TS EAMCET 11-Sep-2020 Shift 2 Solved Paper

Given that, at any instant of time
Rate of change in volume w.r.t. time = rate of change in surface area w.r.t. time
i.e., ‌

dv

dt

=‌

ds

dt

....(i)
Volume of sphere of radius (r),V=‌

4

3

πr3
Differentiating w.r.t, ′t′, we get
‌

dv

dt

=‌

d

dt

(‌

4

3

πr3)=‌

4

3

π‌

d

dt

(r3)
Surface area of sphere of radius (r),
S=4πr2
Differentiating w.r.t. ′t′, we get
Putting the values from Eqs. (ii) and (iii) in Eq. (i), we get
4πr2‌

dr

dt

=8πr‌

dr

dt

or r=2