Chapter 1, Problem 1.90AP

(a)

Interpretation Introduction

Interpretation: The masses, height and diameter of the two metal cylinders. The volume of each cylinder using its dimension and by water displacement method is to be calculated. Also, the volume that allows the greater number of the significance number in the calculated density and the density with the appropriate number of significance number is to be determined.

Concept introduction: The volume of the cylinder is calculated by the formula,

$\text{Volume}=\text{Height}\times \text{π}\times {\left(\text{Radius}\right)}^2$ The density of the cylinder is calculated by the formula,

$\text{Density}=\frac{\text{Mass}}{\text{Volume}}$

The rule for the significant figures in division states that the significant figures in the answer is equal to the least significant figure involved of the number in the division.

To determine: The volume of each cylinder using the dimensions of the cylinder.

Answer to Problem 1.90AP

Solution

The volume of the cylinder A and cylinder B using the dimensions of the cylinder is $\underset{\_}{5.76504c{m}^3}$ and $\underset{\_}{7.827235c{m}^3}$, respectively.

Explanation of Solution

Explanation

Given

The height of the cylinder A is $5.1\text{cm}$.

The height of the cylinder B is $5.9\text{cm}$.

The diameter of the cylinder A is $1.2\text{cm}$. Therefore, the radius is $0.6\text{cm}$.

The height of the cylinder B is $1.3\text{cm}$. Therefore, the radius is $0.65\text{cm}$.

The volume of the cylinder is calculated by the formula,

$\text{Volume}=\text{Height}\times \text{π}\times {\left(\text{Radius}\right)}^2$ (1)

Substitute the value of height and the radius of cylinder A in equation (1).

$\begin{array}{c}\text{Volume}\text{of}\text{cylinder}\text{A}=5.1\text{cm}\times \text{3}\text{.14}\times {\left(0.6\text{cm}\right)}^2\\ =5.76504{\text{cm}}^{\text{3}}\end{array}$

Substitute the value of height and the radius of cylinder B in equation (1).

$\begin{array}{c}\text{Volume}\text{of}\text{cylinder}\text{B}=5.9\text{cm}\times \text{3}\text{.14}\times {\left(0.65\text{cm}\right)}^2\\ =7.827235{\text{cm}}^{\text{3}}\end{array}$

Therefore, the volume of cylinder A and cylinder B is $\underset{\_}{5.76504c{m}^3}$ and $\underset{\_}{7.827235c{m}^3}$, respectively.

(b)

Interpretation Introduction

To determine: The volume of each cylinder using water displacement method.

Answer to Problem 1.90AP

Solution

The volume of the cylinder A and cylinder B using water displacement method is $\underset{\_}{5c{m}^3}$ and $\underset{\_}{8c{m}^3}$, respectively.

Explanation of Solution

Explanation

Given

The original volume of the cylinder A and cylinder B containing water is $25\text{mL}$.

The rise in the level of water on immersing the metallic cylinder in the water containing cylinder is shown in the given figure (Refer question).

The displacement of water in cylinder A is $30\text{mL}$ and that in cylinder B is $33\text{mL}$.

According to the water displacement method, the formula for the volume of the cylinder is given as,

$\text{Volume}\text{of}\text{metal}\text{cylinder}=\left(\begin{array}{c}\text{Volume}\text{of}\text{water}\text{cylinder}\\ -\text{Volume}\text{of}\text{water}\text{cylinder}\end{array}\right)$ (2)

Substitute the volume of the cylinder A before and after the immersion of the metal cylinder I equation (2)

$\begin{array}{c}\text{Volume}\text{of}\text{metal}\text{cylinder}\text{A}=\left(\text{30}\text{mL}-\text{25}\text{mL}\right)\\ =5\text{mL}\end{array}$

Substitute the volume of the cylinder B before and after the immersion of the metal cylinder I equation (2)

$\begin{array}{c}\text{Volume}\text{of}\text{metal}\text{cylinder}\text{B}=\left(\text{33}\text{mL}-\text{25}\text{mL}\right)\\ =8\text{mL}\end{array}$

Therefore, the volume of cylinder A and cylinder B is $\underset{\_}{5c{m}^3}$ and $\underset{\_}{8c{m}^3}$, respectively.

(c)

Interpretation Introduction

To determine: The calculated volume that allows for the greater number of significant figures in the calculated densities.

Answer to Problem 1.90AP

Solution

The density calculated using the volumes of the cylinders calculated using the dimensions of the cylinder gives greater significant figure.

Explanation of Solution

Explanation

The density of the cylinder is calculated by the formula,

$\text{Density}=\frac{\text{Mass}}{\text{Volume}}$

The rule for the significant figures in division states that the significant figures in the answer is equal to the least significant figure involved of the number in the division.

The mass of the cylinder A and cylinder B is $15.560\text{g}$ and $35.536\text{g}$, respectively. Therefore, the masses of both the cylinder have five significant figures. The volume of cylinder A and cylinder B calculated using the dimensions of the cylinder is $5.76504{\text{cm}}^{\text{3}}$ and $7.827235{\text{cm}}^3$, respectively. Therefore, the volume of cinder A and cylinder B have six and seven significant figures, respectively.

Thus, on dividing the mass of the cylinder using the value determined using the dimensions gives the answer with five significant figures.

The volume of cylinder A and cylinder B calculated using the water displacement method is $5{\text{cm}}^{\text{3}}$ and $8{\text{cm}}^3$, respectively. Thus, the volume of the cylinders has one significant figure.

Therefore, on dividing the mass of the cylinders using the value determined using the water displacement method gives the answer with one significant figure.

Hence, the density calculated using the volumes of the cylinders calculated using the dimensions of the cylinder gives greater significant figure.

(d)

Interpretation Introduction

To determine: The densities of the each cylinder with appropriate significant figures.

Answer to Problem 1.90AP

Solution

The density of cylinder A and cylinder B using the volume calculated by dimensions is $\underset{\_}{2.6990g/c{m}^3}$ and $\underset{\_}{4.5400g/c{m}^3}$, respectively.

The density of cylinder A and cylinder B using the volume calculated by water displacement method is $\underset{\_}{3g/c{m}^3}$ and $\underset{\_}{4g/c{m}^3}$, respectively.

Explanation of Solution

Explanation

Given

The mass of the cylinder A and cylinder B is $15.560\text{g}$ and $35.536\text{g}$, respectively.

The volume of cylinder A and cylinder B calculated using the dimensions of the cylinder is $5.76504{\text{cm}}^{\text{3}}$ and $7.827235{\text{cm}}^3$, respectively.

The density of the cylinder is calculated by the formula,

$\text{Density}=\frac{\text{Mass}}{\text{Volume}}$ (3)

The rule for the significant figures in division states that the significant figures in the answer is equal to the least significant figure involved of the number in the division.

The mass of the cylinder A and cylinder B is $15.560\text{g}$ and $35.536\text{g}$, respectively. Therefore, the masses of both the cylinder have five significant figures. The volume of cylinder A and cylinder B calculated using the dimensions of the cylinder is $5.76504{\text{cm}}^{\text{3}}$ and $7.827235{\text{cm}}^3$, respectively. Therefore, the volume of cinder A and cylinder B have six and seven significant figures, respectively.

Thus, on dividing the mass of the cylinder using the value determined using the dimensions gives the answer with five significant figures.

Substitute the value of mass and volume of cylinder A in equation (3).

$\text{Density}\text{of}\text{cylinder}\text{A}=\frac{15.560\text{g}}{5.76504{\text{cm}}^{\text{3}}}=2.6990\text{g}/{\text{cm}}^{\text{3}}$

Substitute the value of mass and volume of cylinder B in equation (3).

$\text{Density}\text{of}\text{cylinder}\text{B}=\frac{35.536\text{g}}{7.827235{\text{cm}}^3}=4.5400\text{g}/{\text{cm}}^{\text{3}}$

Therefore, the density of cylinder A and cylinder B using the volume calculated by dimensions is $\underset{\_}{2.6990g/c{m}^3}$ and $\underset{\_}{4.5400g/c{m}^3}$, respectively.

The volume of cylinder A and cylinder B calculated using the water displacement method is $5{\text{cm}}^{\text{3}}$ and $8{\text{cm}}^3$, respectively. Thus, the volume of the cylinders has one significant figure.

Therefore, on dividing the mass of the cylinders using the value determined using the water displacement method gives the answer with one significant figures.

Substitute the value of mass and volume of cylinder A in equation (3).

$\text{Density}\text{of}\text{cylinder}\text{A}=\frac{15.560\text{g}}{5{\text{cm}}^{\text{3}}}=3\text{g}/{\text{cm}}^{\text{3}}$

Substitute the value of mass and volume of cylinder B in equation (3).

$\text{Density}\text{of}\text{cylinder}\text{B}=\frac{35.536\text{g}}{8{\text{cm}}^3}=4\text{g}/{\text{cm}}^{\text{3}}$

Therefore, the density of cylinder A and cylinder B using the volume calculated by water displacement method is $\underset{\_}{3g/c{m}^3}$ and $\underset{\_}{4g/c{m}^3}$, respectively.

Conclusion

1. a. The volume of the cylinder A and cylinder B using the dimensions of the cylinder is $\underset{\_}{5.76504c{m}^3}$ and $\underset{\_}{7.827235c{m}^3}$, respectively.
2. b. The volume of the cylinder A and cylinder B using water displacement method is $\underset{\_}{5c{m}^3}$ and $\underset{\_}{8c{m}^3}$, respectively.
3. c. The density calculated using the volumes of the cylinders calculated using the dimensions of the cylinder gives greater significant figure.
4. d. The density of cylinder A and cylinder B using the volume calculated by dimensions is $\underset{\_}{2.6990g/c{m}^3}$ and $\underset{\_}{4.5400g/c{m}^3}$, respectively.

   The density of cylinder A and cylinder B using the volume calculated by water displacement method is $\underset{\_}{3g/c{m}^3}$ and $\underset{\_}{4g/c{m}^3}$, respectively