A hollow hemisphere has uniform thickness. Its inner radius is r, and outer radius is R. If R=2r, find the ratio of outer to inner curved surface areas.

Correct — A. Each curved surface of a hemisphere is 2π × (its radius)².

Outer curved surface = 2πR²
Inner curved surface = 2πr²
Divide: 2πR² ÷ 2πr² = R² ÷ r²

Substitute R = 2r: (2r)² ÷ r² = 4r² ÷ r² = 4
Ratio outer : inner = 4 : 1 → option (a)

Why the others are wrong

• (b)
  7:4 — 7 : 4 needs R² : r² = 7 : 4, that is R ≈ 1.32r. With R = 2r the squares are 4r² and r², so the ratio is 4 : 1.
• (c)
  6:5 — 6 : 5 needs R ≈ 1.10r, a shell barely thicker than a skin. Here the thickness R − r equals r itself, and the areas scale as 2², not as 6⁄5.
• (d)
  2:5 — 2 : 5 puts the outer surface below the inner one, impossible when the outer radius is the larger. It would need R ≈ 0.63r, yet R = 2r.

Areas scale as the square of lengths. Both curved surfaces use the same formula, 2πr², so the 2π cancels and the ratio of areas is the ratio of radii, squared.

Radii 2 : 1 give areas 4 : 1. The uniform thickness describes the object, not the ratio: here it equals r, because R − r = 2r − r.

Key facts

• Curved surface area of a hemisphere = 2πr².
• If lengths are in the ratio k : 1, areas are k² : 1 and volumes k³ : 1.
• The flat rim of a hollow hemisphere is a ring of area π(R² − r²).

Common traps

• Stopping at the radius ratio 2 : 1 and forgetting to square it.
• Treating the thickness as a third length in the formula, when the curved surfaces depend on R and r alone.

No directly related past PYQ was found.