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**(a) **State four factors on which the resistance of a conductor depends.

**(b)**A piece of wire of resistance 20Ω is drawn out so that its length increases to twice its original length. Calculate the resistance of the wire in the new situation.

(a) Four factors on which the resistance of a conductor depends are as follows:

**1. **Nature of the material of the conductor.

**2.** Length of the conductor.

**3.** Area of the cross-section of a conductor.

**4.** Resistivity or temperature of a conductor.

(b) Given,

Resistance, **R** = 20 Ω

** $R=\frac{\rho \times L}{A}=20\Omega $**

Here, R = Resistance, ρ = Resistivity, L = Length and A = Area

Let, the Length of wire is **L **and when it is drawn out, its **new length **will be **2L**.

In both situations, the volume of wire remains constant.

So, as the length of the wire gets doubled **(2L),** the area of the cross-section of wire becomes half **$\left(\frac{A}{2}\right)$ **because, from the equation of resistance, we know that, the length and area of the wire are inversely proportional to each other.

Now,

New Length = 2L, and New Area =** $\frac{A}{2}$**

Then, the new resistance of wire will be-

**${R}_{new}=\frac{\rho \times 2L}{\frac{A}{2}}$**

**${R}_{new}=\frac{\rho \times 2L\times 2}{A}$**

**${R}_{new}=\frac{\rho \times 4L}{A}$**

**${R}_{new}=4\frac{\rho \times L}{A}$**

**${R}_{new}=4\times 20$ $(\because old\ resistnce\ of\ wire,\ R=\frac{\rho \times L}{A}=20)$${R}_{new}=80\Omega $**

Thus, the resistance of the wire in the new situation will be **80Ω.**