Discussion

Question:

Radius of a spherical balloon, of radii 30 cm, increases at the rate of 2 cm per second. Then its curved surface area increases by:

Explanation:

Curved surface area of a sphere = 4πr²

^{$\frac{d r}{d t}$ = 2 $\frac{c m}{s e c}$}

^{∴ $\frac{d}{d t}$}4πr² = 4 × π × 2 × r × ^{$\frac{d r}{d t}$}

^{= 8π × 30 × 2 = 480π}

∴ Curved surface area of the balloon increases by 480π cm² per second.

Hence, option (b).

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