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Q11 CQ

Expert-verifiedFound in: Page 437

Book edition
9th Edition

Author(s)
Raymond A. Serway, John W. Jewett

Pages
1624 pages

ISBN
9781133947271

**How would you determine the density of an irregularly shaped rock?**

The buoyant force can be thought of as the weight of so many grams of water, which is that number of cubic centimeters of water, which is the volume of the submerged rock. This volume with the actual rock mass tells us its density.

When an object is partially or fully submerged in a fluid, the fluid exerts on the object an upward force called the buoyant force. According to Archimedes’s principle, the magnitude of the buoyant force is equal to the weight of the fluid displaced by the object:

$B={\rho}_{Fluid}g{V}_{disp}$

WhereV${V}_{disp}$ the volume of fluid is displaced and ${\rho}_{Fluid}$ is the density of the fluid.

- Use a balance to determine its mass. Then partially fill a graduated cylinder with water. Immerse the rock in the water and determine the volume of water displaced.
- Divide the mass by the volume and you have the density. It may be more precise to hang the rock from a string, measure the force required to support it under water, and subtract to find the buoyant force.
- The buoyant force can be thought of as the weight of so many grams of water, which is that number of cubic centimeters of water, which is the volume of the submerged rock. This volume with the actual rock mass tells us its density.

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