BITSAT 2018 Slot 2 Solved Paper

Let depth of water at time t be h and the radius of the base of water level be r.
From similar ΔOMP and ΔONQ, we have

OM

ON

=

PM

QN

⇒

h

10

=

r

5

⇒r=

5h

10

⇒r=

h

2

We have, V=

1

3

πr2h=

1

3

π(

h

2

)2h
⇒V=

πh3

12

On differentiating both sides, we get

dV

dt

=

π

12

.3h2

dh

dt

=

πh2

4

dh

dt

Given,

dV

dt

=3cm3∕sec when h=4cm, so we get
3=

π×42

4

dh

dt

dh

dt

=

3

4π

cm∕sec
Hence, the water level is rising at

3

4π

cm∕sec.
∴ Option (b) is correct.