1. Problem: If volume and surface area of a sphere are numerically equal, find its radius. Step 1: Let radius be $r$. Step 2: Volume of sphere = $\frac{4}{3}\pi r^3$. Step 3: Surface area of sphere = $4\pi r^2$. Step 4: Given volume = surface area, so $$\frac{4}{3}\pi r^3 = 4\pi r^2$$ Step 5: Cancel $4\pi r^2$ (assuming $r \neq 0$), we get $$\frac{r}{3} = 1 \implies r = 3$$ Answer: Radius is 3 units. 2. Problem: Find ratio of volume of cone to cylinder with same radius and height. Step 1: Volume of cone = $\frac{1}{3}\pi r^2 h$. Step 2: Volume of cylinder = $\pi r^2 h$. Step 3: Ratio = $\frac{\frac{1}{3}\pi r^2 h}{\pi r^2 h} = \frac{1}{3}$. Answer: Ratio is 1 : 3. 3. Problem: Height of conical tent to accommodate 11 persons, each needs 4 m² ground and 20 m³ air. Step 1: Total ground area = $11 \times 4 = 44$ m². Step 2: Base area of cone = $\pi r^2 = 44$. Step 3: Total air volume = $11 \times 20 = 220$ m³. Step 4: Volume of cone = $\frac{1}{3}\pi r^2 h = 220$. Step 5: Substitute $\pi r^2 = 44$ into volume: $$\frac{1}{3} \times 44 \times h = 220 \implies \frac{44h}{3} = 220 \implies h = \frac{220 \times 3}{44} = 15$$ Answer: Height is 15 m. 4. Problem: Find radius if surface area of sphere is $676 \pi$ cm². Step 1: Surface area = $4\pi r^2 = 676 \pi$. Step 2: Cancel $\pi$: $$4 r^2 = 676 \implies r^2 = 169 \implies r = 13$$ Answer: Radius is 13 cm. 5. Problem: Diameter decreased by 25%, find percentage decrease in volume. Step 1: Original diameter = $d$, new diameter = $0.75 d$. Step 2: Radius changes from $r = \frac{d}{2}$ to $0.75 r$. Step 3: Volume proportional to $r^3$, so volume ratio: $$\left(0.75\right)^3 = 0.421875$$ Step 4: Volume decreases by: $$(1 - 0.421875) \times 100 = 57.8125\%$$ Answer: Volume decreases by approximately 57.81%. 6. Problem: 100 jugs of 3850 cm³ water emptied into conical flask, height of water level 75 cm, find diameter of water level. Step 1: Total volume = $100 \times 3850 = 385000$ cm³. Step 2: Volume of cone = $\frac{1}{3} \pi r^2 h = 385000$, height $h=75$ cm. Step 3: Solve for $r^2$: $$\frac{1}{3} \pi r^2 \times 75 = 385000 \implies 25 \pi r^2 = 385000 \implies r^2 = \frac{385000}{25 \pi} = \frac{15400}{\pi}$$ Step 4: Calculate $r$: $$r = \sqrt{\frac{15400}{\pi}} \approx \sqrt{4903.32} \approx 70\text{ cm}$$ Step 5: Diameter = $2r = 140$ cm. Answer: Diameter is 140 cm. 7. Problem: Radius of hemispherical balloon increases from 6 cm to 12 cm, find ratio of surface areas. Step 1: Surface area of hemisphere = $2\pi r^2$. Step 2: Ratio = $\frac{2\pi (6)^2}{2\pi (12)^2} = \frac{36}{144} = \frac{1}{4}$. Answer: Ratio is 1 : 4. 8. Problem: Cone 8.4 cm high, radius 2.1 cm melted into sphere, find diameter of sphere. Step 1: Volume of cone = $\frac{1}{3} \pi r^2 h = \frac{1}{3} \pi (2.1)^2 (8.4)$. Step 2: Calculate volume: $$= \frac{1}{3} \pi \times 4.41 \times 8.4 = \frac{1}{3} \pi \times 37.044 = 12.348 \pi$$ Step 3: Volume of sphere = $\frac{4}{3} \pi R^3 = 12.348 \pi$. Step 4: Cancel $\pi$ and solve for $R$: $$\frac{4}{3} R^3 = 12.348 \implies R^3 = \frac{12.348 \times 3}{4} = 9.261$$ Step 5: $R = \sqrt[3]{9.261} \approx 2.1$ cm. Step 6: Diameter = $2R = 4.2$ cm. Answer: Diameter is 4.2 cm. 9. Assertion: Total surface area of cone with radius $\frac{r}{2}$ and slant height $2l$ is $\pi r \left(\frac{r}{4} + 2l\right)$. Reason: Total surface area of cone is $\pi r (l + r)$. Step 1: Total surface area formula is $\pi r (l + r)$. Step 2: Substitute radius $\frac{r}{2}$ and slant height $2l$: $$\pi \times \frac{r}{2} \times \left(2l + \frac{r}{2}\right) = \pi \frac{r}{2} \left(2l + \frac{r}{2}\right) = \pi r \left(l + \frac{r}{4}\right)$$ Step 3: Given assertion is $\pi r \left(\frac{r}{4} + 2l\right)$ which is incorrect order. Step 4: Reason is true, assertion is false. Answer: (d) A is false but R is true. 10. Assertion: Volume of largest cone cut from cube edge $r$ is $\frac{1}{3} \pi r^3$. Reason: Volume of cone is $\frac{1}{3} \pi r^2 h$. Step 1: Largest cone inside cube has base radius $\frac{r}{2}$ and height $r$. Step 2: Volume = $\frac{1}{3} \pi \left(\frac{r}{2}\right)^2 r = \frac{1}{3} \pi \frac{r^2}{4} r = \frac{1}{12} \pi r^3$. Step 3: Assertion says $\frac{1}{3} \pi r^3$, which is incorrect. Step 4: Reason is true. Answer: (d) A is false but R is true. 11. Problem: Sphere radius 3 cm melted into balls diameter 0.6 cm, find number of balls. Step 1: Volume of big sphere = $\frac{4}{3} \pi 3^3 = 36 \pi$. Step 2: Volume of small ball = $\frac{4}{3} \pi (0.3)^3 = \frac{4}{3} \pi 0.027 = 0.036 \pi$. Step 3: Number of balls = $\frac{36 \pi}{0.036 \pi} = 1000$. Answer: 1000 small balls. 12. Problem: Curved surface area of cone 308 cm², slant height 14 cm, find total surface area. Step 1: Curved surface area = $\pi r l = 308$. Step 2: $\pi r \times 14 = 308 \implies r = \frac{308}{14 \pi} = \frac{22}{\pi} \approx 7$ cm. Step 3: Total surface area = curved surface area + base area = $308 + \pi r^2 = 308 + \pi \times 7^2 = 308 + 49 \pi$. Step 4: Calculate $49 \pi \approx 153.94$. Step 5: Total surface area $= 308 + 153.94 = 461.94$ cm². Answer: Approximately 462 cm². 13. Problem: Canvas needed for conical tent height 3.5 m, radius 12 m. Step 1: Slant height $l = \sqrt{r^2 + h^2} = \sqrt{12^2 + 3.5^2} = \sqrt{144 + 12.25} = \sqrt{156.25} = 12.5$ m. Step 2: Curved surface area = $\pi r l = \pi \times 12 \times 12.5 = 150 \pi$. Step 3: Canvas required = $150 \pi \approx 471.24$ m². Answer: Approximately 471.24 m². 14. Problem: Diameters equal, slant heights ratio 7:4, find ratio of curved surface areas. Step 1: Radius equal, say $r$. Step 2: Curved surface area ratio = $\pi r l_1 : \pi r l_2 = l_1 : l_2 = 7 : 4$. Answer: Ratio is 7 : 4. 15. Problem: Radius and height ratio 4:3, base area 154 cm², find curved surface area. Step 1: Let radius $r = 4x$, height $h = 3x$. Step 2: Base area $= \pi r^2 = 154$. Step 3: Using $\pi = \frac{22}{7}$: $$\frac{22}{7} (4x)^2 = 154 \implies \frac{22}{7} 16 x^2 = 154 \implies 22 \times 16 x^2 = 154 \times 7$$ $$352 x^2 = 1078 \implies x^2 = \frac{1078}{352} = 3.0625 \implies x = 1.75$$ Step 4: Slant height $l = \sqrt{r^2 + h^2} = \sqrt{(4x)^2 + (3x)^2} = 5x = 5 \times 1.75 = 8.75$ cm. Step 5: Curved surface area = $\pi r l = \frac{22}{7} \times 7 \times 8.75 = 22 \times 8.75 = 192.5$ cm². Answer: Curved surface area is 192.5 cm². 16. Problem: Cloth 2.5 m wide for conical tent radius 7 m, height 24 m. Step 1: Slant height $l = \sqrt{7^2 + 24^2} = \sqrt{49 + 576} = \sqrt{625} = 25$ m. Step 2: Curved surface area = $\pi r l = \frac{22}{7} \times 7 \times 25 = 22 \times 25 = 550$ m². Step 3: Length of cloth = $\frac{\text{area}}{\text{width}} = \frac{550}{2.5} = 220$ m. Answer: 220 m of cloth required. 17. Problem: Circumference 44 m, height 10 m, canvas width 2 m, find length of canvas. Step 1: Radius $r = \frac{44}{2\pi} = \frac{44}{2 \times \frac{22}{7}} = 7$ m. Step 2: Slant height $l = \sqrt{r^2 + h^2} = \sqrt{7^2 + 10^2} = \sqrt{49 + 100} = \sqrt{149} \approx 12.21$ m. Step 3: Curved surface area = $\pi r l = \frac{22}{7} \times 7 \times 12.21 = 22 \times 12.21 = 268.62$ m². Step 4: Length of canvas = $\frac{268.62}{2} = 134.31$ m. Answer: Approximately 134.31 m. 18. Problem: Volumes of spheres ratio 64:27, sum of radii 21 cm, find radii. Step 1: Volume ratio $= \left(\frac{r_1}{r_2}\right)^3 = \frac{64}{27} = \left(\frac{4}{3}\right)^3$. Step 2: So, $\frac{r_1}{r_2} = \frac{4}{3}$. Step 3: Let $r_1 = 4x$, $r_2 = 3x$. Step 4: Sum $= 4x + 3x = 7x = 21 \implies x = 3$. Step 5: Radii are $r_1 = 12$ cm, $r_2 = 9$ cm. Answer: Radii are 12 cm and 9 cm. 19. (i) Volume of cylinder radius 6 cm, height 8 cm: $$V = \pi r^2 h = 3.14 \times 6^2 \times 8 = 3.14 \times 36 \times 8 = 904.32 \text{ cm}^3$$ (ii) Volume of sphere equals cylinder volume: $$\frac{4}{3} \pi r^3 = 904.32 \implies r^3 = \frac{904.32 \times 3}{4 \times 3.14} = \frac{2712.96}{12.56} = 216$$ $$r = \sqrt[3]{216} = 6 \text{ cm}$$ Answer: (i) 904.32 cm³, (ii) radius 6 cm. 20. (i) Surface area of 2 hemispheres radius 4.2 m: Surface area of one hemisphere = $2 \pi r^2 = 2 \times \frac{22}{7} \times (4.2)^2 = 2 \times \frac{22}{7} \times 17.64 = 110.88$ m². For 2 hemispheres = $2 \times 110.88 = 221.76$ m². (ii) Volume of one pillar (cylinder + hemisphere): $$V = \pi r^2 h + \frac{2}{3} \pi r^3$$ (iii) Volume of hemispherical dome radius 7 m: $$V = \frac{2}{3} \pi r^3 = \frac{2}{3} \times \frac{22}{7} \times 7^3 = \frac{2}{3} \times \frac{22}{7} \times 343 = 718.67 \text{ m}^3$$ (iv) Lateral surface area of 4 pillars (without dome): One pillar lateral area = $2 \pi r h = 2 \times \frac{22}{7} \times 1.4 \times 14 = 123.2$ m². For 4 pillars = $4 \times 123.2 = 492.8$ m². Answer: (i) 221.76 m², (ii) $\pi r^2 h + \frac{2}{3} \pi r^3$, (iii) 718.67 m³, (iv) 492.8 m².