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+0.5r2sin⁡2θ-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+0.5r22cos⁡2θ-πr2(-)sin⁡2θ

Simplify to get

dAdθ=-r2+r2cos⁡2θ+πr2sin⁡2θ

Set dAdθ=0

-r2+r2cos⁡2θ+πr2sin⁡2θ=0

Factor out r2 from the first two terms.

-r2(1-cos⁡2θ)+πr2sin⁡2θ=0

Simplify.

-r22sin2⁡θ+πr2sin⁡2θ=0

Factor out r2.

r2-2sin2⁡θ+2πsin⁡θcos⁡θ=0

Factor out 2sin⁡θ.

2r2sin⁡θ-sin⁡θ+πcos⁡θ=0

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

So, -sin⁡θ+πcos⁡θ=0

Add sin⁡θ to both the sides.

-sin⁡θ+sin⁡θ+πcos⁡θ=0+sin⁡θ

Simplify.

πcos⁡θ=sin⁡θ

Divide by cos⁡θ, and simplify.

tan⁡θ=π Using tan2⁡x+1=sec2⁡x

We get

cos⁡θ=11+π2

This means height of centre above water surface rcos⁡θ=r1+π2

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

Conclusion:

The height of centre above water surface is r1+π2