A cone whose height always equal to its diameter is increasing in volume at the ran `40 (cm^3)/sec` .At what rate is the radius increasing when its circular base area is `1m^2?` (A) 1 (B)0.001 (C) 2 (D) 0.002A. 1 mm/secB. 0.001 cm/secC. 2 mm/secD. 0.002 cm/sec

A cone whose height always equal to its diameter is increasing in volu

1.	A cone whose height always equal to its diameter is increasing in volume at the ran `40 (cm^3)/sec` .At what rate is the radius increasing when its circular base area is `1m^2?` (A) 1 (B)0.001 (C) 2 (D) 0.002A. 1 mm/secB. 0.001 cm/secC. 2 mm/secD. 0.002 cm/sec
Answer» Correct Answer - D Let h be the height, r be the radius of the base and V be the volume of the cone at time t. Then, `V=(1)/(3)pir^(2)h` `implies V=(2)/(3)pi r^(3)" "[because h = 2r]` `implies (dV)/(dt)=2pir^(2)(dr)/(dt)` `implies 40=2(10^(4))(dr)/(dt)" "[because pir^(2)=1m^(2)" (given)"=10^(4) cm^(2) and (dV)/(dt)=40cm^(3)//sec]`

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A cone whose height always equal to its diameter is increasing in volume at the ran `40 (cm^3)/sec` .At what rate is the radius increasing when its circular base area is `1m^2?` (A) 1 (B)0.001 (C) 2 (D) 0.002A. 1 mm/secB. 0.001 cm/secC. 2 mm/secD. 0.002 cm/sec

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