Question

If a wire of resistance $\mathrm{R}=10\mathrm{ohm}$ is stretched to double of its length, then new resistance will be ______.

Answer 1

Step 1: Given data

Resistance of the wire $\mathrm{R}=10\mathrm{ohm}$.

Wire is stretched to double of its length.

Step 2: Formula used

1. The formula for the resistance of a wire is, $\mathrm{R}=\mathrm{\rho }\frac{\mathrm{l}}{\mathrm{A}}$ . Here, $\mathrm{\rho }$ is the resistivity of a material, $\mathrm{l}$ is the length of wire, $\mathrm{A}$ is the cross-sectional area.
2. For the solution, we have to find the new cross section area of the wire, then we find the new resistance.

Step 3: Determine the new cross-section area (A') of the wire.

The volume of wire remains uniform even after stretching the wire. So,

${\mathrm{V}}_{\mathrm{i}}={\mathrm{V}}_{\mathrm{f}}$

$\mathrm{LA}=\mathrm{L}\text{'}\mathrm{A}\text{'}$

From the given, the wire of resistance is stretched to double of its length.$\mathrm{L}\text{'}=2\mathrm{L}$. So,

$\mathrm{LA}=2\mathrm{LA}\text{'}$

$\mathrm{A}\text{'}=\frac{\mathrm{A}}{2}$

Step 4: Find the new resistance (R').

We know that, initial resistance, $\mathrm{R}=\mathrm{\rho }\frac{\mathrm{l}}{\mathrm{A}}$. Then new resistance will be:

$\mathrm{R}\text{'}=\mathrm{\rho }\frac{\mathrm{L}\text{'}}{{\mathrm{A}}^{\text{'}}}$

$=\frac{\mathrm{\rho }\left(2\mathrm{L}\right)}{{\displaystyle \frac{\mathrm{A}}{2}}}$

$=\frac{4\mathrm{\rho L}}{\mathrm{A}}$

$=4\mathrm{R}$

$=4\left(10\right)$

$=40\mathrm{ohm}$

Hence, If a wire of resistance $\mathrm{R}=10\mathrm{ohm}$ is stretched to double of its length, then new resistance will be $40\mathrm{ohm}$.