Problem Calculate the resistance of an aluminum cylinder that has a length of 10.0 cm and a cross-sectional area of 2.00 x 10-4 m². Repeat the calculation for a cylinder of the same dimensions and made of glass having a resistivity of 3.0 x 1010 N•m.

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**Example 27.2: The Resistance of a Conductor**

**Problem:**  
Calculate the resistance of an aluminum cylinder that has a length of 10.0 cm and a cross-sectional area of 2.00 x 10⁻⁴ m². Repeat the calculation for a cylinder of the same dimensions and made of glass having a resistivity of 3.0 x 10¹⁰ Ω·m.

**Solution:**

From Equation 27.11 and Table 27.1, we can calculate the resistance of the aluminum cylinder as follows.

\[
R = \frac{\ell}{A} \cdot \rho = (2.82 \times 10^{-8} \, \Omega \cdot \text{m}) \left( \frac{0.100 \, \text{m}}{2.00 \times 10^{-4} \, \text{m}^2} \right)
\]

\[R = \boxed{\text{Enter a number}} \, \Omega\]

**Note:** The number differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully.

Similarly, for glass, we find the following:

\[
R = \frac{\ell}{A} \cdot \rho = (3.0 \times 10^{10} \, \Omega \cdot \text{m}) \left( \frac{0.100 \, \text{m}}{2.00 \times 10^{-4} \, \text{m}^2} \right)
\]

\[R = \boxed{\phantom{0}} \, \Omega\]

As you might guess from the large difference in resistivities, the resistances of identically shaped cylinders of aluminum and glass differ widely. The resistance of the glass cylinder is 18 orders of magnitude greater than that of the aluminum cylinder.
Transcribed Image Text:**Example 27.2: The Resistance of a Conductor** **Problem:** Calculate the resistance of an aluminum cylinder that has a length of 10.0 cm and a cross-sectional area of 2.00 x 10⁻⁴ m². Repeat the calculation for a cylinder of the same dimensions and made of glass having a resistivity of 3.0 x 10¹⁰ Ω·m. **Solution:** From Equation 27.11 and Table 27.1, we can calculate the resistance of the aluminum cylinder as follows. \[ R = \frac{\ell}{A} \cdot \rho = (2.82 \times 10^{-8} \, \Omega \cdot \text{m}) \left( \frac{0.100 \, \text{m}}{2.00 \times 10^{-4} \, \text{m}^2} \right) \] \[R = \boxed{\text{Enter a number}} \, \Omega\] **Note:** The number differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. Similarly, for glass, we find the following: \[ R = \frac{\ell}{A} \cdot \rho = (3.0 \times 10^{10} \, \Omega \cdot \text{m}) \left( \frac{0.100 \, \text{m}}{2.00 \times 10^{-4} \, \text{m}^2} \right) \] \[R = \boxed{\phantom{0}} \, \Omega\] As you might guess from the large difference in resistivities, the resistances of identically shaped cylinders of aluminum and glass differ widely. The resistance of the glass cylinder is 18 orders of magnitude greater than that of the aluminum cylinder.
**Exercise 27.2**

**Hints: Getting Started | I'm Stuck**

What if the same amount of aluminum is used to make a cylinder with **4.0** times the original length. What will be the resistance of the resulting cylinder (in units of μΩ)?

[Input Box] μΩ
Transcribed Image Text:**Exercise 27.2** **Hints: Getting Started | I'm Stuck** What if the same amount of aluminum is used to make a cylinder with **4.0** times the original length. What will be the resistance of the resulting cylinder (in units of μΩ)? [Input Box] μΩ
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