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MHT CET202619 April 2026Evening ShiftMathematicsApplication of DerivativesActual

If the rate of increase of surface area of a spherical balloon is 5 , cm ^2/ sec and rate of increase of volume of a spherical balloon is 10 , cm ^3/ sec , then the radius of the balloon at that time is...

Options

  1. A3 cm
  2. B5 cm
  3. C6 cm
  4. D4 cm

Correct answer

D. 4 cm

Step-by-step solution

Let r be the radius, S be the surface area, and V be the volume of the spherical balloon. S = 4 r^2 Differentiating with respect to time t : dS dt = 8 r dr dt = 5 V = 4 3 r^3 Differentiating with respect to time t : dV dt = 4 r^2 dr dt = 10 Dividing dV dt by dS dt : 4 r^2 dr dt 8 r dr dt = 10 5 r 2 = 2 r = 4 cm Answer: 4 cm

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