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Answer :

Given :

Inner radius Of hemispherical bowl = 5cm

Thickness of the bowl = 0.25 cm

Outer radius (r) Of hemispherical bowl = (5 + 0.25) cm = 5.25 cm

We also know that,

Outer CSA of hemispherical bowl

= \(2 {\pi} {r}^2\)

= \(2 × \frac{22}{7} × {5.25}^2\)

= \(173.25 {cm}^2\)

Therefore, the outer curved surface area Of the bowl is 173.25 \({cm}^{2}\)

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