Q1. Volume of a cube with side \(3\) cm?

Ans: \(V=3^3=27\) cc.

Q2. Volume formula of a cuboid?

Ans: \(V=lbh.\)

Q3. Units for volume of liquid commonly used?

Ans: litre (L) and millilitre (ml); \(1\ \text{L}=1000\) ml.

Q4. 1 cc equals how many ml?

Ans: \(1\ \text{cc}=1\ \text{ml}.\)

Q5. Volume of cuboid \(l=10\), \(b=5\), \(h=2\) (cm)?

Ans: \(V=10\times5\times2=100\) cc.

Q6. c.s.a. of a cylinder with \(r=7\) cm, \(h=10\) cm (use \(\pi=\tfrac{22}{7}\))?

Ans: \(2\pi rh=2\times\tfrac{22}{7}\times7\times10=440\) sq.cm.

Q7. t.s.a. of cube side \(4\) cm?

Ans: \(6a^2=6\times4^2=96\) sq.cm.

Q8. Volume of cube side \(2\) cm?

Ans: \(2^3=8\) cc.

Q9. If \(V=1000\) cc, liters = ?

Ans: \(1000\ \text{cc}=1\) L.

Q10. t.s.a. of cylinder with \(r=0.5\) m, \(h=2\) m (use \(\pi=3.14\))?

Ans: \(2\pi r(h+r)=2\times3.14\times0.5(2+0.5)=7.85\) m\(^2\).

Q11. How many cc in \(5\) litres?

Ans: \(5\times1000=5000\) cc.

Q12. Volume of cuboid \(l=20\), \(b=10.5\), \(h=8\) cm (from Practice Set)?

Ans: \(V=20\times10.5\times8=1680\) cc.

Q13. Number of \(1\)-cm cubes in \(2\times2\times2\) cube?

Ans: \(2^3=8\) cubes.

Q14. Volume formula of cylinder?

Ans: \(V=\pi r^2 h.\)

Q15. If radius of cylinder is \(5\) cm and height \(10\) cm, volume (use \(\pi=3.14\))?

Ans: \(V=3.14\times5^2\times10=785\) cc.

Q16. Relationship \(1\) m \(=\) ? cm used for converting volumes.

Ans: \(1\) m \(=100\) cm so convert before computing volume in cc.

Q17. c.s.a. of cylinder with \(r=3\), \(h=7\) (use \(\pi=\tfrac{22}{7}\))?

Ans: \(2\pi rh=2\times\tfrac{22}{7}\times3\times7=2\times22\times3=132\) sq.cm.

Q18. A cuboid measures \(1\) m \(\times\) 40 cm \(\times\) 50 cm — capacity in litres?

Ans: convert \(1\) m = 100 cm, \(V=100\times40\times50=200000\) cc = \(200\) L.

Q19. A cube of side \(10\) cm has volume = ? (in litres)

Ans: \(V=10^3=1000\) cc = \(1\) L.

Q20. Number of \(40\) cm cube boxes in a \(6\) m × \(4\) m × \(4\) m warehouse?

Ans: Warehouse \(=600\times400\times400\) cc; box \(=40^3\) cc → count \(= (600/40)\times(400/40)\times(400/40)=15\times10\times10=1500\).

Q1. Find volume of box \(20\times10.5\times8\) cm.

Ans: \(V=20\times10.5\times8=1680\) cc.

Q2. A cuboid soap bar has volume \(150\) cc, length \(10\) cm and breadth \(5\) cm. Find thickness.

Ans: \(h=V/(lb)=150/(10\times5)=3\) cm.

Q3. How many bricks \(25\times15\times10\) cm to build wall \(6\) m × \(2.5\) m × \(0.5\) m?

Ans: Convert wall to cm: \(600\times250\times50\) cc; brick \(=25\times15\times10\) cc. Number \(= (600\times250\times50)/(25\times15\times10)= (600/25)\times(250/15)\times(50/10)=24\times(16.\overline{6})\times5\). Exact integer: compute volumes: wall \(=7,500,000\) cc, brick \(=3750\) cc → \(7,500,000/3750=2000\) bricks.

Q4. Rainwater tank \(10\) m × \(6\) m × \(3\) m capacity in litres?

Ans: \(V=10\times6\times3=180\) m\(^3\) = \(180\times1000=180000\) L.

Q5. c.s.a. and t.s.a. of cylinder \(r=7\) cm \(h=10\) cm?

Ans: c.s.a. \(=2\pi rh=2\times\tfrac{22}{7}\times7\times10=440\) cm\(^2\); t.s.a. \(=2\pi r(h+r)=2\times\tfrac{22}{7}\times7(10+7)=2\times22\times17=748\) cm\(^2\).

Q6. A closed cylinder of diameter \(50\) cm and height \(45\) cm: find t.s.a. (\(\pi=3.14\)).

Ans: \(r=25\) cm; \(t.s.a.=2\pi r(h+r)=2\times3.14\times25(45+25)=157\times70=10990\) cm\(^2\) (compute \(2\times3.14\times25=157\)).

Q7. Curve area given \(660\) cm\(^2\) and \(h=21\) cm. Find base area and radius?

Ans: c.s.a. \(=2\pi rh=660\Rightarrow r=660/(2\pi h)=660/(2\times\tfrac{22}{7}\times21)=660/(2\times22\times3)=660/132=5\) cm. Area of base \(=\pi r^2=\tfrac{22}{7}\times5^2= \tfrac{22}{7}\times25=78.57\) approx.

Q8. Sheet \(3.3\) m × \(3\) m → how many pipes length \(30\) cm radius \(3.5\) cm? (textbook)

Ans: Sheet area \(=330\times300=99,000\) cm\(^2\). Curved area per pipe \(=2\pi r h=2\times\tfrac{22}{7}\times3.5\times30=660\) cm\(^2\). Number \(=99000/660=150\).

Q9. Volume of cylinder \(r=5\) cm \(h=10\) cm?

Ans: \(V=\pi r^2h=3.14\times25\times10=785\) cc.

Q10. Drum capacity \(70.4\) L and \(h=56\) cm → find radius (use \(\pi=\tfrac{22}{7}\)).

Ans: \(V=70.4\times1000=70400\) cc. \(r^2=V/(\pi h)=70400/( \tfrac{22}{7}\times56)=70400/(22\times8)=70400/176=400\Rightarrow r=20\) cm.

Q11. How many discs (d=1.4, t=0.2) from cylinder (R=4.2,H=16)?

Ans: Volume big \(=\pi(4.2)^2\times16\), small \(=\pi(0.7)^2\times0.2\). Ratio cancels \(\pi\): \(n=(4.2^2\times16)/(0.7^2\times0.2)=(17.64\times16)/(0.49\times0.2)=282.24/0.098=2880\).

Q12. How much iron for rod length \(90\) cm diameter \(1.4\) cm?

Ans: radius \(=0.7\) cm, volume \(=\pi r^2 h=\pi\times0.7^2\times90\approx3.1416\times0.49\times90\approx138.5\) cc.

Q13. Tank interior diameter \(1.6\) m and depth \(0.7\) m — water capacity?

Ans: \(r=0.8\) m, \(V=\pi r^2 h=3.1416\times0.8^2\times0.7\approx1.407\) m\(^3\)=1407 L (approx).

Q14. Cylinder circumference \(132\) cm, \(h=25\) cm → volume?

Ans: \(2\pi r=132\Rightarrow r=132/(2\pi)=132/(2\times3.1416)\approx21\). Use exact: if \(\pi=\tfrac{22}{7}\), \(r=132/(44/7)=21\). \(V=\pi r^2 h=\tfrac{22}{7}\times21^2\times25=22\times63\times25=34650\) cc.

Q15. c.s.a. for \(r=2.5\) cm, \(h=7\) cm?

Ans: \(2\pi rh=2\times3.14\times2.5\times7=109.9\) sq.cm.

Q16. t.s.a. of cylinder closed (r=4.2,h=14) with \(\pi=\tfrac{22}{7}\)?

Ans: \(t.s.a.=2\pi r(h+r)=2\times\frac{22}{7}\times4.2(14+4.2)=2\times22\times0.6\times18.2\) — compute numeric ≈ \(2\times22\times0.6\times18.2\approx480.96\) sq.cm.

Q17. If curved surface area is 660 cm² and \(h=21\) cm, find \(r\).

Ans: \(2\pi rh=660\Rightarrow r=660/(2\pi h)=660/(2\times\frac{22}{7}\times21)=5\) cm.

Q18. Area required to make open cylinder diameter 28 cm and height 20 cm?

Ans: sheet area = curved area \(= \pi d h = \pi\times28\times20\). With \(\pi\approx3.14\) → \(=3.14\times560\approx1758.4\) sq.cm. Lid of height 2 cm approx area \(= \pi[(r+2)^2-r^2]=\pi( (14+2)^2-14^2)=\pi(256-196)=\pi\times60\approx188.4\) sq.cm.

Q19. How many pipes from sheet (another route)?

Ans: same as Q8 → 150 pipes.

Q20. Quick: cylinder with r=10.5 cm h=8 cm volume?

Ans: \(V=\pi r^2 h=\tfrac{22}{7}\times10.5^2\times8= \tfrac{22}{7}\times110.25\times8=22\times15.75\times8=2772\) cc.

Q1. A fish tank: \(1\) m × \(40\) cm × \(50\) cm. How many litres?

Ans: convert \(1\) m = 100 cm → \(V=100\times40\times50=200000\) cc \(=200\) L.

Q2. Warehouse \(6\) m×\(4\) m×\(4\) m; boxes side \(40\) cm. How many?

Ans: warehouse \(=600\times400\times400\) cc = \(96,000,000\) cc? (check) — easier: along dimensions: \(600/40=15,\;400/40=10,\;400/40=10\). Count \(=15\times10\times10=1500\) boxes.

Q3. Tray holds \(5\) L (\(=5000\) cc), breadth \(40\) cm and height \(2.5\) cm. Find length.

Ans: \(l\times40\times2.5=5000\Rightarrow l=5000/(40\times2.5)=5000/100=50\) cm.

Q4. c.s.a. & t.s.a. of cylinder \(r=7\), \(h=10\) done earlier — show steps.

Ans: c.s.a. \(=2\pi rh=2\times\frac{22}{7}\times7\times10=2\times22\times10=440\) cm². t.s.a. \(=440+2\pi r^2=440+2\times\frac{22}{7}\times49=440+308=748\) cm².

Q5. Paint both sides (inside & outside) of closed cylinder r=0.5 m h=2 m at ₹80 per m² — cost?

Ans: t.s.a. \(=2\pi r(h+r)=7.85\) m²; both sides painted means twice t.s.a. area to paint? The problem in textbook painted both internal and external including lid → area to paint \(=2\times t.s.a.=15.7\) m². Cost = \(15.7\times80=₹1256\).

Q6. Sheet \(3.3\) m × \(3\) m → pipes problem (worked): show steps.

Ans: sheet area \(=3.3\times3=9.9\) m² \(=9900\) cm²? careful: 3.3 m = 330 cm, 3 m =300 cm ⇒ sheet = 99000 cm². Per pipe curved area \(=2\pi rh=2\times\tfrac{22}{7}\times3.5\times30=660\) cm². Count \(=99000/660=150\).

Q7. Cylinder volume for r=5 h=10 — show full steps.

Ans: \(V=\pi r^2 h=3.14\times25\times10=785\) cc.

Q8. Drum radius found from capacity 70.4 L and height 56 cm — show algebra.

Ans: \(V=70400=\pi r^2\times56\Rightarrow r^2=70400/(\pi\times56)\). With \(\pi=\tfrac{22}{7}\), \(r^2=70400/( \tfrac{22}{7}\times56)=70400/(22\times8)=70400/176=400\Rightarrow r=20\) cm.

Q9. Cylinder to discs problem — algebraic cancellation of \(\pi\).

Ans: \(n=\dfrac{\pi R^2 H}{\pi r^2 h}=\dfrac{R^2 H}{r^2 h}\). Substitute \(R=4.2,H=16,r=0.7,h=0.2\): \(n=(4.2^2\times16)/(0.7^2\times0.2)=2880\).

Q10. Cylinder with circumference 132 and height 25; find volume using exact numbers.

Ans: \(2\pi r=132\Rightarrow r=132/(2\pi)\). With \(\pi=\tfrac{22}{7}\), \(r=21\). Then \(V=\pi r^2 h=\tfrac{22}{7}\times21^2\times25=34650\) cc.

Q11. Convert \(1\) m³ to litres.

Ans: \(1\) m³ = \(100\) cm × \(100\) cm × \(100\) cm = \(1,000,000\) cc = \(1000\) L.

Q12. If c.s.a.=660 and \(h=21\), show base area.

Ans: \(r=5\) cm → base area \(=\pi r^2=\tfrac{22}{7}\times25=78.57\) cm² approx.

Q13. Cylinder open at one side diameter 28 height 20: area of sheet required and lid height 2 (approx)?

Ans: sheet area \(=\pi d h=\pi\times28\times20\approx1758.4\) cm². Lid (rim) approx area \(\approx\pi((14+2)^2-14^2)=\pi\times60\approx188.4\) cm².

Q14. Explain using unit cubes how \(V=lbh\) for cuboid.

Ans: Fill cuboid with \(1\)-cm³ cubes: along length \(l\) you place \(l\) cubes, breadth \(b\) and height \(h\) layers → total \(=l\times b\times h\).

Q15. A cube side \(10\) cm → t.s.a. and volume?

Ans: t.s.a. \(=6\times10^2=600\) cm²; \(V=10^3=1000\) cc \(=1\) L.

Q16. A cylindrical tank painted inside & outside including lid: why area doubled?

Ans: painting both inner and outer surfaces doubles the area that must be covered compared to painting only outer surface (for closed tank t.s.a. already counts both circular faces and curved surface once; painting both sides means painting both inside and outside surfaces, so double).

Q17. How many rods of length 90 cm and dia 1.4 can be produced from iron of volume X? (method)

Ans: compute volume of single rod = \(\pi(0.7)^2\times90\). Divide available iron volume by rod volume → integer count.

Q18. Prove c.s.a. of cylinder equals area of rectangle with sides circumference and height.

Ans: Unwrap curved surface → rectangle with length equal to circumference \(2\pi r\) and breadth \(h\) → area \(=2\pi r h.\)

Q19. If a cylinder and a prism have equal base area and height, what about volumes?

Ans: volumes equal because \(V=\) base area \(\times\) height for any prism/cylinder.

Q20. Euler's formula: state it and verify for cube.

Ans: \(F+V=E+2\). Cube: \(F=6,V=8,E=12\). \(6+8=14\) and \(12+2=14\) verified.

Ex.1. How many litres will a tank contain if \(l=1\) m, \(b=40\) cm, \(h=50\) cm?

Solution: convert \(1\) m = \(100\) cm. \(V=100\times40\times50=200000\) cc = \(200000/1000=200\) L.

Ex.2. Warehouse \(6\) m × \(4\) m × \(4\) m; boxes side \(40\) cm – how many boxes?

Solution: convert dimensions to cm: warehouse \(600\times400\times400\) cc. box \(=40^3=64000\) cc? (No: \(40^3=64000\)). Number \(= (600\times400\times400)/64000=(96,000,000)/64,000=1500\).

Ex.3. Tray filled with \(5\) L liquid; \(b=40\) cm, \(h=2.5\) cm → length?

Solution: \(5\) L = \(5000\) cc. \(l=5000/(40\times2.5)=5000/100=50\) cm.

Q1. Volume of box \(20\) cm × \(10.5\) cm × \(8\) cm?

Ans: \(V=20\times10.5\times8=1680\) cc.

Q2. Cuboid soap bar volume \(150\) cc, \(l=10\), \(b=5\) → thickness?

Ans: \(h=150/(10\times5)=3\) cm.

Q3. Number of bricks (brick \(25\times15\times10\) cm) for wall \(6\) m×\(2.5\) m×\(0.5\) m?

Ans: wall \(=600\times250\times50=7,500,000\) cc. brick \(=25\times15\times10=3750\) cc. number \(=7,500,000/3750=2000\) bricks.

Q4. Rainwater tank \(10\) m × \(6\) m × \(3\) m capacity (in litres)?

Ans: \(V=10\times6\times3=180\) m\(^3\)=\(180\times1000=180000\) L.

Q1(1). For \(r=7\) cm, \(h=10\) cm: c.s.a. and t.s.a.?

Ans: c.s.a. \(=2\pi rh=2\times\frac{22}{7}\times7\times10=440\) cm². t.s.a. \(=2\pi r(h+r)=748\) cm².

Q2. t.s.a. of closed drum \(d=50\) cm, \(h=45\) cm (\(\pi=3.14\)).

Ans: \(r=25\). t.s.a. \(=2\pi r(h+r)=2\times3.14\times25(45+25)=157\times70=10990\) cm².

Q3. If c.s.a.=660 cm² and \(h=21\) cm, find base area & radius.

Ans: \(2\pi rh=660\Rightarrow r=660/(2\pi h)=5\) cm. Base area \(=\pi r^2=\tfrac{22}{7}\times25\approx78.57\) cm².

Q4. Sheet for open cylinder \(d=28\) cm \(h=20\) cm (one open side) — sheet area & lid of height \(2\) cm?

Ans: sheet area (curved) \(=\pi d h=\pi\times28\times20\approx1758.4\) cm². Lid rim area ≈ \(\pi[(14+2)^2-14^2]=\pi\times60\approx188.4\) cm².

Q1(1). Volume cylinder \(r=10.5\) cm, \(h=8\) cm?

Ans: \(V=\pi r^2 h=\tfrac{22}{7}\times10.5^2\times8=2772\) cc.

Q2. Iron rod length \(90\) cm, dia \(1.4\) cm — how much iron needed?

Ans: \(r=0.7\) cm, \(V=\pi r^2 h\approx3.1416\times0.49\times90\approx138.5\) cc.

Q3. Tank interior dia \(1.6\) m and depth \(0.7\) m: water held?

Ans: \(r=0.8\) m, \(V=\pi r^2 h\approx3.1416\times0.64\times0.7\approx1.407\) m³ = \(1407\) L approx.

Q4. Cylinder circumference \(=132\) cm and \(h=25\) cm → volume?

Ans: with \(\pi=\tfrac{22}{7}\), \(r=21\), \(V=\tfrac{22}{7}\times21^2\times25=34650\) cc.

Q: State Euler's formula and verify for cube, cuboid, triangular prism, triangular pyramid, pentagonal pyramid, hexagonal prism.

Ans: Euler: \(F+V=E+2\). Example cube: \(F=6,V=8,E=12\) → \(6+8=14\) and \(12+2=14\). (Verify each solid similarly by counting faces, vertices and edges.)