Electricity: Resistivity, Wire Stretching & Temperature Class 10

Why are household electrical cables made of copper or aluminium, while the glowing heating coils of electric toasters and room heaters are made of nichrome alloy? Why does stretching a metallic wire to double its length cause its electrical resistance to jump four times, rather than merely doubling?

In CBSE Class 10 Science, Chapter 11 (Electricity), the relationship between a conductor's geometric dimensions and its material properties is governed by the resistance formula: $R = \rho \frac{l}{A}$

Board examinations frequently test two major competencies: distinguishing between Resistance ($R$) and Resistivity ($\rho$), and solving wire stretching and reshaping numericals.

In this master guide, we break down the physics of resistivity and solve the most common board numericals.

What You Will Learn

• The 4 physical factors governing electrical resistance: $R = \rho \frac{l}{A}$
• Definition of Electrical Resistivity ($\rho$) and its SI unit ($\Omega \cdot \text{m}$)
• Resistance vs. Resistivity: What changes and what remains constant?
• The Wire Stretching Theorem: Conservation of volume ($V = A \times l$)
• Step-by-step solution: Stretching a wire to double its length ($R' = 4R$)
• Wire compression / doubling upon itself ($R' = R/4$)
• Why Alloys (Nichrome, Constantan) are used in electrical heating appliances

1. Resistance vs. Resistivity

    Parameter               RESISTANCE (R)                              RESISTIVITY (ρ)
    --------------------------------------------------------------------------------------------------
    Definition              Opposition offered by a SPECIFIC conductor  Inherent RESISTANCE PER UNIT LENGTH 
                            to the flow of electric current             and unit cross-sectional area
    SI Unit                 Ohm (Ω)                                     Ohm-metre (Ω·m)
    Depends on              Length (l), Area (A), Material, Temp        MATERIAL and TEMPERATURE ONLY!
    Changes with shape?     YES! (Stretching changes R)                 NO! (ρ is completely CONSTANT!)
    --------------------------------------------------------------------------------------------------

Important: <u>If a metallic wire is cut into two halves, its resistance halves ($R/2$), but its RESISTIVITY ($ho$) REMAINS COMPLETELY UNCHANGED! Resistivity is a characteristic material property that does NOT depend on the dimensions (length or thickness) of the wire!</u>

2. The Wire Stretching Theorem (Conservation of Volume)

When a wire of length $l_1$, cross-sectional area $A_1$, and initial resistance $R_1 = \rho \frac{l_1}{A_1}$ is stretched:

• The total volume of metal remains strictly constant: $\mathbf{V = A_1 \times l_1 = A_2 \times l_2}$
• Therefore, if the wire is stretched so that its length increases by a factor of $n$ ($l_2 = n \times l_1$): $A_1 \times l_1 = A_2 \times (n \, l_1) \implies \mathbf{A_2 = \frac{A_1}{n}}$
• The cross-sectional area decreases by the same factor $n$!

    Original Wire:  Length l, Area A  ───>  Resistance R = ρ (l / A)
    Stretched Wire: Length 2l, Area A/2 ─>  Resistance R' = ρ (2l / (A/2)) = 4 × [ρ (l/A)] = 4R!

Solved Example 1: Stretching a Wire to Double Length (CBSE 3-Mark Classic)

Problem: A wire of resistance $4\ \Omega$ is stretched to double its original length. Calculate its new resistance.

Solution:

1. Initial State:
   • Initial length $= l_1$, Initial area $= A_1$.
   • Initial resistance: $R_1 = \rho \frac{l_1}{A_1} = \mathbf{4\ \Omega}$.
2. Stretched State:
   • New length: $l_2 = 2l_1$.
   • Since volume is conserved ($A_1 l_1 = A_2 l_2$): $A_2 = \frac{A_1 l_1}{l_2} = \frac{A_1 l_1}{2l_1} = \mathbf{\frac{A_1}{2}}$
3. Calculate New Resistance ($R_2$): $R_2 = \rho \frac{l_2}{A_2} = \rho \frac{2l_1}{\frac{A_1}{2}} = \rho \frac{2l_1 \times 2}{A_1} = 4 \times \left(\rho \frac{l_1}{A_1}\right)$
4. Substitute $R_1 = 4\ \Omega$: $R_2 = 4 \times R_1 = 4 \times 4 = \mathbf{16\ \Omega}$
5. Therefore, <u>the new resistance of the stretched wire is $16\ \Omega$</u>.

Solved Example 2: Wire Doubled on Itself (Folding in Half)

Problem: A cylinder of wire of resistance $20\ \Omega$ is doubled on itself. What is the new resistance?

Solution:

1. "Doubling on itself" means folding the wire in half:
   • New length: $l_2 = \mathbf{\frac{l_1}{2}}$.
   • Cross-sectional area doubles (two strands side by side): $A_2 = \mathbf{2A_1}$.
2. Calculate New Resistance ($R_2$): $R_2 = \rho \frac{l_2}{A_2} = \rho \frac{\frac{l_1}{2}}{2A_1} = \frac{1}{4} \times \left(\rho \frac{l_1}{A_1}\right) = \frac{R_1}{4}$ $R_2 = \frac{20}{4} = \mathbf{5\ \Omega}$
3. Therefore, <u>the new resistance is $5\ \Omega$</u>.

3. Why are Alloys Used in Heating Elements?

Electrical heating appliances (electric toasters, irons, geysers, hair dryers) use heating coils made of alloys like Nichrome (Nickel + Chromium + Manganese + Iron) rather than pure metals (like copper):

1. Higher Electrical Resistivity: <u>The resistivity of an alloy is generally much higher than that of its constituent pure metals, producing intense Joule heating ($H = I^2Rt$)!</u>
2. Resistance to High-Temperature Oxidation (Burning): Pure metals oxidize and melt quickly when red-hot. Alloys do not oxidize or burn easily even at scorching red-hot temperatures ($> 800^\circ ext{C}$).
3. High Melting Point: Ensures the element does not melt during continuous operation.

4. Summary and Examination Tips

Exam Tip: Remember the shortcut: when a wire is stretched by a factor $n$, the new resistance is $R' = n^2 R$! (If stretched 3 times $\implies R' = 3^2 R = 9R$; if stretched 4 times $\implies R' = 16R$).

Common Mistake: Assuming area remains unchanged when a wire is stretched. You cannot stretch a wire without making it thinner! The area decreases in exact proportion to length expansion ($V = A \times l = \text{constant}$).