Mensuration

Curved surface area of a hemisphere = 2πr² = 2 × 3.14159 × 11² = 2 × 3.14159 × 121 = 760.57 sq cm. This is half the total surface area of a sphere (4πr²), covering only the curved portion without the base.

Area = 361 sq ft, so side = sqrt(361) = 19 ft. Perimeter of square = 4 × 19 = 76 ft. Cost at Rs 72 per foot = 76 × 72 = Rs 5472. Option B is correct. Options A (Rs 6352), C (Rs 4846), and D (Rs 6389) are calculation errors.

Volume = πr²h = 25872 cm³ (converting litres). Given h = 3r, we have πr²(3r) = 25872. Using π ≈ 22/7 and solving, r² = 196, so base area = πr² = 22/7 × 196 = 616 cm². The key is relating height to radius.

Area of a circle = πr². When diameter is doubled, radius also doubles. New area = π(2r)² = 4πr², which is 4 times the original area. This follows from the square relationship between radius and area.

Volume of first cylinder: π × 4² × 6 = π × 16 × 6 = 96π. Volume of second cylinder: π × 5² × 4 = π × 25 × 4 = 100π. Total volume = 196π. When recast into a disc of thickness 1 cm: π × r² × 1 = 196π, so r² = 196, and r = 14 cm. The volume is conserved during the recasting process.

Let r=4x, h=3x. Slant height l = √(16x² + 9x²) = 5x. Curved surface area = πrl = π(4x)(5x) = 20πx². Total surface area = πr(l+r) = π(4x)(5x+4x) = 36πx². Ratio = 20:36 = 5:9.

Area of square village = side² = (12 km)² = 144 km². To convert km² to hectares: 1 km² = 100 hectares. So 144 km² = 144 × 100 = 14400 hectares. Option A is correct.

Area of circle = πr² = (22/7) × 14² = 616 cm². Since rectangle area equals circle area, Length × 22 = 616, giving Length = 28 cm. The calculation uses π ≈ 22/7 for convenience.

Volume of a prism = base area × height (lateral edge). If base remains same and lateral edges are halved, new volume = base area × (height/2) = (base area × height)/2 = original volume/2. Reduction = (original - new)/original = 50%. The volume is directly proportional to height.

For internally tangent circles with radii R and r, we have R² + r² = 116 and R - r = 6. Solving gives R = 10, r = 4, so sum = 14 cm. Option A (16 cm) would satisfy R² + r² = 116 but not R - r = 6.

Let radius r = 5k and height h = 7k. Volume V = πr²h = π(5k)²(7k) = 175πk³ = 550 cm³. Solving: k³ = 550/(175π) ≈ 1, so k ≈ 1. Thus r ≈ 5 and h ≈ 7. Curved surface area = 2πrh = 2π(5)(7) = 70π ≈ 220 cm². Using π = 22/7: 70 × 22/7 = 220. Option C is correct.

For an open cylinder (no top or bottom), the curved surface area = 2πrh = 2 × (22/7) × 7 × 15 = 660 sq.cm. This curved surface equals the area of the rectangular sheet: length × width. Given width=11 cm, length = 660/11 = 60 cm. This matches the expected answer though not listed in options.

Original volume V = πr²h. After 10% increase: r' = 1.1r, h' = 1.1h. New volume V' = π(1.1r)²(1.1h) = π(1.21r²)(1.1h) = 1.331πr²h = 1.331V. Percentage increase = (1.331 - 1) × 100 = 33.1%.

आयताकार ठोस का आयतन = लंबाई × चौड़ाई × ऊंचाई। चूंकि आधार वर्गाकार है, लंबाई = चौड़ाई = 16 cm। ऊंचाई = 20 cm। आयतन = 16 × 16 × 20 = 5120 cm³। विकल्प C (5120 cm2) सही है, हालांकि इकाई cm² होनी चाहिए cm³।

Volume of sphere = (4/3)πr³ and surface area = 4πr². Setting equal: (4/3)πr³ = 4πr². Cancel 4πr² from both sides: r/3 = 1, so r = 3. Option A is correct.