Q80P - Fundamentals Of Physics

Question

When a metal rod is heated not only its resistance, but also its length and cross-sectional area is changed. The relation $\mathit{R}\mathbf{=}\mathit{\rho }\mathit{L}\mathbf{/}\mathit{A}$ suggests that all three factors should be taken into account in measuring r at various temperatures. If the temperature changes by $\mathbf{1}\mathbf{.}{\mathbf{0}}^{\mathbf{0}}\mathit{C}$, what percentage changes in (a) L, (b) A, and (c) R occur for a copper conductor? (d) What conclusion do you draw? The coefficient of linear expansion is $\mathbf{1}\mathbf{.}\mathbf{70}\mathit{x}{\mathbf{10}}^{\mathbf{-}\mathbf{5}}{\mathbf{K}}^{\mathbf{-}\mathbf{1}}$.

Step-by-Step Solution

Answer

Percentage change in

1. Length of the metal rod is 0.0017%.   
2. Area of the metal rod is 0.0034%.
3. Resistance of the metal rod is 0.43%.
4.  The conclusion drawn here is the change in the length and area has a very small effect on resistance than the changes in the resistivity. This can be seen from the fact that fractional change in the length and area is much smaller than the fractional change in resistivity.

Step 1: The given data

1. For copper, the coefficient of linear expansion, ${\alpha }_L=1.70x{10}^{-5 }{K}^{-1}$
2. Change in temperature, $∆T=1^0\mathrm{C}$

Step 2: Understanding the concept of the percentage change

By using the coefficient of thermal expansions and the relation between resistivity and the temperature, we can calculate the percentage change in length, area, and resistance of the rod.

 

Formulae:

1. Percentage change in length $\frac{∆L}{L}\times 100={\alpha }_L∆T$
2. Percentage change in length, $\frac{∆A}{A}\times 100={\alpha }_A∆T$
3. Change in resistivity with temperature $\frac{\rho -{\rho }_0}{{\rho }_0}\times 100=\alpha ∆T$
4. Resistance of rod or a wire  $R=\rho \frac{t}{A}$

Step 3: a) Calculation of the percentage change in length

Using the given values in equation (i), we can get the change in the percentage of the length as follows:

$$\begin{gathered} \frac{∆L}{L}=1.70\times {10}^{-5}\times 1 \\ =1.7\times {10}^{-5} \\ \frac{∆L}{L}\times 100=0.0017\% \end{gathered}$$

Since the value of $∆T$ is the difference in the temperature, it would be the same for Kelvin as well as degree Celsius. So units of coefficient of expansion and temperature are consistent.

Hence, the percentage change in the length of the rod is 0.0017%.

Step 4: b) Calculation of the percentage change in area

Using ${\alpha }_A=2{\alpha }_L$ and equation (ii), the percentage change in area can be given as follows: 

$$\begin{gathered} \frac{∆A}{A}=2\times 1.7\times {10}^{-5}\times 1 \\ =3.4\times {10}^{-5} \\ \frac{∆A}{A}\times 100=0.0034\% \end{gathered}$$

Hence, the percentage change in the area of the rod is 0.0034%.

Step 5: c) Calculation of the percentage change in resistance

The value of resistance depends on length, resistivity, and area of the conductor. So a very small change in the value of these would affect the value of resistance as well.

We can write the change in the resistance as follows:

$∆R=\frac{\partial R}{\partial \rho }∆\rho +\frac{\partial R}{\partial L}∆L+\frac{\partial R}{\partial A}∆A.....................\left(a\right)$

Now, the above terms can be calculated using equation (iv) as follows:

$\frac{aR}{a\rho }=\frac{L}{A}$

$$\begin{gathered} \frac{\partial R}{\partial L}=\frac{\rho }{A} \\ =\frac{R}{L} \end{gathered}$$

And,

$$\begin{gathered} \frac{\partial R}{\partial A}=\frac{\rho L}{A^2} \\ =-\frac{R}{A} \end{gathered}$$

($\alpha$ Is the temperature coefficient of resistance for copper.)

Now, substituting the above value sin equation (a), we ge the change in resistance equation as follows:

$$\begin{gathered} \frac{∆R}{R}=\frac{∆\rho }{\rho }+\frac{∆L}{L}-\frac{∆A}{A} \\ =\left(\alpha +{\alpha }_L-2{\alpha }_L\right)∆T \left(from equation \left(i\right),\left(ii\right) and \left(iii\right)\right) \\ =\left(\alpha -{\alpha }_L\right)∆T \\ =\left(4.3\times {10}^{-3} {/}^0 C-1.7\times {10}^{-5} {/}^0 C\right)\left(1.0^{0}C\right) \\ =4.3\times {10}^{-3} \\ \frac{∆R}{R}\times 100=0.43\% \end{gathered}$$

Hence, the value of the percentage change in resistance is 0.43%.

Step 6: d) Calculation of the reason to conclude the case

The changes in the length and area have a very small effect on resistance than the changes in the resistivity. This can be seen from the fact that fractional change in the length and area is much smaller than the fractional change in resistivity.