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Chapter 1: Problem 109

The density of an irregularly shaped object was determined as follows. The mass of the object was found to be \(28.90 \mathrm{g}=\) 0.03 \(\mathrm{g} .\) A graduated cylinder was partially filled with water. The reading of the level of the water was \(6.4 \mathrm{cm}^{3} \pm 0.1 \mathrm{cm}^{3}\) The object was dropped in the cylinder, and the level of the water rose to $9.8 \mathrm{cm}^{3} \pm 0.1 \mathrm{cm}^{3} .$ What is the density of the object with appropriate error limits? (See Appendix \(1.5 . )\)

Short Answer

Expert verified
The density of the irregularly shaped object is \(8.50 \frac{g}{cm^3} \pm 0.47 \frac{g}{cm^3}\).

Step by step solution

01

Identify the given values

From the problem, we have the following data: - Mass of the object: \(m = 28.90g \pm 0.03g\) - Initial volume of water: \(V_1 = 6.4cm^3 \pm 0.1cm^3\) - Final volume of water (after object drops in): \(V_2 = 9.8cm^3 \pm 0.1cm^3\)
02

Calculate the volume of the object

To calculate the volume of the object, find the difference between the final and initial water volumes in the graduated cylinder: \(V_{object} = V_2 - V_1\)
03

Find the error of the volume

Since both initial and final volume readings have the same error of \(0.1cm^3\), the error in the volume of the object is the sum of the errors: \(Error_{volume} = Error_{V1} + Error_{V2}\)
04

Calculate the density of the object

Now, we can use the mass and volume of the object to find its density using the formula: \(density = \dfrac{mass}{volume}\)
05

Determine the error limits in density

To estimate the error limits on the calculated density, use the following formula: \(Error_{density} = density \times \sqrt{(\dfrac{Error_{mass}}{mass})^2+(\dfrac{Error_{volume}}{volume})^2}\) Now, by substituting the given values and calculating the results:
06

Calculate the volume of the object

\(V_{object} = 9.8 cm^3 - 6.4 cm^3 = 3.4 cm^3\)
07

Find the error of the volume

\(Error_{volume} = 0.1cm^3 + 0.1cm^3 = 0.2cm^3\)
08

Calculate the density of the object

\(density = \dfrac{28.90g}{3.4 cm^3} = 8.50 \frac{g}{cm^3}\)
09

Determine the error limits in density

\(Error_{density} = 8.50 \frac{g}{cm^3} \times \sqrt{(\dfrac{0.03g}{28.90g})^2+(\dfrac{0.2cm^3}{3.4cm^3})^2} \approx 0.47 \frac{g}{cm^3}\)
10

Write the final result with error limits

The density of the object with appropriate error limits is: \(8.50 \frac{g}{cm^3} \pm 0.47 \frac{g}{cm^3}\)

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