#### To determine

**To find:**

the rate at which water level rising at the instant the cone is completely submerged dx/dt.

**Solution:**the rate at which water level rising at the instant the cone is completely submerged is r2R2-r2 cm/s

#### Explanation

**1) Formula:**

i. Volume of cone

V=πr2.h3

ii. Volume of cylinder

V=πr2h

**2) Given:**

i. r=radius of cone in cm

ii. h=height of cone in cm

iii. H=initial height of water

iv. R=radius of cylinder

3) **Calculation:**

Assume the axes of the cone and the cylinder are parallel.

Let H denote the initial height of the water.

When the cone has been dropping for t seconds, the water level has risen x centimeters, so the tip of the cone is x+1t cm below the water line.

To find dx/dt when x+t=h( when the cone is completely submerged).

Using the similar triangle,

r1x+t=rh

⇒r1=rh(x+t)

volume of water and cone at time t=original volume of water+volume of submerged part of cone

πR2H+x=πR2H+13πr12(x+t)

πR2H+πR2x=πR2H+13πr2h2x+t3

subtracting πR2H from both side,

πR2x=13πr2h2x+t3

Canceling π from both side

R2x=13r2h2x+t3

multiplying by 3h2 on both side,

3h2R2x=r2x+t3

Differentiating above equation implicitly with respect to t

3h2R2.dxdt=r2[3x+t2+3x+t2

dxdt=r2x+t2h2R2-r2x+t2

dxdt|x+t=h=r2h2h2R2-r2h2=r2R2-r2

Thus, the water level is rising at a rate of r2R2-r2 cm/s at the instant the cone is completely submerged.

**Conclusion:**

The rate at which water level rising at the instant the cone is completely submerged is

r2R2-r2 cm/s