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JEE Main 2022 Mathematics question

MathematicsApplication of DerivativesJEE Main 2022

The surface area of a balloon of spherical shape being inflated, increases at a constant rate. If initially, the radius of balloon is 3 units and after 5 seconds, it becomes 7 units, then its radius after 9 seconds is :

Answer: (1)

Solution

, so dS/dt = dr/dt is constant, i.e. r dr/dt = k. Integrating, = kt + C. At t = 0,

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Question ID VK-024132