The radius of a cylinder is increasing at the rate of `5 cm min^(-1)`, so that its volume is constant. When its radius is 5 cm and height is 3 cm, then the rate of decreasing of its height is

Options

A. `6 cm min^(-1)`
B. `3 cm min^(-1)`
C. `4 cm min^(-1)`
D. `5 cm min^(-1)`

Answer

A

Explanation

Given `(dr)/(dt)=5 cm min^(-1)`, radius (`r)=5` cm and height `(h)=3cm`
`:'` Volume of cylinder `V=pir^(2)h`…………….i
`implies (dV)/(dt)=2pi r (dr)/(dt) . h+pir^(2)(dh)/(dt)`…………..ii
For volume to be constant `(dv)/(dt)=0`
`2pir(dr)/(dt).h=-pir^(2)(dh)/(dt)`
`implies2(dr)/(dt).h=4(dh)/(dt)implies2.5.3=5(dh)/(dt)`
`implies(dh)/(dt)=6cm//min`