Practicing Success
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 : |
1 mm/sec 2 mm/sec 0.01 mm/sec 0.002 cm/sec |
0.002 cm/sec |
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 |