A hemispherical bowl made of iron has inner diameter 84 cm. Find the cost of tin plating it on the inside at the rate of Rs. 21 per 100 cm2 (take \(\pi = \frac{{22}}{7}\)) correct to two places of decimal.

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SSC CGL 2022 Tier-I Official Paper (Held On : 02 Dec 2022 Shift 1)
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  1. Rs. 2,328.48
  2. Rs. 2,425.48
  3. Rs. 2,425.60
  4. Rs. 2,355.48

Answer (Detailed Solution Below)

Option 1 : Rs. 2,328.48
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Detailed Solution

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Given:

Inner Diameter = 84 cm

Cost of tin plating it on the inside = Rs. 21 per cm2.

Concept Used:

Curved Surface area of Hemisphere = 2π r2

Calculation:

Inner Diameter = 84 cm

Inner Radius = 84/2 cm = 42 cm

Curved Surface area of Hemisphere = 2π r2

⇒ 2 × 22/7 × 422

⇒ 2 × 22 × 6 × 42

⇒ 11088 cm2

The cost of tin-plating 100 cm2 of the bowl = Rs. 21

The cost of tin-plating 1 cm2 of the bowl = Rs. 21/100

The cost of tin-plating 11088 cmarea of the bowl = (21/100) × 11088 = Rs. 2,328.48

Thus, the cost of tin-plating is Rs. 2,328.48.

∴ Option 1 is the correct answer.

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