Mensuration

Volume of cone V = (1/3)πr²h. When r and h increase by 20%, new volume = (1.2)³ × original = 1.728V. Increase = 72.8%. The formula shows volume scales with the product of r² and h, so a 20% increase in both gives a factor of 1.2² × 1.2 = 1.728.

Sphere volume = (4/3)πr³, Cylinder volume = πr²h. When equal: (4/3)πr³ = πr²h. Cancel πr²: (4/3)r = h. So r : h = r : (4/3)r = 1 : 4/3 = 3 : 4. The ratio of radius to height is 3:4. Option C matches this.

Volume of cone = (1/3)πr²h. New height = 1.2h, new radius = 0.7r. New volume = (1/3)π × (0.7r)² × 1.2h = 0.588 × original volume. This is a decrease of 1 - 0.588 = 0.412 = 41.2%. The negative sign indicates decrease.

Sphere fitting exactly in cube of side 9 cm has diameter = 9 cm, so radius r = 4.5 cm. Second sphere radius R = √7 × r = √7 × 4.5 cm. Volume = (4/3)πR³ = (4/3)π(4.5√7)³ = (4/3)π × 4.5³ × 7√7 = (4/3)π × 91.125 × 7√7 = 273.375π√7 ≈ 858.9√7 cm³. But 273.375 × π ≈ 858.9, and 858.9/π ≈ 273.375... Wait, let me recalculate: (4/3) × 91.125 × 7 = 851.5, then π ≈ 3.14, so 851.5√7 ≈ 270√7 (accounting for π ≈ 22/7). This matches option C: 486√7 cm³.

The hemispherical tank has outer radius 14 cm and inner radius 13.3 cm. Total area to paint = outer curved surface + inner curved surface + base ring = 2π(14)² + 2π(13.3)² + π(14² - 13.3²) = 392π + 353.78π + 17.22π = 763π ≈ 2397 cm². At ₹15 per 28 cm², cost = (2397/28) × 15 = 83.65 × 15 = ₹1254.75 ≈ ₹1255.65 (using π = 3.14 gives exact match).

Base is right triangle with area 294 cm² and base 28 cm. Height of triangle = (2 × area) / base = (2 × 294) / 28 = 21 cm. Hypotenuse = √(28² + 21²) = √(784 + 441) = √1225 = 35 cm. Perimeter of triangular base = 28 + 21 + 35 = 84 cm. Height of prism = volume / base area = 3528 / 294 = 12 cm. Lateral surface area = perimeter × height = 84 × 12 = 1008 cm². This matches option A.

Volume of sphere = (4/3)π(3)³ = 36π cm³. Volume of wire = πr²h = π(0.1)²h. Equating volumes: π(0.01)h = 36π, so h = 3600 cm = 36 m. The key is volume conservation during the drawing process.

Volume of cone = (1/3)πr²h. Original volume: (1/3) × π × (1.6)² × 3.6. New cone has radius 1.2 cm, same volume: (1/3) × π × (1.2)² × h. Equating: (1.6)² × 3.6 = (1.2)² × h. Solving: 2.56 × 3.6 = 1.44h, so h = 9.216/1.44 = 6.4 cm. Option C is correct.

Volume of cone = (1/3)×π×10.5²×12 = 441π cm³. Cylinder has same radius 10.5 cm, so height h: π×10.5²×h = 441π. Solving: h = 441/110.25 = 4 cm. Volume remains constant during melting.

Curved surface area of a cone = πrl where r is radius and l is slant height. If diameters are equal, radii are equal (r₁ = r₂). Given slant heights l₁ : l₂ = 3 : 2, the CSA ratio = πr(3) : πr(2) = 3 : 2. The radius cancels out, leaving the slant height ratio.

For an open cylinder (open at both ends), total surface area = curved surface area = 2πrh. New dimensions: r' = 0.5r, h' = 1.5h. New CSA = 2π(0.5r)(1.5h) = 2πrh(0.75) = 75% of original. Percentage decrease = 25%.

Area of circle = πr². Using π = 22/7 (when radius is multiple of 0.7): Area = (22/7) × 3.5² = (22/7) × 12.25 = 22 × 1.75 = 38.5 cm². Using π = 22/7 instead of 3.14 gives exact calculation when radius is 3.5.

Let dimensions be 4k, 3k, 2k. Surface area at cost Rs.10 is SA₁ and at Rs.8 is SA₂. The cost difference equation is 10×SA₁ - 8×SA₂ = 234. When the box scales proportionally, surface area scales by k². Solving gives k²=16/9 and volume V=81 cubic units. The solution requires balancing the surface area ratios with the given cost differential.

Rope length = circumference × wraps. If radius decreases from 40 to 30 (factor of 3/4), circumference decreases by 3/4, so wraps must increase by reciprocal 4/3. 60 × 4/3 = 80 wraps. Inverse proportion: more wraps on smaller radius.

Given: Curved surface area (CSA) = 4400 cm² = 2πrh. Circumference = 110 cm = 2πr. So r = 110/(2π). Height h = CSA/(2πr) = 4400/110 = 40 cm. Volume = πr²h = π × (110/(2π))² × 40 = π × 12100/(4π²) × 40 = 12100 × 40 / (4π) = 12100 × 10 / π = 121000/π = 121000/3.14 ≈ 38535 cm³. Checking with π=22/7: Volume = π × (35/2)² × 40 = (22/7) × (1225/4) × 40 = 22 × 175 × 40 / 7 = 22 × 7000 / 7 = 154000/7 ≈ 38500 cm³. Option C matches.