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A hemispherical bowl has 3.5 cm radius. It is to be painted inside as well as outside. Find the cost of painting it at the rate of ₹15 per 10 cm². (Use ∏ = 22/7)

A₹431

B₹331

C₹231

D₹531

Answer:

C. ₹231

Read Explanation:

Radius (r = 3.5) cm

A hemispherical bowl painted inside and outside, so total surface area:

Total SA=2×(2πr2)=4πr2\text{Total SA} = 2 \times (2\pi r^2) = 4\pi r^2
Compute area

4×227×(3.5)24 \times \frac{22}{7} \times (3.5)^2

(3.5)2=12.25(3.5)^2 = 12.25

=4×227×12.25= 4 \times \frac{22}{7} \times 12.25

=4×22×1.75= 4 \times 22 \times 1.75

=154 cm2= 154 \text{ cm}^2

Cost calculation

Rate = ₹15 per 10 cm²
So per 1 cm² = ₹1.5

Cost=154×1.5=231\text{Cost} = 154 \times 1.5 = 231

Final Answer: ₹231


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