UPSC NDA Math Model Paper 7

At any instant t, let r be the radius, V be the volume and S be the surface area of the balloon.
Given: dS∕dt=2cm2∕sec
As we know that, surface area of sphere is given by: 4πr2 where r is the radius of the sphere. i.e S=4πr2
Now 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 dS/dt = cm‌2/sec in the above equation we get
⇒2=8πr.

dr

dt

⇒

dr

dt

=

1

4πr

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 dr/dt = 1/4πr in the above equation we get
⇒

dV

dt

=4πr2.

1

4πr

=r By substituting r = 6 cm in the above equation we get,
⇒[

dV

dt

]r=6=6‌cm3∕sec