Q) The internal and external radii of a hollow hemisphere are 5√2 cm and 10 cm respectively. A cone of height 5√7 cm and radius 5√2 cm is surmounted on the hemisphere as shown in the figure.

Find the total surface area of the object in terms of π. (Use √2 = 1.4)

(Q 27 B – 30/4/2 – CBSE 2026 Question Paper)

Ans:

Step 1: We have following dimensions:

For Hemisphere: Internal radius, r₁ = 5√2 cm,

external radius, r₂ = 10 cm

For Cone: height of cone = 5√7 cm

Base radius = internal radius of Hemisphere = 5√2 cm

∴ slant height of cone, L = √(r ² + h ²)

= √[(5√2) ²+(5√7) ²] = √(50+175)

= √225 = 15 m

Step 2: Curved Surface Area of conical top = π r₁ L

= π (5√2)(15) = 75√2 π cm ²

Step 3: Surface area of base ring

= Base area of hemisphere – base area of conical top

= π (r₂ ² – r₁ ²) = π [(10) ² – (5√2) ²]

= π [(100 – 50] = 50 π cm ²

Step 4: Outer Curved Surface area of hemisphere = 2 π r₁  ²

= 2 π (10)2 = 200 π cm ²

Step 5: Inner Curved Surface area of hemisphere

= 2 π r₂ ²  = 2 π (5√2)2

= 2 π (50) = 100 π cm ²

Step 6: Total Surface area of the object = CSA of conical top + SA of base ring + Outer CSA of hemisphere + Inner CSA of hemisphere

= 75√2 π + 50 π + 200 π + 100 π

= (350 + 75√2)π = (350 + 75 x 1.4) π

= (350 + 105) π = 455 π

Therefore, the total surface area of the object is 455 π cm ²

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