Q27P - Fundamentals Of Physics

Question

Two conductors are made of the same material and have the same length. Conductor A is a solid wire of diameter . Conductor B is a hollow tube of outside diameter 1.0 mm and inside diameter 1.0mm. What is the resistance ratio ${\mathit{R}}_{\mathit{A}}\mathit{/}{\mathit{R}}_{\mathit{B}}$, measured between their ends?

Step-by-Step Solution

Answer

The resistance ratio $R_A\mathit{/}{R}_B$ measured between the ends of the two conductors is 3.

Step 1: Identification given data

1. Diameter of wire A, ${\mathrm{D}}_{\mathrm{A}}=1\mathrm{mm}$
2. Outer and inner diameters of wire B,${\mathrm{D}}_{\mathrm{Bo}}=2.0 \mathrm{mm}, \mathrm{and} {\mathrm{D}}_{\mathrm{Bi}}=1.0 \mathrm{mm}$

Step 2: Significance of resistance

The resistivity is the opposition to the flow of current. The resistance of the wire is directly proportional to its length and inversely proportional to the area of the cross-section. The constant of proportionality is a characteristic of that material called resistivity.

We have to use the formula of resistance for wire A and wire B and take their ratio to find the required resistance ratio.

Formulae:

The resistance of the material, $R=\frac{pL}{A}$                                                             …(i)

Here, R is resistance, p is resistivity, L is the length, A and is the area of the cross-section of the wire.

 

The cross-sectional area of the wire, $A=\pi r^2$                                                   …(ii)

 A is the area of the cross-section of the wire, r is the area of cross-section.

Step 3: Determining the ratio of the resistances

For wire A, the resistance of the wire can be calculated using equation (ii) in equation (i) as follows:

 $R_A=\frac{pL}{\pi r_A^2}$                                                                                                         …(iii)

For wire B, the resistance of the wire can be calculated using equation (ii) in equation (i) as follows:

$R_B=\frac{pL}{\pi \left(r_{B0}^2-r_{Bi}^2\right)}$                                                                                               …(iv)

But, the values of the radius of wire A, inner radius of B and outer radius of B can be calculated as follows:

$$\begin{gathered} {\mathrm{r}}_{\mathrm{A}}=\frac{{\mathrm{D}}_{\mathrm{A}}}{2} \\ =\frac{1\mathrm{mm}}{2} \\ =0.50\mathrm{mm} \end{gathered}$$

$$\begin{gathered} {\mathrm{r}}_{\mathrm{B}}=\frac{{\mathrm{D}}_{\mathrm{Bi}}}{2} \\ =\frac{1.0\mathrm{mm}}{2} \\ =0.50\mathrm{mm} \end{gathered}$$

$$\begin{gathered} {\mathrm{r}}_{\mathrm{Bo}}=\frac{{\mathrm{D}}_{Bo}}{2} \\ =\frac{2.0\mathrm{mm}}{2} \\ =1.0\mathrm{mm} \end{gathered}$$

Dividing equation (iii) by equation (iv) and using the above values, we can get the required value of the resistance ratio as follows:

$$\begin{gathered} \frac{{\mathrm{R}}_{\mathrm{A}}}{{\mathrm{R}}_{\mathrm{B}}}=\frac{{\displaystyle \frac{\frac{\mathrm{pL}}{{\mathrm{\pi r}}_{\mathrm{A}}^2}}{\mathrm{pL}}}}{\mathrm{\pi }\left({\mathrm{r}}_{\mathrm{Bo}}^2-{\mathrm{r}}_{\mathrm{Bi}}^2\right)} \\ =\frac{\left({\mathrm{r}}_{\mathrm{Bo}}^2-{\mathrm{r}}_{\mathrm{Bi}}^2\right)}{{\mathrm{r}}_{\mathrm{A}}^2} \\ =\frac{{\left(1.00\mathrm{mm}\right)}^2-{\left(0.50\mathrm{mm}\right)}^2}{{\left(0.50\mathrm{mm}\right)}^2} \\ =3.0 \end{gathered}$$

Therefore, the resistance ratio $R_A/R_B$ measured between the ends of the two conductors is 3.0 .