TS EAMCET 11-Sep-2020 Shift 2 Solved Paper

Given that,
Rate of increase in, ‌

dV

dt

=4πcc/sec
Volume, V=288πcc
As we know, volume of a sphere of radius (r),
V=‌

4

3

πr3.....(i)
Differentiating above expression w.r.t. ' t′ on both sides, we get
‌

dV

dt

=‌

d

dt

(‌

4

3

πr3)=‌

4

3

π3r2‌

dr

dt

‌

dV

dt

=4πr2‌

dr

dt

...(ii)
From Eq. (i) (288π)=‌

4

3

πr3 or r=6cm
Putting this value in Eq. (ii), we get
4π=4π(6)2‌

dr

dt

or ‌

dr

dt

=‌

1

36