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Q59P

Expert-verifiedFound in: Page 769

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**A cylindrical metal rod is 1.60 m**** long and 5.50 m**** in diameter. The resistance between its two ends (at ${\mathbf{20}}{\mathbf{\xb0}}$****) is ${\mathbf{1}}{\mathbf{.}}{\mathbf{09}}{\mathbf{\times}}{{\mathbf{10}}}^{\mathbf{-}\mathbf{3}}{\mathit{\Omega}}$ ****. (a) What is the material? (b) A round disk,**** 2.00 cm in diameter and 1.00 mm**** thick, is formed of the same material. What is the resistance between the round faces, assuming that each face is an equi-potential surface?**

a) The material is silver.

b) The resistance between the round faces of the disk is $5.16\times {10}^{-8}\Omega $ .

The given data can be listed below as:

- Length of the rod is, L = 1.60 m .
- Diameter of the rod is, $d=5.50mm=5.5\times {10}^{-3}m$ .
- Resistance of the rod is, $R=1.09\times {10}^{-3}\Omega $ .
- Diameter of the round disk is, $d=2.00cm=0.02m$ .
- Thickness of the disk is, $t=1.0mm=1.00\times {10}^{-3}m$ .
- Material and hence the resistivity of the disk is same as that of rod.

**Here, we need to use the equation relating to resistance and resistivity. By using that equation, we can find the type of material and then the resistance between the faces of the disk.**

Formulae:

The cross-sectional area of the circle is,

$A=\mathrm{\pi}\frac{{\mathrm{d}}^{2}}{4}$ … (i)

The resistance of a material is,

$R=\rho \frac{L}{A}$ … (ii)

The resistivity of a material is,

$\rho =\frac{RA}{L}$ … (iii)

Area of cross section of the rod can be calculated using the given data in equation (i) as follows:

$A=3.14\times {\left(\frac{5.5\times {10}^{-3}m}{4}\right)}^{2}\phantom{\rule{0ex}{0ex}}=23.74\times {10}^{-6}{m}^{2}$

Using value of area and all given values in equation (iii), we can get the resistivity of the material as follows:

role="math" localid="1661418391967" $\rho =\frac{23.74\times {10}^{-6}{m}^{2}\times 1.09\times {10}^{-3}\Omega}{1.60m}\phantom{\rule{0ex}{0ex}}=1.62\times {10}^{-8}\Omega .m$

This is the resistivity of the silver.

Therefore, the material of the rod is silver.

As we know, that rod and disk are of same material. Thus, the resistivity of the disk is given by, $\rho =1.62\times {10}^{-8}\Omega .m$ .

Distance between two faces is the thickness of the disk.

Thus, the value of the length is given by, $L=1\times {10}^{-3}m$ .

Now, the area of the round disk can be calculated using the diameter value in equation (i) as follows:

$A=3.14\times \frac{{\left(0.02m\right)}^{2}}{4}\phantom{\rule{0ex}{0ex}}=3.14\times {10}^{-4}{m}^{2}\phantom{\rule{0ex}{0ex}}$

Using value of area and all the given values in formula of equation (ii), we can get the value of the resistance between the two ends of the rod as follows:

$R=\frac{1.62\times {10}^{-8}\Omega .m\times 1\times {10}^{-3}m}{3.14\times {10}^{-4}{m}^{2}}\phantom{\rule{0ex}{0ex}}=5.16\times {10}^{-8}\Omega $

Hence, the value of the resistance is $5.16\times {10}^{-8}\Omega $ .

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