Numericals based on Resistivity

Given: $l = 2\text{ m}$, $A = 10^{-6}\text{ m}^2$, and $\rho = 1.7 \times 10^{-8}\text{ }\Omega\text{m}$. The formula for resistance is $R = \rho \frac{l}{A}$.

Given: $l = 2\text{ m}$, $A = 1.7 \times 10^{-6}\text{ m}^2$, and $R = 2 \times 10^{-2}\text{ }\Omega$. Using the resistance formula rearranged for resistivity: $\rho = \frac{R \cdot A}{l}$.

Given: Charge $Q = 3\text{ C}$, Time $t = 12\text{ s}$. Electric current is the rate of flow of charge: $I = \frac{Q}{t}$.

Given: Potential difference $V = 12\text{ V}$, Resistance $R = 4\text{ }\Omega$. According to Ohm’s Law, $I = \frac{V}{R}$.

Resistivity is a fundamental property of the material and its temperature. It does not depend on the physical dimensions (length or area of cross-section) of the wire. Therefore, the resistivity remains completely unchanged when the wire is stretched.

When a wire is stretched, its volume remains constant. Volume = Area $\times$ Length ($V = A \cdot L$).
If new length $L’ = 2L$, the new area $A’$ must become $A/2$ to maintain the same volume.

Similar to the previous problem, stretching a wire to $n$ times its original length makes its new resistance $n^2$ times the original resistance ($R’ = n^2 R$).

Here, $n = 3$, so the new resistance will be $3^2 = 9$ times the original.

Given $d = 0.5\text{ mm} = 5 \times 10^{-4}\text{ m}$. Radius $r = 2.5 \times 10^{-4}\text{ m}$.
Area $A = \pi r^2 = \pi (2.5 \times 10^{-4})^2 \approx 1.96 \times 10^{-7}\text{ m}^2$.
Using $R = \rho \frac{l}{A} \implies l = \frac{R \cdot A}{\rho}$:

If the diameter is doubled, the area ($A \propto d^2$) becomes 4 times larger. Since resistance is inversely proportional to area ($R \propto 1/A$), the new resistance will be $1/4$th of the original.

When a wire is folded double on itself, its length becomes half ($L’ = L/2$) and its cross-sectional area becomes double ($A’ = 2A$).

Given $l = 10\text{ km} = 10^4\text{ m}$. Diameter $d = 20\text{ mm} = 2 \times 10^{-2}\text{ m}$.
Radius $r = 10^{-2}\text{ m}$.
Area $A = \pi r^2 = 3.14 \times (10^{-2})^2 = 3.14 \times 10^{-4}\text{ m}^2$.

Given: $l = 1.0\text{ m}$, $R = 23\text{ }\Omega$, $\rho = 1.84 \times 10^{-6}\text{ }\Omega\text{m}$. Rearranging $R = \rho \frac{l}{A}$ gives $A = \rho \frac{l}{R}$.

When a wire is drawn out (stretched), its volume remains constant. If length is doubled ($n=2$), the new resistance is $n^2$ times the original resistance.

Given $\frac{L_1}{L_2} = \frac{4}{9}$ and $\frac{R_1}{R_2} = \frac{4}{1}$.
We know $R = \rho \frac{L}{\pi r^2}$. Since the material is the same, $\rho$ is constant. Therefore, $r^2 \propto \frac{L}{R}$.

Taking the square root on both sides: $\frac{r_1}{r_2} = \frac{1}{3}$. The ratio is 1:3.

Since $\rho$ and $A$ are constant, $R \propto L$. Therefore, the ratio of their resistances is $\frac{R_1}{R_2} = \frac{L_1}{L_2} = \frac{3}{1}$.
According to Ohm’s law, $I = \frac{V}{R}$. Since $V$ is constant, current is inversely proportional to resistance ($I \propto 1/R$).

The ratio of current flowing through them is 1:3.

Resistance $R \propto \frac{L}{r^2}$. We compare the two wires:

Since $R_A / R_B = 0.75$, wire A has less resistance than wire B.

Given $l = 1\text{ km} = 1000\text{ m}$. Radius $r = 1\text{ mm} = 10^{-3}\text{ m}$.
Area $A = \pi r^2 = 3.14 \times (10^{-3})^2 = 3.14 \times 10^{-6}\text{ m}^2$.

When a wire is folded double on itself, its length becomes half ($L’ = L/2$) and its cross-sectional area becomes double ($A’ = 2A$).

Since the wires are “similar” (same material and area of cross-section), resistance is directly proportional to length ($R \propto L$).

Given $\rho = 64 \times 10^{-6}\text{ }\Omega\text{m}$, $l = 1.98\text{ m}$, $R = 7\text{ }\Omega$. First, calculate Area ($A = \pi r^2$):

Now, finding radius ($r$):

Given $l = 1.0\text{ m}$, $d = 0.4\text{ mm} \implies r = 0.2\text{ mm} = 2 \times 10^{-4}\text{ m}$, and $R = 2.0\text{ }\Omega$.
Area $A = \pi r^2 = 3.14 \times (2 \times 10^{-4})^2 = 3.14 \times 4 \times 10^{-8} = 12.56 \times 10^{-8}\text{ m}^2$.