Answer

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**Hint:**Resistance depends on the length, type, thickness, and temperature of the material. Resistivity which is also known as specific resistance of the material is the measure of the resistance of the material. We prefer materials with low resistance for making connecting cords and materials with high resistance to make heating devices.

**Formula used:**

$R \propto \dfrac{l}{A}$ where R stands for the resistance of the material, l stands for the length of the material, and A stands for the area of the material

$R = \rho \dfrac{l}{A}$ where $\rho $ stands for the resistivity of the material.

When$\dfrac{l}{A} = 1$ ,

\[ \Rightarrow \]$R = \rho $

**Step by step Solution:**

The resistance of a material is directly proportional to the length of the material and it is inversely proportional to the area of cross-section. There are other factors also which affect the resistance which is the type of material and the temperature of the material.

List of factors that affect the resistance of a conductor

1) Type of the material

2) Length of the material

3) Thickness/ Area of the material

4) Temperature of the material.

-Material A is the best for making connecting cords.

For making connecting cords materials with low resistance is the best, this will avoid loss of current due to heating.

-Material D is best for making heating devices.

For making heating devices we need materials with very high resistance since it will induce more heat in the material. Since material E has a resistivity in the range of${10^{ - 4}}\Omega - m$, it falls under the category of semiconductors hence material E cannot be used for making heating devices. This makes material D best among the given materials to make a heating device.

-Let ${\rho _1}$ be the resistivity of material A and ${\rho _2}$ be the resistivity of the material D.

The resistance of the material A is given by ${R_1}$.

We have to find the resistance of the material D ${(R_2)} $ . To find the resistance of the material D, we should use the formula for resistance which involves resistivity. The formula is given by

$R = \rho \dfrac{l}{A}$

Using the formula for two resistances, ${R_1}$ and ${R_2}$, and then taking their ratio gives us,

$\dfrac{{{R_1}}}{{{R_2}}} = \dfrac{{{\rho _1}\dfrac{l}{A}}}{{{\rho _2}\dfrac{l}{A}}}$

We know that the length and thickness are same for the material. Hence we get $\dfrac{{\dfrac{l}{A}}}{{\dfrac{l}{A}}} = 1$

Therefore, $\dfrac{{{R_1}}}{{{R_2}}} = \dfrac{{{\rho _1}}}{{{\rho _2}}}$ (Since $\dfrac{{\dfrac{l}{A}}}{{\dfrac{l}{A}}} = 1$ )

${R_2} = \dfrac{{{R_1}{\rho _2}}}{{{\rho _1}}}$$\Omega $

From the equation of resistance that includes resistivity in it, we have deduced an expression for the resistance of the material. That is shown in the above steps. From the ratio of resistances, we have to write an equation for the resistance to be found which in this case is${R_2}$. After finding the equation substitute the values of${R_1},{\rho _1},{\rho _2}$

${R_1} = 2\Omega $ (Given in the question)

${\rho _1} = 1.6 \times {10^{ - 8}}\Omega - m$ (Given in the question)

${\rho _2} = 96 \times {10^{ - 8}}\Omega - m$ (Given in the question)

${R_2} = \dfrac{{2 \times 96 \times {{10}^{ - 8}}}}{{1.6 \times {{10}^{ - 8}}}}\Omega $

${R_2} = 120\Omega $

**Note:**

Keep in mind that the order of resistivity of conductors is in the range ${10^{ - 8}}$$\Omega m$.The resistivity of semiconductors (example: Silicon, Germanium, Carbon, Graphite etc.) are in the range ${10^{ - 5}} - {10^3}\Omega m$ and the resistivity of bad conductors are in the range ${10^{14}} - {10^{16}}\Omega m$. Pay attention to power while solving problems.

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