Indian Institute of Foreign Trade 2009 Solved Paper

Question : 100 of 122

Marks: +1, -0

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=4π×R^2

{dS}/{dt}=4π×(2R)×{dR}/{dt}

It is given that

{dR}/{dt}=2

.
Hence,

{dS}/{dt}=4π×(2×30)×2=480π

Hence, option B is the correct answer.

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