Physics

Physics

Problem

## Explorations 69. Archimedes' Principle A spherical buoy has a radius of 1 m and a density one-fourth that of seawater. By Archimedes' Principle, the weight of the displaced water will equal the weight of the buoy. - Let x=x= the depth to which the buoy sinks. - Let d=d= the density of seawater. - Let r=r= the radius of the circle formed where buoy, air, and water meet. See the figure below. ![](https://storage.googleapis.com/diraq-contents/question-bank/question-images/0f12be284813/2025_11_21_209c1aea55442890e1f4g-11.jpg) Notice in the figure that r2=1−(1−x)2=2x−x2r^{2}=1-(1-x)^{2}=2 x-x^{2}, and recall from geometry that the volume of submerged spherical cap is V=πx6⋅(3r2+x2)V=\frac{\pi x}{6} \cdot\left(3 r^{2}+x^{2}\right). (a) Verify that the volume of the buoy is 4π/34 \pi / 3. (b) Use your result from (a) to establish the weight of the buoy as πd/3\pi d / 3. (c) Prove the weight of the displaced water is πd⋅x(3r2+x2)/6\pi d \cdot x\left(3 r^{2}+x^{2}\right) / 6 (d) Approximate the depth to which the buoy will sink.

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