#### To determine

**To show:**

The center of the disk should be positioned at height r1+π2 above the surface of the liquid

#### Answer

Height above water surface is r1+π2

#### Explanation

**1) Concept:**

Exposed wetted area = area of the disk - area of arc OAB + area of triangle OAB - area of circle with center O and radius rcosθ

2) **Calculation:**

Exposed wetted area= area of the disk - area of arc OAB + area of triangle OAB - area of circle with center O and radius rcosθ

That is,

A=πr2-r2θ+0.5r2sin2θ-πrcosθ2

To maximize this area, let us find critical points of A. So differentiate it with respect to θ and equate it to zero.

dAdθ=0-r21+0.5r22cos2θ-πr2(-)sin2θ

Simplify to get

dAdθ=-r2+r2cos2θ+πr2sin2θ

Set dAdθ=0

-r2+r2cos2θ+πr2sin2θ=0

Factor out r2 from the first two terms.

-r2(1-cos2θ)+πr2sin2θ=0

Simplify.

-r22sin2θ+πr2sin2θ=0

Factor out r2.

r2-2sin2θ+2πsinθcosθ=0

Factor out 2sinθ.

2r2sinθ-sinθ+πcosθ=0

Here,r2 cannot be zero, and θ cannot be zero and

So, -sinθ+πcosθ=0

Add sinθ to both the sides.

-sinθ+sinθ+πcosθ=0+sinθ

Simplify.

πcosθ=sinθ

Divide by cosθ, and simplify.

tanθ=π Using tan2x+1=sec2x

We get

cosθ=11+π2

This means height of centre above water surface rcosθ=r1+π2

Therefore, the height of centre above water surface is r1+π2

**Conclusion:**

The height of centre above water surface is r1+π2