MHT CET 2017 Math Solved Paper

(a) Let r be the radius of spherical ball
∵ Volume of spherical ball V =

4

3

πr3 ...(i)
Now, 288π=

4

3

πr3
⇒ r3=72×3 = 8×27
r = 6
After differentiating eq. (i) w.r.t. ‘t’, we get

dV

dt

=4πr2

dr

dt

⇒ 4π=4πr2

dr

dt

[∵

dV

dt

=4πcm3∕s]
⇒ I=(6)2

dr

dt

⇒

dr

dt

=

1

36

∵ Surface area of spherical ball, s = 4πr2
Afterdifferentiating on both sides, w.r.t. ‘t’, we get

ds

dt

=4×2πr

dr

dt

⇒

ds

dt

= 8×π×6×

1

36

Hence,

ds

dt

=

4π

3

cm2∕s