UPSC NDA Math Model Paper 4

volume of spherical ball =288π
volume of spherical ball v=

4

3

πr3
⇒288π=

4

3

πr3
⇒r=6
The volume of spherical ball is increasing at the rate of 4πcm∕sec

dv

dt

=4π

dv

dt

=4πr2

dr

dt

4π=4πr2

dr

dt

dr

dt

=

1

r2

Area of spherical ball =A=4πr2

dA

dt

=4π×2r

dr

dt

dA

dt

=8πr.

1

r2

dA

dt

=8π.

1

r

dA

dt

=8π.

1

6

dA

dt

=

4

3

π
Hence, the volume of spherical ball is increasing at the rate of 4π cc/sec then the rate of change of its surface area when the volume is 288π cc is

4

3

πr3