MHT CET202615 April 2026Evening ShiftMathematicsApplication of DerivativesActual
The surface area of a spherical ball is increasing at the rate of 4 cm ^2 /second. The rate at which the radius is increasing when the surface area is 16 cm ^2 is
Options
- A0.5 cm/second
- B0.25 cm/second
- C0.125 cm/second
- D1 cm/second
Correct answer
B. 0.25 cm/second
Step-by-step solution
Let S be the surface area and r be the radius of the spherical ball. S = 4 r^2 Differentiating with respect to time t : dS dt = 8 r dr dt Given dS dt = 4 cm ^2 /s. When S = 16 cm ^2 : 4 r^2 = 16 r^2 = 4 r = 2 cm. Substituting the values into the derivative equation: 4 = 8 (2) dr dt 4 = 16 dr dt dr dt = 4 16 = 0.25 cm/s.