UPSC NDA Math Model Paper 6

At any instant t, let r be the radius, V be the volume and S be the surface area of the balloon.
Given: dV/dt = 20cm‌3/sec
As we know that, volume of sphere is given by:

4

3

πr3 where r is the radius of the sphere.
i.e V=

4

3

πr3
Now differentiating V with respect to t we get
⇒

dV

dt

=

dV

dr

.

dr

dt

......(By chain Rule)
As, V=

4

3

πr3
⇒

dV

dr

=

d

dr

(

4

3

πr3)=

4

3

π.3r2 =4πr2
By substituting dV/dr in dV/dt we get
⇒

dV

dt

=4πr2.

dr

dt

Now substitute dV/dt = 20 cm‌3/sec in the above equation we get
⇒20=4πr2.

dr

dt

⇒

dr

dt

=

5

πr2

As we know that, S=4πr2
By differentiating S with respect to t we get
⇒

dS

dt

=

dS

dr

.

dr

dt

.....(By chain Rule)
⇒

dS

dr

=

d(4πr2)

dr

=8πr
By substituting the value of dS/dr in dS/dt we get
⇒

dS

dt

=8πr.

dr

dt

By substituting

dr

dt

=

5

πr2

in the above equation we get
⇒

dS

dt

=8πr.

5

πr2

=

40

r

By substituting r = 8 cm in the above equation we get
⇒[

dS

dt

]r=8=

40

8

=5‌cm2∕sec