Mensuration

The volume of water displaced equals the volume of the spherical ball. Volume of cylinder = πr²h = π × 16² × 9 = 2304π cm³. For a sphere, V = (4/3)πr³, so r³ = (2304 × 3)/4 = 1728, giving r = 12 cm. Surface area = 4πr² = 4π × 144 = 576π cm².

First find the total height of the cone using similar triangles. Since current oil surface has radius 6 cm (from πr² = 36π) and is 3 cm from top, and top radius is 9 cm, we get total height = 9 cm. Current oil height from vertex = 6 cm, giving volume = (1/3)π(6²)(6) = 72π cm³. At drain rate of 1 cm³/hour, time = 72π hours.

Volume of prism = Base area × height = 250 cm³. Since height = 10 cm, base area = 250/10 = 25 cm². For a square base, side = √25 = 5 cm. Total surface area = 2(base area) + (perimeter × height) = 2(25) + (4×5×10) = 50 + 200 = 250 cm². Option C is correct.

Square area = 12cm², so side = √12 = 2√3 cm. Diagonal of square = side×√2 = 2√3×√2 = 2√6 cm. This equals the circle's radius. Circle area = πr² = π(2√6)² = π(4×6) = 24π cm². The subject 'English Language' is incorrect - this is Mathematics.

For a right triangle, circumradius R = hypotenuse/2, so hypotenuse = 24 cm. Inradius r = (a+b-c)/2 where a,b are legs. Given r=3, we have a+b-24=6, so a+b=30. Using a²+b²=24²=576 and (a+b)²=900, we get 2ab=324, so ab=162. Area = ab/2 = 81 cm².

The volume of a pyramid is (1/3) × base area × height. The base is a square of side 80 cm, so base area = 6400 cm². Setting 80000 = (1/3) × 6400 × h and solving: h = (80000 × 3) / 6400 = 240000/6400 = 37.5 cm.

Volume of cone = (1/3)πr²h = (1/3)π(3.2)²(14.4) = (1/3)π × 10.24 × 14.4 = 49.152π. Volume of cylinder = πR²H = πR²(19.2). Equating: πR²(19.2) = 49.152π. R² = 49.152/19.2 = 2.56. R = 1.6. Diameter = 2R = 3.2 cm.

CSA of cone = πrl = 4664. Given r = 28, π = 22/7, so l = 4664 × 7/(22 × 28) = 53 cm. Height h = √(l² - r²) = √(2809 - 784) = √2025 = 45 cm. Volume = (1/3)πr²h = (1/3)(22/7)(784)(45) = 116160 cm³. Total surface area = πr(l + r) = (22/7)(28)(81) = 7128 cm². Ratio = 116160 : 7128 = 140 : 27 after simplification.

The curved surface area of a right circular cone is πrl, where r is the radius and l is the slant height. Given r = 6 cm and l = 11 cm, CSA = π × 6 × 11 = 66π cm². This formula only gives the area of the curved surface, not including the base. The total surface area would additionally include πr² for the base.

Volume of sphere = (4/3)πr³ = (4/3)π(13)³. Cone has diameter 13 cm, so radius = 6.5 cm. Let height be h. Volume of cone = (1/3)πr²h = (1/3)π(6.5)²h. Equating volumes: (4/3)π(13)³ = (1/3)π(6.5)²h. Cancel (1/3)π: 4(13)³ = (6.5)²h. Note 6.5 = 13/2, so (6.5)² = (13)²/4. Substituting: 4(13)³ = ((13)²/4)h. Dividing both sides by (13)²/4: h = 4 × 4 × 13 = 16 × 13 = 208 cm.

Given: height h=2.1 cm, radius r=3h=6.3 cm. Volume of cylinder V=πr²h. V=π×(6.3)²×2.1=π×39.69×2.1=83.349π≈261.95 cm³. Using π≈3.1416, we get V≈261.95 cm³.

Volume of sphere = (4/3)π × 9³ = 972π cm³. This equals volume of wire = π × 2² × h = 4πh. So h = 972π ÷ 4π = 243 cm. The length is 243 cm. Option D is correct.

Sector area formula: A = (θ/360) × πr². Given A=308 cm², θ=45°. So 308 = (45/360) × πr² = (1/8) × πr². πr² = 2444, r² = 784, r = 28 cm. Option D correct. Option A (14cm) and C (7cm) are too small, B (21cm) doesn't satisfy the equation.

When a hemisphere is melted and recast into a cylinder, the volume remains constant. Volume of hemisphere = (2/3)πr³ = (2/3)π(21)³ = 2π(21)² × 7. For a cylinder with radius R and height h: πR²h = 2π(21)² × 7. The curved surface area (2πRh) to total surface area (2πR(R+h)) ratio is 2:5, giving h=R. This means R=21 cm from volume conservation.

Outer radius R = 6 + 0.37 = 6.37 cm. Outer surface area = 2πR² = 2 * (22/7) * (6.37)² = 255.0548 cm² (approximately).