle where they stick does not touch the air. Remove it from both surfaces, or add only curved surfaces from the start.
Two cubes of volume 64 cm³ joined end to end make a cuboid 8 × 4 × 4, because the edge is 4 cm. The surface in question 1 is 160 cm². Count the new outer surface, not 2 × 6a² of two separate cubes.
Name the joined face firstActivity
The end square of a cube is 4 × 4 = 16 cm². Two such squares hide, 32 cm² in all. Two separate cubes had area 2 × 96 = 192. 192 − 32 = 160. The cuboid formula 2(lb+bh+hl) gives the same answer.
Worked exampleExample
Question: Two cubes of volume 64 cm³ are joined end to end. Find the surface area.
Formula: Surface of a cuboid = 2(lb + bh + hl).
Substitution: Edge 4 cm. Size 8, 4, 4. 2(32 + 16 + 32) = 2 × 80 = 160.
Answer: 160 cm².
10-second revision
- A hidden square is removed twice
- The 8 × 4 × 4 surface is 160 cm²
- This subtraction is not done for volume
Board tip · BSEBBoard tip
In a BSEB answer show one full path, either 192 − 32 or 2(lb+bh+hl).
Board tip · CBSEBoard tip
On CBSE do not leave 192 cm² of the two separate cubes as the final answer.
Check your understandingall correct = mastery ★
1
The surface after joining two cubes of volume 64 cm³ is —
💡 Hint:
8 × 4 × 4.
Explanation:
160 cm².
2
The square where they join is counted in the surface of both cubes.
Explanation:
False — it is not outside.
3
The edge of a cube of volume 64 cm³ is ______ cm.
💡 Hint:
4³ = 64.
Explanation:
4 cm.
4
The outer surface of a hemisphere on a cylinder is —
💡 Hint:
Drop the joint circle.
Explanation:
2πrh + 2πr².
5
Write the surface of one cube of edge 4 cm and of two such cubes joined.
Model answer (board format)
One cube is 6 × 16 = 96 cm². Joined, the surface is 160 cm².
Check your answer:
A hemisphere on a cone, and a hollow vessel
शंकु पर अर्धगोला और खोखला बर्तन · Exercise 12.1 questions 2 and 3
Split the height, then add curved surfacesNotes
In question 3 the toy is 15.5 cm tall and both radii are 3.5 cm. The hemisphere takes 3.5 cm, so the cone height is 12 cm.
Total surface = πrl + 2πr² = πr(l + 2r) = 214.5 cm². Question 2 is hollow: inner surface = curved hemisphere + curved cylinder. Diameter 14 cm and total height 13 cm give a cylinder height of 6 cm.
For a hollow solid, the inner surfaceActivity
Inner surface of the vessel = 2πr² + 2πrh = 2πr(r + h). r = 7, h = 6. 2 × (22/7) × 7 × 13 = 572 cm². Do not find the outer surface; the question asks for the inside.
Worked exampleExample
Question: r = 3.5 cm, cone height 12 cm, l = 12.5 cm. Total surface of the toy. π = 22/7.
Formula: πr(l + 2r).
Substitution: (22/7) × 3.5 × (12.5 + 7) = 11 × 19.5 = 214.5.
Answer: 214.5 cm².
10-second revision
- Cone height = total − r
- The toy is 214.5 cm²
- The hollow vessel is 572 cm²
Board tip · BSEBBoard tip
In a BSEB answer write the line 15.5 − 3.5 = 12 first.
Board tip · CBSEBoard tip
On CBSE do not turn the outer surface of the hollow vessel into the answer.
Check your understandingall correct = mastery ★
1
The total surface of the toy is —
💡 Hint:
πr(l+2r).
Explanation:
214.5 cm².
2
In a total height of 15.5 cm the cone height is also 15.5 cm.
Explanation:
False — the hemisphere takes 3.5 cm.
3
The inner surface of the hollow vessel is ______ cm².
💡 Hint:
2πr(r+h), h = 6.
Explanation:
572 cm².
4
With diameter 14 cm and total height 13 cm, the cylinder height is —
💡 Hint:
13 − 7.
Explanation:
6 cm.
5
The base circle is added separately in the total surface of the toy.
Explanation:
False — it is hidden in the joint.
6
r = 7 cm, cylinder height 6 cm. Write the inner-surface calculation.
Model answer (board format)
Formula 2πr(r+h). 2×(22/7)×7×(7+6) = 44×13 = 572 cm².
Check your answer:
Capsule, tent and cavity
कैप्सूल, तंबू और गुहा · Exercise 12.1 questions 6, 7 and 8
What is covered, and what is notNotes
In a capsule the two hemispheres make one sphere. If the whole length is 14 mm and the diameter is 5 mm, the cylinder length is 14 − 5 = 9 mm. Surface = 2πrh + 4πr² = 220 mm².
Tent canvas does not cover the floor. Canvas = 2πrh + πrl. With r = 2 m, h = 2.1 m and l = 2.8 m the area is 44 m² and the cost is 44 × 500 = ₹22000. In a conical cavity the inner curved cone is added and the top circle is open.
The cavity surface in three piecesActivity
Cylinder height 2.4 cm, diameter 1.4 cm, and a cone of the same size is removed. l = 2.5 cm. Total surface of what remains = curved cylinder + bottom base + curved cone = 2πrh + πr² + πrl = 17.6 cm².
Worked exampleExample
Question: A tent has r = 2 m, cylinder height 2.1 m and cone slant height 2.8 m. Canvas and cost at ₹500 per m². π = 22/7.
Formula: Area = πr(2h + l). Cost = area × 500.
Substitution: (22/7) × 2 × (4.2 + 2.8) = (44/7) × 7 = 44. Cost = 22000.
Answer: 44 m² and ₹22000.
10-second revision
- Capsule 220 mm²
- No floor on the tent
- Cavity surface 17.6 cm²
Board tip · BSEBBoard tip
In a BSEB answer write both the area and the rupees for the tent.
Board tip · CBSEBoard tip
On CBSE do not add the floor πr² to the canvas. The question itself forbids it.
Check your understandingall correct = mastery ★
1
The capsule surface is —
💡 Hint:
2πr(h+2r).
Explanation:
220 mm².
2
The tent canvas also covers the floor.
Explanation:
False — the question leaves the floor out.
3
The tent canvas is ______ m².
💡 Hint:
πr(2h+l).
Explanation:
44 m².
4
The surface of the solid left after a conical cavity is about —
💡 Hint:
2πrh + πr² + πrl.
Explanation:
17.6 cm².
5
A capsule is 14 mm long and 5 mm across. Write the cylinder length and the surface. π = 22/7.
Model answer (board format)
Cylinder = 14 − 2×2.5 = 9 mm. Surface = 2×(22/7)×2.5×(9+5) = 220 mm².
Check your answer:
A hemisphere on a cube, and both ends
घन पर अर्धगोला और दोनों सिरे · Exercise 12.1 questions 4, 5 and 9
Remove the base, add the curveNotes
In question 4 the cube edge is 7 cm. The greatest hemisphere takes that edge as diameter, so r = 3.5 cm. Surface = cube surface − base circle + curved hemisphere = 6a² + πr² = 332.5 cm².
In question 5 the depression is inside, but the formula takes the same shape: 6l² + πr². In question 9 a hemisphere is scooped from each end of a cylinder. Surface = 2πrh + 4πr² = 374 cm², because both ends are now inner curves. r = 3.5 cm, h = 10 cm.
Diameter and edge are the same numberActivity
If the edge is 7 cm, the diameter is 7 cm and the radius is 3.5 cm. πr² = (22/7)×(49/4) = 77/2 = 38.5. 6×49 = 294. 294 + 38.5 = 332.5. Do not leave the radius as 7.
Worked exampleExample
Question: Cylinder h = 10 cm, r = 3.5 cm, a hemisphere scooped from each end. Total surface. π = 22/7.
Formula: 2πr(h + 2r).
Substitution: 2 × (22/7) × 3.5 × (10 + 7) = 22 × 17 = 374.
Answer: 374 cm².
10-second revision
- Greatest diameter = the edge
- Cube + hemisphere = 332.5 cm²
- Both ends = 374 cm²
Board tip · BSEBBoard tip
In a BSEB answer add 6a² and πr² separately to reach 332.5.
Board tip · CBSEBoard tip
On CBSE do not leave the flat circle of a scooped end in the outer surface.
Check your understandingall correct = mastery ★
1
The surface of a 7 cm cube with the greatest hemisphere is —
💡 Hint:
294 + 38.5.
Explanation:
332.5 cm².
2
The radius of the greatest hemisphere is 7 cm.
Explanation:
False — the diameter is 7 cm, so the radius is 3.5 cm.
3
The total surface in question 9 is ______ cm².
💡 Hint:
2πr(h+2r).
Explanation:
374 cm².
4
Edge 7 cm. Write the calculation of 6a² + πr². r = 3.5 cm, π = 22/7.
Model answer (board format)
6×49 = 294. πr² = (22/7)×(49/4) = 38.5. The sum is 332.5 cm².
Check your answer:
Add the volumes
आयतन जोड़ो · NCERT 12.3 · Exercise 12.2 questions 1 and 2
The space addsNotes
The circle removed from the surface is not removed from the volume. For a cone and a hemisphere the volume is (1/3)πr²h + (2/3)πr³.
In question 1, r = 1 cm and h = r. The volume is π cm³. In question 2 the model is 12 cm long and each cone is 2 cm high, so the cylinder keeps 8 cm. Diameter 3 cm gives r = 1.5 cm. The volume of air is 66 cm³.
Cut the cones out of the lengthActivity
Total length = cylinder + two cones. 12 = h + 2 + 2, so h = 8. Volume = πr²h + 2×(1/3)πr²×2 = πr²(8 + 4/3) = 21π = 66 cm³, when π = 22/7 and r² = 2.25.
Worked exampleExample
Question: A cone with r = 1 cm, h = 1 cm, and a hemisphere of the same radius. Volume in terms of π.
Formula: (1/3)πr²h + (2/3)πr³.
Substitution: (1/3)π + (2/3)π = π.
Answer: π cm³.
10-second revision
- Question 1 has volume π cm³
- Cylinder height 12 − 4 = 8
- The air is 66 cm³
Board tip · BSEBBoard tip
In a BSEB answer do not turn π cm³ into 22/7; the question asks for a form in π.
Board tip · CBSEBoard tip
On CBSE do not write the total length of the model as the cylinder height.
Check your understandingall correct = mastery ★
1
The volume of a cone and hemisphere with r = h = 1 cm is —
💡 Hint:
1/3 + 2/3.
Explanation:
π cm³.
2
The joint circle is subtracted in the volume.
Explanation:
False — that belongs only to the surface.
3
The cylinder height in the model is ______ cm.
💡 Hint:
12 − 4.
Explanation:
8 cm.
4
The volume of air inside the model is —
💡 Hint:
21π and π = 22/7.
Explanation:
66 cm³.
5
The heights of the two cones are cut out of the total length.
Explanation:
True — 2 + 2 = 4.
6
r = 1.5 cm, cylinder 8 cm, two cones of height 2 cm each. Find the volume. π = 22/7.
Model answer (board format)
πr²(8 + 4/3) = (22/7)×2.25×(28/3) = 21×22/7 = 66 cm³.
Check your answer:
Holes, syrup and water left
गड्ढा, रस और बचा पानी · Exercise 12.2 questions 3 to 8 · summary
Space that is not solid is subtractedNotes
The wood in the pen stand = cuboid − four cones. In a gulab jamun the syrup is 30 percent of the volume. Lead shots fill a quarter of the water in the cone, and the number of shots is 100.
The mass of the pole = volume × 8 g, and there π = 3.14. The cylinder of water keeps what remains after the cone and the hemisphere are removed. The glass vessel calculates to about 346.5 cm³, so the child figure 345 cm³ is not exact.
One shot volume firstActivity
Cone r = 5 cm, h = 8 cm. A quarter of the water = (1/4)×(1/3)πr²h = 1100/21 cm³. One shot of r = 0.5 cm has volume 11/21 cm³. Number = (1100/21) ÷ (11/21) = 100.
Worked exampleExample
Question: A cone of r = 5 cm and h = 8 cm is full of water. Shots of r = 0.5 cm spill a quarter of the water. Find the number. π = 22/7.
Formula: Number = (water spilled) / (one shot).
Substitution: Water spilled = 1100/21 cm³. One shot = 11/21 cm³. The quotient is 100.
Answer: 100 shots.
10-second revision
- 100 shots
- Syrup is 30 percent of the volume
- Mass = volume × 8 g
Board tip · BSEBBoard tip
In a BSEB answer show the line where 1100/21 and 11/21 cancel.
Board tip · CBSEBoard tip
On CBSE do not accept 345 cm³ as exact without the check.
Check your understandingall correct = mastery ★
1
The number of lead shots is —
💡 Hint:
1100/11.
Explanation:
100.
2
The syrup in a gulab jamun equals the whole volume.
Explanation:
False — about 30 percent.
3
The mass of iron is the volume multiplied by ______ g.
💡 Hint:
The mass of 1 cm³.
Explanation:
8 g.
4
The child figure of 345 cm³ —
💡 Hint:
The calculation is about 346.5.
Explanation:
345 is not exact.
5
A pen stand is 15 × 10 × 3.5 cm. To start the wood volume, write the cuboid volume. Four cones are subtracted later.
Model answer (board format)
Cuboid = 15 × 10 × 3.5 = 525 cm³. Wood = 525 − the volume of four cones.
Check your answer:
❓ Full question bank — with answers and explanations — 65 questions
Pick a type. The 35 lesson checks are separate — each lesson has as many as its topic needs. All correct earns mastery ★.
BSEB modelpractice · not annual60 tried · 0 correctMultiple choiceMCQ120 tried · 0 correctTrue / falseTrue / false80 tried · 0 correctFill in the blanksFill80 tried · 0 correctMatchMatch20 tried · 0 correctAssertion–reasonAssertion–reason50 tried · 0 correctBalancingBalancing40 tried · 0 correctIdentifyIdentify20 tried · 0 correctShort and longVSA, SA, LA120 tried · 0 correctCBSE-stylenot a PYQ60 tried · 0 correct
No question is marked as a verified past paper. The BSEB set is a model for practice. CBSE items are CBSE-style, not a copy of any year’s paper.
Multiple choice
0/12Pick one option. A wrong try brings a hint.
1
The surface after joining the two cubes is —
💡 Hint:
8 × 4 × 4.
Explanation:
160 cm².
2
The toy surface is —
💡 Hint:
πr(l+2r).
Explanation:
214.5 cm².
3
The inner surface of the hollow vessel is —
💡 Hint:
2πr(r+h).
Explanation:
572 cm².
4
The tent canvas is —
💡 Hint:
No floor.
Explanation:
44 m².
5
The capsule surface is —
💡 Hint:
Two hemispheres make one sphere.
Explanation:
220 mm².
6
The surface in question 9 is —
💡 Hint:
2πr(h+2r).
Explanation:
374 cm².
7
The cone and hemisphere volume is —
💡 Hint:
1/3 + 2/3.
Explanation:
π cm³.
8
The air in the model is —
💡 Hint:
Cylinder 8 cm.
Explanation:
66 cm³.
9
The lead shots number —
💡 Hint:
A quarter of the water.
Explanation:
100.
10
The cube with a hemisphere has surface —
💡 Hint:
6a² + πr².
Explanation:
332.5 cm².
11
The cavity surface is —
💡 Hint:
Three pieces.
Explanation:
17.6 cm².
12
The mass of 1 cm³ of iron is —
💡 Hint:
Question 6.
Explanation:
8 g.
True or false
0/81
The joint circle is counted in the outer surface.
Explanation:
False — it does not touch the air.
2
The joint is not subtracted when volumes are added.
Explanation:
True.
3
Tent canvas covers the floor.
Explanation:
False.
4
The cone volume uses l.
Explanation:
False — it uses h.
5
The volume in question 1 is π cm³.
Explanation:
True.
6
The number of shots is 100.
Explanation:
True.
7
The syrup equals the whole gulab jamun.
Explanation:
False — 30 percent.
8
345 cm³ is the exact volume of the glass vessel.
Explanation:
False — the calculation is about 346.5 cm³.
Fill in the blanks
0/81
The two joined cubes have surface ______ cm².
💡 Hint:
160.
Explanation:
2(32+16+32).
2
The cone height of the toy is ______ cm.
💡 Hint:
12.
Explanation:
15.5 − 3.5.
3
The tent area is ______ m².
💡 Hint:
44.
Explanation:
πr(2h+l).
4
The model cylinder is ______ cm tall.
💡 Hint:
8.
Explanation:
12 − 4.
5
There are ______ shots.
💡 Hint:
100.
Explanation:
1100/11.
6
The capsule is ______ mm².
💡 Hint:
220.
Explanation:
2πr(h+2r).
7
Question 9 is ______ cm².
💡 Hint:
374.
Explanation:
22 × 17.
8
1 cm³ of iron is ______ g.
💡 Hint:
8.
Explanation:
Question 6.
Match
0/21
Match the solid with its volume.
Column B: A. (2/3)πr³ · B. πr²h · C. (1/3)πr²h · D. (4/3)πr³
1. Cylinder
2. Cone
3. Sphere
4. Hemisphere
Explanation:
Cylinder πr²h, cone one third, sphere 4/3, hemisphere 2/3.
2
Match the question with its answer.
Column B: A. 100 shots · B. 160 cm² · C. 214.5 cm² · D. 44 m²
1. 12.1 item 1
2. 12.1 item 3
3. 12.1 item 7
4. 12.2 item 5
Explanation:
Cubes 160, toy 214.5, tent 44, shots 100.
Assertion–reason
0/5Check both statements, then see whether the reason explains the assertion.
1
Assertion (A): Two joined cubes have surface 160 cm².
Reason (R): The area of the two hidden squares leaves the surface.
Explanation:
Both are true and R explains A.
2
Assertion (A): The joint circle must also be subtracted in the volume.
Reason (R): Volume is the space inside, and joining does not erase that space.
Explanation:
A is false. R is true.
3
Assertion (A): The tent canvas is 44 m².
Reason (R): The floor area is included in it.
Explanation:
A is true. R is false.
4
Assertion (A): The toy surface is 214.5 cm².
Reason (R): The volume of a sphere is (4/3)πr³.
Explanation:
Both are true, but R does not explain this surface.
5
Assertion (A): The number of shots is 100.
Reason (R): The spilled water is 100 times the volume of one shot.
Explanation:
Both are true and R explains A.
Coefficient practice
0/4These chemistry lines are not a result of this maths chapter. They are only practice in filling coefficients. A blank means 1.
1
This is coefficient practice and not a result of this chapter. In the water line, Fill coefficients (blank = 1).
H2 + O2 → H2O
Explanation: 2H2 + O2 → 2H2O
2
This is coefficient practice and not a result of this chapter. In the line of H2 and O2, Fill coefficients (blank = 1).
H2 + O2 → H2O
Explanation: 2H2 + O2 → 2H2O
3
This is coefficient practice and not a result of this chapter. In this coefficient line, Fill coefficients (blank = 1).
H2 + O2 → H2O
Explanation: 2H2 + O2 → 2H2O
4
This is coefficient practice and not a result of this chapter. In the line with blank coefficients, Fill coefficients (blank = 1).
H2 + O2 → H2O
Explanation: 2H2 + O2 → 2H2O
Classify
0/21
Place each job as surface or volume.
Tent canvas
Air in the model
The iron pole demand, in grams
The outer measure of two cubes
Explanation:
Canvas and the cubes are surfaces. Air is volume. The pole asks for mass.
2
Place each question in its exercise.
Two cubes, surface 160
The solid of π cm³
100 shots
Canvas 44 m²
Explanation:
Surface questions are in 12.1. Volume questions are in 12.2.
Very short answer
0/51
Write the curved surface and the volume of a cone.
Model answer (board format)
Curved surface πrl, volume (1/3)πr²h.
Check your answer:
2
Write the answer of the two-cube question.
Model answer (board format)
160 cm².
Check your answer:
3
Why is the floor not included for the tent?
Model answer (board format)
The question says the base is not covered with canvas. The area is 44 m².
Check your answer:
4
Write the volume of question 1 in terms of π.
Model answer (board format)
π cm³.
Check your answer:
5
Write the number of shots and the volume of one shot.
Model answer (board format)
One shot is 11/21 cm³, and the number is 100.
Check your answer:
Short answer
0/41
Find the toy surface. r = 3.5, l = 12.5, π = 22/7.
Model answer (board format)
πr(l+2r) = (22/7)×3.5×19.5 = 214.5 cm².
Check your answer:
2
Find the inner surface of the hollow vessel. r = 7, h = 6.
Model answer (board format)
2πr(r+h) = 2×(22/7)×7×13 = 572 cm².
Check your answer:
3
Find the air in the model. r = 1.5, cylinder 8, two cones of height 2.
Model answer (board format)
πr²(8+4/3) = 21π = 66 cm³.
Check your answer:
4
Find the surface in question 9. r = 3.5, h = 10.
Model answer (board format)
2πr(h+2r) = 2×(22/7)×3.5×17 = 374 cm².
Check your answer:
Long answer
0/31
A hemisphere of r = 3.5 cm sits on a cube of edge 7 cm. Write the full surface calculation. Then say how the volumes would add if asked. π = 22/7.
Model answer (board format)
Surface = 6×49 + (22/7)×(49/4) = 294 + 38.5 = 332.5 cm². Volume = 7³ + (2/3)πr³ = 343 + (2/3)×(22/7)×(343/8). The joint is removed from the surface, not from the volume.
Check your answer:
2
Shots of r = 0.5 cm spill a quarter of the water in a cone of r = 5 cm and h = 8 cm. Find the number. π = 22/7.
Model answer (board format)
Cone = (1/3)×(22/7)×25×8 = 4400/21 cm³. A quarter = 1100/21. One shot = (4/3)×(22/7)×(1/8) = 11/21. The number is 100.
Check your answer:
3
Tent r = 2 m, h = 2.1 m, l = 2.8 m. Write the area and the cost at ₹500 per m². Why is there no floor?
Model answer (board format)
Area = (22/7)×2×(4.2+2.8) = 44 m². Cost = 44×500 = ₹22000. The question says the base is not covered with canvas.
Check your answer:
BSEB model paper · practice
0/6This model set is for practice. It is not a question from any year’s annual examination.
1
The cylinder length in the capsule is —
💡 Hint:
14 − 5.
Explanation:
9 mm.
2
The cuboid volume of the pen stand is —
💡 Hint:
15×10×3.5.
Explanation:
525 cm³.
3
The tent costs ______ rupees.
💡 Hint:
22000.
Explanation:
44 × 500.
4
Find the cavity surface. r = 0.7 cm, h = 2.4 cm, l = 2.5 cm. π = 22/7.
Model answer (board format)
2πrh + πr² + πrl = πr(2h + r + l) = (22/7)×0.7×(4.8+0.7+2.5) = 2.2×8 = 17.6 cm².
Check your answer:
5
A cylinder of r = 60 cm and h = 180 cm is full of water. A hemisphere of r = 60 cm and a cone of h = 120 cm are submerged. Write the remaining water as a multiple of π.
Model answer (board format)
Cylinder 648000π. Cone 144000π. Hemisphere 144000π. Remaining 360000π cm³.
Check your answer:
6
Assertion (A): The volume in question 1 is π cm³.
Reason (R): Writing it as 22/7 cm³ is what the question demands.
Explanation:
A is true. R is false, the question asks for a form in π.
CBSE-style questions
0/6These are competency-based practice questions. They are not a copy of any year’s paper.
1
A student writes 192 cm² for the two cubes. The correct surface is —
💡 Hint:
32 cm² is hidden.
Explanation:
160 cm².
2
The cone height of the toy is wrongly written as 15.5 cm. The correct height is —
💡 Hint:
Cut off the hemisphere.
Explanation:
12 cm.
3
Assertion (A): The glass vessel volume of 345 cm³ is not exact.
Reason (R): The neck and the sphere add to about 346.5 cm³.
Explanation:
Both are true and R explains A.
4
Assertion (A): The floor πr² must be added to the tent canvas.
Reason (R): The question writes that the base will not be covered with canvas.
Explanation:
A is false. R is true.
5
A student puts l = 12.5 into the cone volume. For r = 3.5 and h = 12, write the correct volume formula and the reason.
Model answer (board format)
The volume is (1/3)πr²h, not l. l belongs only in the curved surface πrl. Here h = 12 cm.
Check your answer:
6
Syrup is 30 percent. Take one gulab jamun as 25 cm³. About how much syrup is in 45 pieces?
Model answer (board format)
Syrup in one is 0.3 × 25 = 7.5 cm³. In 45 it is 45 × 7.5 = 337.5 cm³.
Check your answer:
🏛️ Board exam corner — Bihar Board (BSEB)— CBSE
Switch board with BSEB | CBSE above. The lessons follow the same NCERT chapter.
BSEB model · practiceBoard tip
This page has no verified annual-exam question, because no source page has been added. The model set below is practice in the board pattern.
🏛️ Model questions on one page →
The verified label will be used only when a source page for the question is available.
CBSE-style · competency-based
These are case and assertion-reason practice items. Do not treat them as past CBSE questions.
🔁 Spaced review — today’s questions
Wrong questions return soon; correct ones return after a few days.
🧠 What you learned + equation sheet
What you learned
- A joined circle is not counted twice in the outer surface.
- Volumes add; a depression or a cavity is subtracted.
- The curved surface of a cone is πrl, and l = √(r² + h²).
- A hemisphere has curved surface 2πr² and volume (2/3)πr³.
- The canvas of a tent does not cover the floor.
- In Exercises 12.1 and 12.2, π = 22/7 when nothing else is written.
| What | Keep this |
|---|---|
| Cube / cuboid | 6a² और 2(lb+bh+hl); आयतन a³ और lbh |
| Cylinder | वक्र 2πrh; आयतन πr²h |
| Cone | वक्र πrl; आयतन (1/3)πr²h |
| Sphere | 4πr²; आयतन (4/3)πr³ |
| Hemisphere | वक्र 2πr²; आयतन (2/3)πr³ |
| Joint | पृष्ठ से हटाओ, आयतन से नहीं |
The notes are original writing. The textbook was used only for activity order and numbers. “Verified” will be used only on a question that has a source page.