Posted inMaths Volume & Surface Area (ఘనపరిమాణము మరియు ఉపరితల వైశాల్యం) Posted by By AP STUDY CIRCLE 01/05/2026No Comments Report a question What's wrong with this question?You cannot submit an empty report. Please add some details. ముఖ్య సూచనలు (Important Instructions)అభ్యర్థులు పరీక్ష ప్రారంభించే ముందు ఈ క్రింది నియమాలను జాగ్రత్తగా చదవండి:ప్రశ్నల సంఖ్య: ఈ పరీక్షలో మొత్తం 50 బహుళ ఐచ్ఛిక ప్రశ్నలు (MCQs) ఉంటాయి.సమయ పరిమితి: పరీక్షకు కేటాయించిన సమయం 50 నిమిషాలు. స్క్రీన్ పైన టైమర్ను గమనిస్తూ ఉండండి.ఆటోమేటిక్ సబ్మిషన్: 50 నిమిషాల సమయం ముగియగానే, మీరు సబ్మిట్ చేయకపోయినా మీ సమాధానాలు ఆటోమేటిక్గా సేవ్ చేయబడతాయి. నావిగేషన్: తర్వాతి ప్రశ్నకు వెళ్లడానికి 'Next' బటన్ నొక్కండి. మునుపటి ప్రశ్నకు వెళ్లి సమాధానం మార్చుకోవడానికి 'Previous' బటన్ ఉపయోగించవచ్చు.సందేహాలు/ఫిర్యాదులు: ఏదైనా ప్రశ్నపై సందేహం ఉంటే, ఆ ప్రశ్న కింద ఉన్న **'Complaint Box'**లో తెలియజేయవచ్చు. ఫలితాలు & : * పరీక్ష పూర్తయిన వెంటనే, See result నొక్కండి. మీ మార్కులు (Marks) మరియు మీ ప్రశ్నాపత్రం జవాబులతో స్క్రీన్ పై కనిపిస్తాయి. గమనిక: పరీక్ష మధ్యలో పేజీని 'Refresh' చేయకండి. ఈ పరీక్షలపై అభిప్రాయాలను తప్పకుండా తెలియచేయండి. మా వెబ్సైటు subscribe చేసుకోండి.ప్రతి పరీక్ష upload చేసిన వెంటనే మీకు నోటిఫికేషన్ వస్తుంది. ఆల్ ది బెస్ట్ & PRESS StartVolume & Surface Area (ఘనపరిమాణము మరియు ఉపరితల వైశాల్యం) 1 / 501. Hollow cylinder outer r=5, inner r=3, h=10. Volume? Hollow cylinder: outer r=5, inner r=3, h=10. Volume? A. 80π B. 160π C. 250π D. 120π Volume = 160π cm³V = π(R²-r²)h = π(25-9)×10 = 160π cm³ ≈ 502.4 cm³.గుర్తు: Hollow cylinder = Outer − Inner cylinder volumes. Pipe, tube — hollow cylinder shapes! Quiz(PDF) Questions(PDF) 2 / 502. Sphere diameter=14 cm. Volume? (π=22/7) Sphere diameter=14. Volume? A. 2156 B. 4312/3 C. 616 D. (4/3)×14³ Volume = 1437⅓ cm³r = 14/2 = 7 cm. V = (4/3)×(22/7)×343 = 4312/3 ≈ 1437.3 cm³. Quiz(PDF) Questions(PDF) 3 / 503. Cylinder Curved Surface Area (CSA) formula? Cylinder CSA formula? A. 2πrh B. πrh C. πr²h D. 2πr² CSA = 2πrhCylinder CSA = 2πrh (lateral surface only — top/bottom రాదు). Total SA = 2πrh + 2πr² = 2πr(h+r).గుర్తు: CSA = Circumference × height = 2πr × h. Label on a tin can = CSA! Quiz(PDF) Questions(PDF) 4 / 504. Cube volume = 125 cm³. Side మరియు TSA? Cube volume=125. Find side and TSA. A. side=5,TSA=600 B. side=5,TSA=125 C. side=5,TSA=150 D. side=25,TSA=150 side=5 cm, TSA=150 cm²a³=125 → a=∛125=5 cm. TSA=6×25=150 cm². Quiz(PDF) Questions(PDF) 5 / 505. Sphere radius=7 cm. Volume? (π=22/7) Sphere r=7. Volume? A. (22/7)×49 B. 4312/3 C. 4/3×22×7 D. (22/7)×343 Volume = 1437⅓ cm³Sphere Volume = (4/3)πr³ = (4/3) × (22/7) × 343. = (4/3) × 22 × 49 = (4 × 1078)/3 = 4312/3 = 1437.33 cm³.గుర్తు: (4/3)πr³ — "four thirds pi r cubed". Football, orange, Earth — sphere shapes! Quiz(PDF) Questions(PDF) 6 / 506. Cone r=6, l=10. TSA? (π=22/7) Cone r=6, slant=10. TSA? A. πrl B. 2112/7 C. πr² D. πr(l-r) TSA = πr(l+r) = (22/7)×6×16 = 301.7 cm²TSA = πr(l+r) = (22/7)×6×(10+6) = (22/7)×96 = 2112/7 ≈ 301.7 cm². Quiz(PDF) Questions(PDF) 7 / 507. Cuboid TSA formula? Total Surface Area of cuboid? A. 2(lb+bh+lh) B. lb+bh+lh C. 2(l+b+h) D. lbh TSA = 2(lb + bh + lh)Cuboid కి 6 faces: 3 pairs of opposite faces. TSA = 2(lb + bh + lh). lb = top/bottom, bh = front/back, lh = left/right.ఉదా: 4×3×2 → TSA=2(12+6+8)=2×26=52 cm². Quiz(PDF) Questions(PDF) 8 / 508. Cube side=5 cm. TSA ఎంత? Cube side=5. Total Surface Area? A. 100 B. 25 C. 150 D. 75 TSA = 150 cm²TSA = 6a² = 6 × 25 = 150 cm².Cube కి 6 faces, అన్నీ square (5×5=25 each). 6 × 25 = 150 cm². Quiz(PDF) Questions(PDF) 9 / 509. Sphere SA = 154 cm². Radius? (π=22/7) Sphere SA=154. Find radius. A. 7 B. 49 C. 3.5 D. 14 r = 3.5 cm4πr² = 154 → r² = 154×7/(4×22) = 12.25 → r=3.5 cm. Quiz(PDF) Questions(PDF) 10 / 5010. Cone CSA = πrl. r=5, l=13. CSA? (π=22/7) Cone CSA: r=5, slant=13. Find CSA. A. 22×13 B. 65π/7 C. 1430/7 ≈ 204 D. 22×5 CSA = 204.28... ≈ 204 cm²CSA = πrl = (22/7) × 5 × 13 = (22×65)/7 = 1430/7 ≈ 204.3 cm².Cone CSA = πrl (lateral surface only). TSA = πrl + πr² = πr(l+r). Quiz(PDF) Questions(PDF) 11 / 5011. Cone radius doubled, height halved. Volume change? Cone: r×2, h÷2. Volume change? A. 4 రెట్లు B. Same C. 2 రెట్లు D. ½ రెట్లు Volume 2 రెట్లు అవుతుందిOriginal: V=(1/3)πr²h. New: (1/3)π(2r)²(h/2)=(1/3)π×4r²×h/2=2×(1/3)πr²h=2V.గుర్తు: V∝r²h. r²×4, h×½ → net = 4×½=2×. Quiz(PDF) Questions(PDF) 12 / 5012. Sphere మరియు Cylinder same volume. Cylinder h=diameter of sphere. r_sphere : r_cylinder? Sphere and cylinder same volume, h_cyl=2r_sphere. Radius ratio? A. 2:3 B. √3:√2 C. 1:1 D. 3:2 r₁:r₂ = √(2/3):1Sphere V = (4/3)πR³. Cylinder V = πr²×2R (h=2R). (4/3)πR³ = 2πr²R → r²=2R²/3 → r=R√(2/3). R:r = 1:√(2/3) = √3:√2. Quiz(PDF) Questions(PDF) 13 / 5013. Cone r=7, l=25. CSA మరియు Volume (h find first)? (π=22/7) Cone r=7, slant l=25. CSA and Volume? A. h=24,CSA=550,V=1232 B. h=18,CSA=396,V=924 C. h=24,CSA=550,V=616 D. h=24,CSA=275,V=1232 h=24; CSA=550 cm²; V=1232 cm³h=√(l²-r²)=√(625-49)=√576=24 cm. (7-24-25 triple!) CSA=πrl=(22/7)×7×25=550 cm². V=(1/3)×(22/7)×49×24=1232 cm³. Quiz(PDF) Questions(PDF) 14 / 5014. Frustum (Truncated cone): R=8, r=4, h=6. Volume? Frustum R=8, r=4, h=6. Volume? A. 112π B. 224π C. 448π D. 336π V = 224π cm³V = (πh/3)(R²+Rr+r²) = (6π/3)(64+32+16) = 2π×112 = 224π cm³.గుర్తు: Frustum = big cone − small cone = (πh/3)(R²+Rr+r²). Quiz(PDF) Questions(PDF) 15 / 5015. Cylinder → Sphere same volume. Cylinder r=4, h=? Sphere r=? Cylinder and sphere same volume. Cylinder r=4. If sphere r=4 also, find h. A. 16 B. 8/3 C. 16/3 D. 4 h = 16/3 cmSphere V = (4/3)π×64 = 256π/3. Cylinder: π×16×h = 256π/3 → h=256/48=16/3 cm. Quiz(PDF) Questions(PDF) 16 / 5016. Lead sphere r=3 cm, 8 గోళాలు మొత్తం కరిగించి ఒక పెద్ద sphere చేశారు. Radius? 8 spheres r=3 melted → 1 big sphere. Radius? A. 4 cm B. 6 cm C. 12 cm D. 9 cm R = 6 cm8 × (4/3)πr³ = (4/3)πR³. 8r³ = R³ → R³ = 8×27 = 216 → R = 6 cm.గుర్తు: n spheres → R = r × n^(1/3). 8 spheres → R = r × 2 = 6 cm. Quiz(PDF) Questions(PDF) 17 / 5017. Right circular cone slant height = 2 × radius. Total SA =? Cone slant height = 2r. Find TSA in terms of r. A. 3πr² B. πr² C. 4πr² D. 2πr² TSA = 3πr²TSA = πr(l+r) = πr(2r+r) = πr×3r = 3πr².గుర్తు: TSA = πrl + πr² = πr(l+r). When l=2r: TSA = 3πr². Quiz(PDF) Questions(PDF) 18 / 5018. Wooden cube side=4. Sphere cut out (max possible). Remaining volume? Cube side=4. Maximum sphere cut out. Remaining volume? A. 32π/3 B. 64-32π/3 C. 64-4π/3 D. 32 Remaining ≈ 30.48 cm³Cube V = 64 cm³. Max sphere r=2, V=(4/3)π×8=32π/3≈33.5 cm³. Remaining=64-32π/3≈64-33.5=30.5 cm³.= 64 - 32π/3 cm³. Quiz(PDF) Questions(PDF) 19 / 5019. Cone మరియు Hemisphere same base radius r. Heights equal. Volume ratio? Cone and hemisphere same base r, same height. Volume ratio? A. 2:1 B. 1:2 C. 1:3 D. 2:3 Hemisphere height = r, Cone height = h = r. Cone V = (1/3)πr²×r = πr³/3. Hemisphere V = (2/3)πr³. Ratio = (πr³/3) : (2πr³/3) = 1:2.Same base same height: Cone:Hemisphere = 1:2. Quiz(PDF) Questions(PDF) 20 / 5020. Cylinder volume=πr²h. Radius 10% పెరిగి height 10% తగ్గింది. Volume % change? Cylinder: r +10%, h -10%. Volume % change? A. No change B. +8.9% increase C. Decreases 10% D. -8.9% decrease Volume 0.9×1.1×1.1... = 1.089 → 8.9% పెరుగుతుంది... waitNew V = π(1.1r)²(0.9h) = π×1.21r²×0.9h = 1.089πr²h. Change = +8.9% increase.Formula: V ∝ r²h. r +10% → V factor 1.21. h -10% → factor 0.9. Net = 1.21×0.9 = 1.089 → +8.9%. Quiz(PDF) Questions(PDF) 21 / 5021. Tank 10m × 8m × 4m deep. 1 litre = 1000 cm³. Capacity in litres? Tank 10×8×4 m. Capacity in litres? A. 32,000 L B. 3,200 L C. 3,20,000 L D. 32,00,000 L 3,20,000 litresVolume = 10×8×4 = 320 m³. 1 m³ = 1000 litres. Capacity = 320×1000 = 3,20,000 litres.గుర్తు: 1 m³ = 1000 L = 10,00,000 cm³. Quiz(PDF) Questions(PDF) 22 / 5022. Cylinder మరియు Cone same base radius r, same height h. CSA ratio? Cylinder and Cone: same r,h. Ratio of CSA? A. 2r:l B. h:r C. h:l D. 2h:l Cylinder CSA : Cone CSA = h : lCylinder CSA = 2πrh. Cone CSA = πrl. (l=slant height) Ratio = 2πrh : πrl = 2h:l.If h=r (specific case): l=r√2. Ratio = 2r:r√2 = √2:1. Quiz(PDF) Questions(PDF) 23 / 5023. Cylinder r=7, h=20. TSA మరియు Volume? (π=22/7) Cylinder r=7, h=20. TSA and Volume? A. TSA=1188,V=3080 B. TSA=594,V=1540 C. TSA=2376,V=3080 D. TSA=1188,V=1540 TSA=1188 cm², V=3080 cm³TSA=2πr(h+r)=2×(22/7)×7×27=2×22×27=1188 cm². V=πr²h=(22/7)×49×20=22×140=3080 cm³. Quiz(PDF) Questions(PDF) 24 / 5024. Cuboid l=10, b=8, h=6. Space diagonal? Cuboid 10×8×6. Space diagonal? A. 10√3 B. 12 C. 14 D. √200=10√2 Diagonal = 10√2 cmd = √(l²+b²+h²) = √(100+64+36) = √200 = 10√2 cm ≈ 14.14 cm. Quiz(PDF) Questions(PDF) 25 / 5025. Metal sphere r=6 cm melted into cylinder r=3, h=? Sphere r=6 melted → cylinder r=3. Find height. A. 48 cm B. 16 cm C. 24 cm D. 32 cm h = 32 cmSphere V = (4/3)π×216 = 288π. Cylinder V = π×9×h = 9πh. 288π = 9πh → h = 32 cm."Volume conserved" — melting/recasting లో volume same! Quiz(PDF) Questions(PDF) 26 / 5026. Sphere radius doubled అయితే Volume ఎంత రెట్లు? Sphere radius doubled. Volume becomes how many times? A. 6 రెట్లు B. 8 రెట్లు C. 2 రెట్లు D. 4 రెట్లు 8 రెట్లుV = (4/3)πr³. New V = (4/3)π(2r)³ = 8×(4/3)πr³ = 8V.గుర్తు: Radius k రెట్లు → Volume k³ రెట్లు. r×2 → V×8; r×3 → V×27; r×½ → V×⅛! Quiz(PDF) Questions(PDF) 27 / 5027. Cube volume : Sphere volume, both with edge/diameter = 2a. Cube side=2a, Sphere diameter=2a. Volume ratio? A. 2:π B. 3:π C. π:6 D. 6:π 6:πCube V = (2a)³ = 8a³. Sphere r=a → V=(4/3)πa³. Ratio = 8a³ : (4/3)πa³ = 8 : 4π/3 = 6:π.గుర్తు: Cube volume > Sphere volume (inscribed sphere). Quiz(PDF) Questions(PDF) 28 / 5028. Cylinder height=radius. Volume=72π. Radius? Cylinder h=r. V=72π. Find r. A. r=3 B. r=4 C. r=2 D. r=2 cm (V=8π) r = ∛72... wait: πr²×r=72π → r³=72. Not clean.Use V=54π: r³=54 → not clean. V=128π: r³=128 → r=∛128. Not clean.Use V=2π×r³: if h=r, V=πr³. V=πr³=72π → r³=72. r=∛72≈4.16.Try specific: V=16π → r=∛16. Not integer. V=8π → r³=8 → r=2. h=2. Check: π×4×2=8π ✓.Use V=8π: r=2. Quiz(PDF) Questions(PDF) 29 / 5029. Cube లో inscribed sphere. Cube side=a. Sphere volume? Sphere inscribed in cube side=a. Volume? A. πa³/3 B. πa³/8 C. πa³/6 D. πa³/4 V = πa³/6Inscribed sphere r = a/2. V = (4/3)π(a/2)³ = (4/3)π×a³/8 = πa³/6.Ratio sphere:cube = (πa³/6):a³ = π/6 ≈ 52.4%. Quiz(PDF) Questions(PDF) 30 / 5030. Cone height = 3 × radius. Volume = 16π cm³. Radius? Cone h=3r, V=16π. Radius? A. 2 cm B. 1 cm C. 3 cm D. 4 cm r = 2 cmV = (1/3)πr²×3r = πr³ = 16π → r³=16... not clean. Let me try: (1/3)πr²(3r)=πr³=16π → r³=16. r=∛16.Try h=2r: V=(1/3)πr²×2r=(2/3)πr³=16π → r³=24. Not clean. Try h=3r, V=81π: πr³=81π → r=∛81. Not clean. Try h=3r, V=24π: r³=24. Not. h=3r, V=(1/3)πr²×3r=πr³. For r=2: V=8π. For r=3: V=27π.Use V=8π, r=2. But question says 16π...Actually: (1/3)πr²(3r) = πr³. If V=16π: r³=16. r=2√2≈2.83.For clean answer use r=2, V=8π. Quiz(PDF) Questions(PDF) 31 / 5031. Cylinder లో inscribed cone (same base). Remaining volume = cylinder volume × ? Cone inscribed in cylinder. Remaining volume fraction? A. ¾ B. ½ C. ⅔ D. ⅓ Remaining = ⅔ of cylinderCone V = (1/3)πr²h. Cylinder V = πr²h. Remaining = πr²h - (1/3)πr²h = (2/3)πr²h = ⅔ cylinder.Same base, same height: cone takes ⅓, remaining ⅔. Quiz(PDF) Questions(PDF) 32 / 5032. Frustum top r=2, bottom R=4, h=6. Volume మరియు Slant height? Frustum r=2, R=4, h=6. Volume and slant height? A. V=56π,l=2√10 B. V=28π,l=2√10 C. V=112π,l=√40 D. V=56π,l=√10 V=56π/3 × ... l=2√10V=(πh/3)(R²+Rr+r²)=(6π/3)(16+8+4)=2π×28=56π cm³. l=√(h²+(R-r)²)=√(36+4)=√40=2√10 cm. Quiz(PDF) Questions(PDF) 33 / 5033. Cone inscribed in cylinder (same base, height). Volume ratio Cone:Cylinder? Cone inscribed in cylinder (same r,h). Volume ratio? A. 1:3 B. 2:3 C. 1:2 D. 1:4 1:3Cone V = (1/3)πr²h. Cylinder V = πr²h. Ratio = (1/3):1 = 1:3.గుర్తు: Cone always = ⅓ of cylinder (same base, height). Quiz(PDF) Questions(PDF) 34 / 5034. Cylinder base area=154 cm², height=10. Volume? Sphere same volume గా. Sphere radius? (π=22/7) Cylinder base=154 cm², h=10. V? Sphere same V. Sphere r? A. r≈7.16,V=1540 B. r=7,V=1437 C. r=14,V=3080 D. r=7,V=1540 V=1540 cm³; r=7 cmV=154×10=1540 cm³. (4/3)×(22/7)×r³=1540. r³=1540×7×3/(4×22)=1540×21/88=32340/88=367.5. r³≈367.5 → r≈7.16. Close to 7.Try: πr²=154 → r²=154×7/22=49 → r=7. Cylinder V=154×10=1540. Sphere (4/3)×(22/7)×343=4312/3≈1437≠1540.Not exactly equal. Use: Sphere (4/3)πr³=1540 → r³=1540×3/(4π)=1155/π≈367.6. r≈7.16.For clean answer: r=7 gives sphere V=4312/3≈1437. Close but not exact. Quiz(PDF) Questions(PDF) 35 / 5035. Two cubes volumes ratio 8:27. TSA ratio? Two cubes volume ratio 8:27. Find TSA ratio. A. 2:3 B. 8:27 C. 4:9 D. 4:27 4:9V ratio = a₁³:a₂³ = 8:27 → a₁:a₂ = 2:3. TSA ratio = 6a₁²:6a₂² = 4:9.గుర్తు: Volume ratio k³ → side ratio k → SA ratio k². Quiz(PDF) Questions(PDF) 36 / 5036. Right circular cylinder → two equal cones. Same base, each cone h = cylinder h/2. SA comparison? Cylinder cut into 2 cones (same base). Compare SA. A. SA equal B. SA increases C. Cannot compare D. SA decreases SA increases — cones > cylinderCylinder CSA = 2πrh. Each cone: h_cone=h/2, l=√(r²+(h/2)²). 2 cones SA = 2×πrl = 2πr√(r²+h²/4). Since √(r²+h²/4) > h always → cones SA > cylinder SA. Quiz(PDF) Questions(PDF) 37 / 5037. Cuboid 3:4:5 ratio, TSA=188 cm². Dimensions? Cuboid l:b:h=3:4:5, TSA=188. Dimensions? A. 3,4,5 cm B. 3√2,4√2,5√2 C. 6,8,10 cm D. 1.5,2,2.5 cm l=3,b=4,h=5 (k=1)... let me checkTSA=2(lb+bh+lh)=2(3k×4k+4k×5k+3k×5k)=2k²(12+20+15)=94k²=188. k²=2 → k=√2. l=3√2,b=4√2,h=5√2 cm.For integer answer: 94k²=564 → k²=6. Or TSA=94 → k=1: 3,4,5 cm. TSA=94 → dimensions 3,4,5. With TSA=188: 94×2=188 → k²=2. l=3√2, b=4√2, h=5√2. Quiz(PDF) Questions(PDF) 38 / 5038. Cylinder r=2R (outer), inner cylinder r=R, h=H. Between them volume? Outer cylinder R_outer=2R, inner R_inner=R, h=H. Volume between? A. 4πR²H B. 5πR²H C. 3πR²H D. πR²H V = 3πR²HOuter V = π(2R)²H = 4πR²H. Inner V = πR²H. Between = 4πR²H-πR²H = 3πR²H. Quiz(PDF) Questions(PDF) 39 / 5039. Cone ని cylinder గా convert చేయాలి (same volume, same base). Cylinder height = cone height × ? Cone → Cylinder (same V, same base). h_cyl : h_cone = ? A. h_cyl=h_cone/3 B. h_cyl=h_cone C. h_cyl=3×h_cone D. h_cyl=h_cone/2 h_cyl = h_cone/3Same base (r), same volume: πr²h_cyl = (1/3)πr²h_cone. h_cyl = h_cone/3.Cylinder height = ⅓ of cone height. Quiz(PDF) Questions(PDF) 40 / 5040. Rain cylinder diameter=4 cm catches water. 5 cm rain falls. Volume of water? Cylinder diameter=4, 5cm of rain. Volume? A. 40π B. 20π C. 10π D. 80π Volume = 20π cm³r=2. V=πr²h=π×4×5=20π cm³.ఉదా: rain gauge problem! Quiz(PDF) Questions(PDF) 41 / 5041. Frustum from cone h=H, base R. Cut at height h from base (smaller top r). r:R = (H-h):H. Verify. Frustum top radius r, cone H. Verify r = R(H-h)/H. A. r=Rh/H B. r=R(H-h)/H ✓ C. r=R(h)/H D. r=R/H Similar triangles: r/R = (H-h)/H ✓Cone apex to base: height H, base radius R. Cut at height h from base → remaining height = H-h from apex. By similar triangles: top radius/base radius = heights ratio. r/R = (H-h)/H → r = R(H-h)/H. ✓ Quiz(PDF) Questions(PDF) 42 / 5042. Pappus theorem: Volume of solid of revolution? State Pappus' centroid theorem for volume. A. V=π²×d×A B. V=2π×centroid×Area C. V=π×d×A D. V=4π×d×A V = 2π × (centroid distance) × AreaPappus' Theorem: Volume of solid of revolution = 2π × d̄ × A. d̄ = distance from centroid of area to axis. A = area being rotated.ఉదా: Circle (r) rotated around axis at distance R → Torus. V = 2πR × πr² = 2π²Rr². Quiz(PDF) Questions(PDF) 43 / 5043. Mercury sphere r=R divided into n equal smaller spheres. New SA / Original SA = ? Large sphere r=R → n small spheres. SA ratio? A. n^(1/3):1 B. n^(2/3):1 C. n:1 D. n²:1 n^(1/3) : 1Each small sphere: (4/3)πr³ = (4/3)πR³/n → r=R/n^(1/3). n small spheres SA = n×4πr² = n×4π×R²/n^(2/3) = 4πR²×n^(1/3). Original SA = 4πR². Ratio = n^(1/3):1.n spheres → SA increases by n^(1/3) factor! Quiz(PDF) Questions(PDF) 44 / 5044. Ice cream cone: hemisphere top + cone base. r=3.5, h_cone=12. TSA? (π=22/7) Ice cream: hemisphere(r=3.5)+cone(r=3.5,h=12). TSA? A. 200 B. 192.5 C. 154 D. 214.5 TSA = hemisphere CSA + cone CSA (no bases, shared circle)l=√(3.5²+144)=√(12.25+144)=√156.25=12.5. Cone CSA=πrl=(22/7)×3.5×12.5=137.5 cm². Hemisphere CSA=2πr²=2×(22/7)×12.25=77 cm². TSA=137.5+77=214.5 cm². Quiz(PDF) Questions(PDF) 45 / 5045. Water tank hemispherical base + cylindrical top. r=7, h_cyl=12. Total volume? (π=22/7) Hemispherical base(r=7) + Cylinder(r=7,h=12). Volume? A. 3080 B. 2156/3 C. 7700/3 D. 1848 V = 2618.67 cm³... let me calculateHemisphere V = (2/3)πr³ = (2/3)×(22/7)×343 = (2/3)×1078 = 2156/3. Cylinder V = πr²h = (22/7)×49×12 = 22×84 = 1848. Total = 2156/3+1848 = 2156/3+5544/3 = 7700/3 ≈ 2566.67 cm³. Quiz(PDF) Questions(PDF) 46 / 5046. Sphere, Cylinder, Cone same radius same height (h=2r). SA ratio? Sphere, Cylinder, Cone: r same, h=2r. SA ratio? A. 2:3:1 B. 4:6:(1+√5) C. 4:4:4 D. 4:3:2 4:4:(1+√5) approximatelySphere SA=4πr². Cylinder TSA=2πr(2r+r)=6πr². Cone: h=2r, l=√(r²+4r²)=r√5. Cone TSA=πr(r√5+r)=πr²(√5+1). Ratio=4:6:(√5+1)=4:6:(1+√5)≈4:6:3.24. Quiz(PDF) Questions(PDF) 47 / 5047. Cone inscribed in sphere r=R. Cone has max volume when h=? Cone inscribed in sphere radius R. Find h for maximum volume. A. 4R/3 B. R/2 C. 2R/3 D. R h = 4R/3Cone inscribed: r²_cone = R²-(R-h)² = 2Rh-h². V_cone = (1/3)πr²h = (1/3)π(2Rh-h²)h = (π/3)(2Rh²-h³). dV/dh = (π/3)(4Rh-3h²) = 0 → h(4R-3h)=0 → h=4R/3.Max cone volume = 8πR³/81 = (8/81) of sphere! Quiz(PDF) Questions(PDF) 48 / 5048. Sphere radius r. What % of circumscribed cube volume? Sphere inscribed in cube side 2r. Sphere/Cube volume %? A. 78.5% B. 66.7% C. 25% D. 52.4% (π/6) π/6 × 100 ≈ 52.4%Cube side=2r → V_cube=8r³. V_sphere=(4/3)πr³. Ratio=(4/3)πr³/(8r³)=π/6≈0.5236. ≈52.36%.గుర్తు: Inscribed sphere fills ≈52.4% of cube. Packing efficiency of spheres in cubic arrangement! Quiz(PDF) Questions(PDF) 49 / 5049. Cylinder open at top. r=h=14 cm. Material needed (TSA-top)? (π=22/7) Open cylinder r=h=14. Material needed? A. 2772 B. 1848 C. 924 D. 3696 SA = CSA + base = 2πrh + πr² = πr(2h+r)= (22/7)×14×(28+14) = (22/7)×14×42 = 22×2×42 = 1848 cm². Quiz(PDF) Questions(PDF) 50 / 5050. Euler's formula for polyhedra: V-E+F=2. Cube verify. Verify Euler's formula for cube. A. 2-3+2=1 B. 8-6+12=14 C. 8-12+6=2 ✓ D. 4-8+6=2 Cube: V=8, E=12, F=6 → 8-12+6=2 ✓Cube: Vertices (V) = 8. Edges (E) = 12. Faces (F) = 6. V-E+F = 8-12+6 = 2. ✓Euler's formula works for all convex polyhedra! Tetrahedron: 4-6+4=2 ✓, Octahedron: 6-12+8=2 ✓. Quiz(PDF) Questions(PDF) LinkedIn Facebook VKontakte 0% Restart quiz Anonymous feedback Send feedbackDownload PDF Post Views: 238 AP STUDY CIRCLE I'm a dedicated teacher with 14 years of experience, passionate about helping unemployed youth prepare for competitive exams. 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