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COMEDK2014MathematicsApplication of Derivatives

A spherical balloon is being inflated at the rate of 35 ~cm / min . When its radius is 7 ~cm , its surface area increases at the rate of

Options

  1. A10 ~cm ² / min
  2. B15 ~cm ² / min
  3. C20 ~cm ² / min
  4. D25 ~cm ² / min

Correct answer

A. 10 ~cm ² / min

Step-by-step solution

Given, d V d t =35 where, V is volume of spherical balloon. aligned & Also, V= 4 3 r³ & d d t ( 4 3 r³ )=35 4 3 3 r² d r d t =35 & d r d t = 35 3 4 3 r² aligned Let S be surface area of sphere, then S=4 r² Taking derivative w.r.t. r d S d t =8 r d r d t =8 r 35 3 4 3 r² Substitute, r=7 d S d t = 2 35 3 3 7 =10 ~cm ² / min

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