Hemispherical bowl painted inside and outside — compute total cost: A hemispherical bowl of radius 3.5 cm is to be painted on the inside as well as the outside. If the rate is ₹ 5 per 10 sq cm, find the total painting cost.

Introduction / Context:The paint is applied to both the inner and outer curved surfaces of a thin hemispherical shell. Neglecting thickness, the inner and outer radii are effectively the same, so the total painted area is twice the curved area of a hemisphere.

Given Data / Assumptions:

• Radius r = 3.5 cm.
• Curved surface area (hemisphere) = 2πr^2.
• Painting both sides ⇒ total curved area = 2 * (2πr^2) = 4πr^2.
• Rate = ₹ 5 per 10 cm2 = ₹ 0.5 per cm2.

Concept / Approach:Compute total area 4πr^2 and multiply by ₹ 0.5 per cm2. Using π = 22/7 makes the arithmetic exact for r = 3.5.

Step-by-Step Solution:4πr^2 = 4π * (3.5)^2 = 4π * 12.25 = 49π cm2With π = 22/7 ⇒ 49π = 49 * 22/7 = 154 cm2Cost = 0.5 * 154 = ₹ 77

Verification / Alternative check:Compute each curved side separately: 2πr^2 ≈ 76.97 cm2; doubled ≈ 153.94 cm2; times ₹ 0.5 ≈ ₹ 76.97 ≈ ₹ 77.

Why Other Options Are Wrong:₹ 50, ₹ 56, ₹ 81 do not match the exact evaluation with π = 22/7.

Common Pitfalls:Accidentally adding the flat circular rim area; using π = 3.14 with premature rounding may slightly change cents but not the integer rupee choice.

Final Answer:₹ 77