Chapter 1, Problem 1.51P

Review. A student is supplied with a stack of copy paper, ruler, compass, scissors, and a sensitive balance. He cuts out various shapes in various sizes, calculates their areas, measures their masses, and prepares the graph of Figure PI.51. (a) Consider the fourth experimental point from the top. How far is it from the best-lit straight line? Express your answer as a difference in vertical-axis coordinate, (b) Express your answer as a percentage, (c) Calculate the slope of the line, (d) State what the graph demonstrates, referring to the shape of the graph and the results of parts (b) and (c). (e) Describe whether this result should be expected theoretically, (f) Describe the physical meaning of the slope.

To determine

The distance of the fourth experimental point from the top from the best –fit straight line.

Answer to Problem 1.51P

The distance of the fourth experimental point from the top from the best fit straight line is $0.015\text{g}$.

Explanation of Solution

Given data: The student is supplied with a stack of copy paper, ruler, compass, scissors, and a sensitive balance. A graph mass versus area is plotted for different sizes of the paper.

Consider the following figure.

Figure (1)

Figure indicates the graph plotted area of pieces versus mass of the pieces of the paper.

The fourth experimental point from the top is a circle. It lies slightly above the best fit line.

From figure (1), the vertical coordinate for the forth experiment is $0.2025\text{g}$ and the vertical coordinate for the point on best fit line at which it touches the best fit line is $0.1875\text{g}$.

So the difference in the vertical axis coordinate is,

$D=P_1-P_2$

Here,

$P_1$ is the vertical coordinate of the fourth experiment.

$P_2$ is the vertical coordinate of the point on best fit line at which it touches the best fit line.

Substitute $0.2025\text{g}$ for $P_1$ and $0.1875\text{g}$ for $P_2$ in above expression.

$\begin{array}{c}d=0.2025\text{g}-0.1875\text{g}\\ =0.015\text{g}\end{array}$

Conclusion:

Therefore the distance of the fourth experimental point from the top from the best fit straight line is $0.015\text{g}$.

To determine

The answer in the form of percentage.

Answer to Problem 1.51P

The answer in the form of percentage is $8\%$.

Explanation of Solution

Given data: The student is supplied with a stack of copy paper, ruler, compass, scissors, and a sensitive balance. A graph mass versus area is plotted for different sizes of the paper.

The expression for the percentage is,

$P=\frac{D}{P_1}\times 100$

Substitute $0.015\text{g}$ for $D$ and $0.1875\text{g}$ for $P_1$ in above expression.

$\begin{array}{c}P=\frac{0.015\text{g}}{0.1875\text{g}}\times 100\\ =8\%\end{array}$

Conclusion:

Therefore the answer in the form of percentage is $8\%$.

To determine

The slope of the line.

Answer to Problem 1.51P

The slope of the line is $5.2\text{g}/{\text{m}}^{\text{2}}$.

Explanation of Solution

Given data: The student is supplied with a stack of copy paper, ruler, compass, scissors, and a sensitive balance. A graph mass versus area is plotted for different sizes of the paper.

The expression for the slope is,

$S=\frac{y_2-y_1}{x_2-x_1}$

Here,

$y_1$ is the first vertical coordinate.

$y_2$ is the second vertical coordinate.

$x_1$ is the first vertical coordinate.

$x_2$ is the second vertical coordinate.

Substitute $0.2\text{g}$ for $y_2$, $0.1\text{g}$ for $y_1$, $392{\text{cm}}^2$ for $x_2$ and $200{\text{cm}}^2$ for $x_1$ in above expression.

$\begin{array}{c}S=\frac{0.2\text{g}-0.1\text{g}}{392{\text{cm}}^2-200{\text{cm}}^2}\\ =5.2\times {10}^{-4}\text{g}/{\text{cm}}^{\text{2}}\times \frac{10000{\text{cm}}^{\text{2}}}{1{\text{m}}^{\text{2}}}\\ =5.2\text{g}/{\text{m}}^{\text{2}}\end{array}$

Conclusion:

Therefore the slope of the line is $5.2\text{g}/{\text{m}}^{\text{2}}$.

To determine

The demonstration from the graph referring to the shape of the graph and the results of part (b) and (c).

Answer to Problem 1.51P

The graph demonstrates that the mass of the cutout is proportional to its area for the shape cuts from this copy paper and the proportionality constant is $5.2\text{g}/{\text{m}}^{\text{2}}\pm 8\%$ where the uncertainty is estimated.

Explanation of Solution

Given data: The student is supplied with a stack of copy paper, ruler, compass, scissors, and a sensitive balance. A graph mass versus area is plotted for different sizes of the paper.

The graph given figure (1) demonstrates that the mass of the cutout in each shape is proportional to its area for the shape cuts from this copy paper and the proportionality constant is $5.2\text{g}/{\text{m}}^{\text{2}}\pm 8\%$ where the $8\%$ uncertainty is estimated in the aerial mss density of each shape.

The value of the slope from part (c) of the question and from the part (b) question the percentage uncertainty is $8\%$.

Conclusion:

Therefore the graph demonstrates that the mass of the cutout is proportional to its area for the shape cuts from this copy paper and the proportionality constant is $5.2\text{g}/{\text{m}}^{\text{2}}\pm 8\%$ where the uncertainty is estimated.

To determine

Whether this result should be expected theoretically or not.

Answer to Problem 1.51P

This result is to be expected theoretically if the paper has thickness and density that are uniform within the experimental uncertainty.

Explanation of Solution

Given data: The student is supplied with a stack of copy paper, ruler, compass, scissors, and a sensitive balance. A graph mass versus area is plotted for different sizes of the paper.

The graph given figure (1) demonstrates that the mass of the cutout in each shape is proportional to its area for the shape cuts from this copy paper and the proportionality constant is $5.2\text{g}/{\text{m}}^{\text{2}}\pm 8\%$ where the $8\%$ uncertainty is estimated in the aerial mss density of each shape.

The value of the slope from part (c) of the question is $5.2\text{g}/{\text{m}}^{\text{2}}$ and from the part (b) question the percentage uncertainty is $8\%$.

Thus this result is expected theoretically when the object is having same aerial mass density and the paper has thickness and density that are uniform within the experimental uncertainty.

Conclusion:

Therefore this result is to be expected theoretically if the paper has thickness and density that are uniform within the experimental uncertainty.

To determine

The physical meaning of the slope.

Answer to Problem 1.51P

The physical meaning of the slope is the aerial density of the paper that represents the mass per unit area.

Explanation of Solution

Given data: The student is supplied with a stack of copy paper, ruler, compass, scissors, and a sensitive balance. A graph mass versus area is plotted for different sizes of the paper.

The graph given figure (1) demonstrates that the mass of the cutout in each shape is proportional to its area for the shape cuts from this copy paper and the proportionality constant is $5.2\text{g}/{\text{m}}^{\text{2}}\pm 8\%$ where the $8\%$ uncertainty is estimated in the aerial mss density of each shape.

The value of the slope from part (c) of the question is $5.2\text{g}/{\text{m}}^{\text{2}}$ and from the part (b) question the percentage uncertainty is $8\%$.

The physical meaning of the slope is the aerial density of the paper that represent the mass per unit area.

Conclusion:

Therefore the physical meaning of the slope is the aerial density of the paper that represents the mass per unit area.