The area of a circle is given by `A=pir^(2)`, where r is the radius . Calculate the rate of increases of area w.r.t. radius.

Answer

`20sqrt(ln2)`

Explanation

Force at `A (lambda q)/(2pi in_(0) r)=100`
`(lambda q)/(2pi in_(0)xx0.2)=100 underset(A)overset(B)(int)dv=-underset(A)overset(B)(int)vec(E).dr`
`V_(B)-V_(A)=(-lambda)/(2pi in_(0))underset(r_(1))overset(r_(2))(int)(dr)/r=(-lambda)/(2pi in_(0))ln r_(2)/r_(1)`
`V_(A)-V_(B)=lambda/(2pi in_(0))ln r_(2)/r_(1)=lambda/(2pi in_(0))ln2`
COME: `K_(A)+U_(A)=K_(B)+U_(B)`
`0+(lambdaQ)/(2pi squarein_(0))ln2=1/2 mv^(2), 20 ln2=1/2xx0.1xxV^(2)`
`V=20 sqrt(ln2)`