MHT CET 2021 Math Solved Paper

Let r be the radius, V be the volume and S be the surface area of the spherical raindrop at time t.
Then, ‌‌V=‌

4

3

πr3 and S=4πr2
The rate at which the raindrop evaporates is ‌

dV

dt

which is proportional to the surface area.
∴‌

dV

dt

∝S‌‌⇒‌

dV

dt

=−kS,‌ where ‌k>0. . . (i)
‌ Now, ‌‌‌V=‌

4

3

πr3‌ and ‌S=4πr2
∴‌‌‌

dV

dt

=‌

4π

3

×3r2‌

dr

dt

=4πr2‌

dr

dt

4πr2‌

dr

dt

=−k(4πr2)‌‌‌ [from Eq. (i)] ‌
‌

dr

dt

=−k⇒dr=−kdt
On integrating, we get
‌∫dr=−k‌∫dt+C
∴‌‌r=−kt+C
‌ Initially, i.e. ‌‌‌ when ‌t=0,r=3
∴‌3=−k×0+C
∴‌C=3
∴‌r=−kt+3
‌ When ‌t=1,r=2‌‌
∴‌2=−k×1+3
∴‌k=1
∴‌r=−t+3
∴‌‌r=3−t, where 0≤t≤3
This is the required expression for the radius of the raindrop at any time t.