Chapter 1, Problem 51P

(a)

To determine

The position of fourth experimental point from the top from the best fit straight line as the difference in y-axis co-ordinate.

Answer to Problem 51P

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

Explanation of Solution

The experimental point is denoted by a circle and the point under consideration is the one which lies near the point $\left(\text{400}{\text{cm}}^{\text{2}}\text{,0}\text{.20}\text{g}\right)$. It can be seen from the graph that the circle is vertically separated by 0.3 spaces.

Write the equation to find mass equivalent from vertical separation.

    $\Delta y=n\left(0.05\text{g}\right)$

Here, $\Delta y$ is the vertical separation and $n$ is the number of unit columns.

Conclusion:

Substitute $0.3$ for $n$ in the above equation to find $\Delta y$.

  $\begin{array}{c}\Delta y=\left(0.3\right)\left(0.05\text{g}\right)\\ =0.015\text{g}\end{array}$

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

(b)

To determine

Find the percentage equivalent of part (a).

Answer to Problem 51P

Percentage equivalent is $8\%$.

Explanation of Solution

The y-coordinate of point on best fit line so closer to the fourth experimental point from the top is $0.20\text{g}$.

Write the expression to calculate the percentage equivalent of vertical difference between the point and curve fit line.

    $p=\frac{\Delta y}{y_{\text{point}}}\times 100\%$

Here, $p$ is the percentage equivalent and $y_{\text{point}}$ is the y-coordinate of point on best fit line so closer to the fourth experimental point from the top

Conclusion:

Substitute $0.015\text{g}$ for $\Delta y$ and $0.20\text{g}$ for $y_{\text{point}}$ in the above equation to find $p$.

  $\begin{array}{c}p=\frac{0.015\text{g}}{0.20\text{g}}\times 100\%\\ =8\%\end{array}$

Therefore, the percentage equivalent is $8\%$.

(c)

To determine

The slope of line.

Answer to Problem 51P

The slope of line is $5.2\times {10}^{-3}\text{g}/{\text{cm}}^{\text{3}}$.

Explanation of Solution

To find the slope, consider the points $\left(\text{0}{\text{cm}}^{\text{2}}\text{,0}\text{g}\right)$ and $\left(600{\text{cm}}^2,3.1\text{g}\right)$ which line passes.

Write the expression to find the slope.

    $m=\frac{y_2-y_1}{x_2-x_1}$

Here, $m$ is the slope, $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ are the points.

Conclusion:

Substitute $3.1\text{g}$ for $y_2$, $0\text{g}$ for $y_1$, $600{\text{cm}}^2$ for $x_2$, and $\text{0}{\text{cm}}^{\text{2}}$ for $x_1$ in the above equation to find $m$.

    $\begin{array}{c}m=\frac{3.1\text{g}-0\text{g}}{600{\text{cm}}^2-\text{0}{\text{cm}}^{\text{2}}}\\ =5.2\times {10}^{-3}\text{g}/{\text{cm}}^{\text{3}}\end{array}$

Therefore, the slope of line is $5.2\times {10}^{-3}\text{g}/{\text{cm}}^{\text{3}}$.

(d)

To determine

The inferences from the shape of graph and results of part (a) and (b).

Answer to Problem 51P

The shape of graph tells that the mass is directly proportional to area and the value of proportionality constant is $\text{5}{\text{.2 g/cm}}^{\text{2}}\pm 8\%$.

Explanation of Solution

The graph is a straight line with mass on y-axis and area on x-axis. The straight line indicates that the quantity plotted on y-axis is linearly proportional to the quantity plotted on x-axis. The proportionality constant will be the slope. It is found that in part (c), the slope is $\text{5}{\text{.2 g/cm}}^{\text{2}}$. The separation between experimental values from the fitting curve shows that there exists some uncertainty in the value of slope. Extend of uncertainty is given by the vertical separation between point and fitting curve. It is found as $8\%$ in part (b).

Therefore, the shape of graph tells that the mass is directly proportional to area and the value of proportionality constant is $\text{5}{\text{.2 g/cm}}^{\text{2}}\pm 8\%$.

(e)

To determine

Check whether the result of part (d) can be expected theoretically or not.

Answer to Problem 51P

The result of part (d) holds theoretically only if the density and thickness of paper is uniform in the limits of experimental uncertainty.

Explanation of Solution

For result in part (d) to hold good by theory also, should satisfy certain conditions. One of the conditions is that the paper must be of uniform thickness, in addition, mass distribution in unit area must be also uniform. Moreover that, both the two mentioned quantities must be within the obtained uncertainty limit.

Therefore, the result of part (d) holds theoretically only if the density and thickness of paper is uniform in the limits of experimental uncertainty.

(f)

To determine

Explain the physical significance of slope.

Answer to Problem 51P

Slope gives the areal density of paper.

Explanation of Solution

In the graph, the mass is plotted along vertical axis and area is plotted long horizontal axis. So the slope actually tells about the mass distribution of paper. In other words, it is called as areal density.

Therefore, the slope of gives the areal density of paper.