Question 100

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:

Solution

It is given that radius, R = 30 cm.

Curved surface area, S= $$4\pi*R^2$$

$$\dfrac{dS}{dt}$$ = $$4\pi*(2R)*\dfrac{dR}{dt}$$

It is given that $$\dfrac{dR}{dt}$$ = 2.

Hence, $$\dfrac{dS}{dt}$$ = $$4\pi*(2*30)*2$$ = $$480\pi$$. Hence, option B is the correct answer.

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