A cone, whose height is always equal to its diameter, is increasing in volume at the rate of $40\, cm^3/sec.$ The rate at which radius is increasing when circular base area is 1 m2 is :
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 0.002 cm/sec
$V=\frac{πR^2h×2}{3}$
$R = h$
Height = 2R
so $\frac{dV}{dt}=\frac{π}{3}\frac{d}{dt}(2R^3)=\frac{6(πR^2)}{3}\frac{dR}{dt}$
$40=\frac{6×10000}{3}×\frac{dR}{dt}$
$\frac{dR}{dt}=\frac{2×3}{3000}=0.002$cm/sec