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MathsMediumMCQ2013 · 25 Apr Online

Q79.A spherical balloon is being inflated at the rate of 35cc/min . The rate of increase in the surface area (in cm2/min.) of the balloon when its diameter is 14 cm, is : JEE Main 2013 (25 Apr Online) JEE Main Previous Year Paper (1) 10 (2) √10 (3) 100 (4) 10√10

What This Question Tests

This question tests the ability to apply derivatives to find the rate of change of one quantity (surface area) with respect to time, given the rate of change of another related quantity (volume) for a spherical object.

Concepts Tested

Rate of changeDerivative of volumeDerivative of surface areaChain rule

Formulas Used

V = (4/3)πr³

A = 4πr²

dV/dt = dV/dr * dr/dt

dA/dt = dA/dr * dr/dt

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