1. 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:
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By: anil on 05 May 2019 02.40 am
It is given that radius, R = 30 cm. Curved surface area, S= $$4pi*R^2$$ $$dfrac{dS}{dt}$$ = $$4pi*(2R)*dfrac{dR}{dt}$$ It is given that $$dfrac{dR}{dt}$$ = 2. Hence, $$dfrac{dS}{dt}$$ = $$4pi*(2*30)*2$$ = $$480pi$$. Hence, option B is the correct answer.
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