Resistance of wire, Resistivity and Conductivity Questions in English

(A) The resistivity of materials is categorized as follows:
$1$. Conductors (Metals): $10^{-8} \Omega \cdot m$ to $10^{-6} \Omega \cdot m$.
$2$. Semiconductors: $10^{-5} \Omega \cdot m$ to $10^6 \Omega \cdot m$.
$3$. Insulators: $10^6 \Omega \cdot m$ to $10^{18} \Omega \cdot m$.
Given the resistivity of the material is $10^8 \Omega \cdot m$,which falls within the range of insulators.
Therefore,the material is an insulator.

(B) The resistance of a wire is given by $R = \frac{\rho l}{A}$,where $\rho$ is resistivity,$l$ is length,and $A$ is the cross-sectional area.
When two wires are connected in series,the total resistance $R_{eq}$ is the sum of individual resistances:
$R_{eq} = R_{1} + R_{2}$
Since the wires have the same radius,their cross-sectional areas are equal $(A_{1} = A_{2} = A)$.
$R_{eq} = \frac{\rho_{1} l_{1}}{A} + \frac{\rho_{2} l_{2}}{A} = \frac{\rho_{1} l_{1} + \rho_{2} l_{2}}{A}$
For the equivalent wire,the total length is $L = l_{1} + l_{2}$ and the area is $A$. The equivalent resistivity $\rho_{eq}$ is defined as:
$R_{eq} = \frac{\rho_{eq} (l_{1} + l_{2})}{A}$
Equating the two expressions for $R_{eq}$:
$\frac{\rho_{eq} (l_{1} + l_{2})}{A} = \frac{\rho_{1} l_{1} + \rho_{2} l_{2}}{A}$
$\rho_{eq} = \frac{\rho_{1} l_{1} + \rho_{2} l_{2}}{l_{1} + l_{2}}$

(C) The resistance of a wire is given by $R = \rho \frac{l}{A}$.
Since the volume $V = A \cdot l$ remains constant,we can write $A = \frac{V}{l}$.
Substituting this into the resistance formula,we get $R = \rho \frac{l}{V/l} = \rho \frac{l^2}{V}$.
Since the resistivity $\rho$ and volume $V$ remain constant,we have $R \propto l^2$.
Given that the length is doubled,$l_2 = 2l_1$.
Therefore,$\frac{R_2}{R_1} = \left( \frac{l_2}{l_1} \right)^2 = \left( \frac{2l_1}{l_1} \right)^2 = 2^2 = 4$.
Thus,$R_2 : R_1 = 4 : 1$.

(B) The resistance of a wire is given by $R = \rho \frac{L}{A}$,where $\rho$ is the resistivity,$L$ is the length,and $A$ is the cross-sectional area.
When the wire is elongated $n$-fold,the new length $L' = nL$.
Since the volume of the wire remains constant,$V = A \times L = A' \times L'$.
Therefore,$A' = \frac{A \times L}{L'} = \frac{A \times L}{nL} = \frac{A}{n}$.
The new resistance $R'$ is given by $R' = \rho \frac{L'}{A'} = \rho \frac{nL}{A/n} = n^2 \left( \rho \frac{L}{A} \right) = n^2 R$.

(C) The mass of the copper wire remains constant. Since $Mass = \text{Volume} \times \text{Density} = (A \times \ell) \times \sigma$,where $A = \pi r^2$ is the cross-sectional area,$\ell$ is the length,and $\sigma$ is the density.
For two wires of the same mass: $\pi r_1^2 \ell_1 \sigma = \pi r_2^2 \ell_2 \sigma \implies \frac{\ell_1}{\ell_2} = \left(\frac{r_2}{r_1}\right)^2$.
Given diameters $d_1 = 1 \ mm$ and $d_2 = 2 \ mm$,the radii ratio is $\frac{r_2}{r_1} = \frac{2}{1} = 2$. Thus,$\frac{\ell_1}{\ell_2} = (2)^2 = 4$.
The resistance $R$ is given by $R = \rho \frac{\ell}{A} = \rho \frac{\ell}{\pi r^2}$.
Therefore,the ratio of resistances is $\frac{R_1}{R_2} = \frac{\ell_1}{\ell_2} \times \left(\frac{r_2}{r_1}\right)^2 = 4 \times (2)^2 = 4 \times 4 = 16$.
Thus,the ratio is $16: 1$.

(A) The resistance $R$ of a wire is given by $R = \rho \frac{\ell}{A}$.
Since the volume $V = A \ell$ remains constant when the wire is stretched,$A = \frac{V}{\ell}$.
Substituting this into the resistance formula,we get $R = \rho \frac{\ell^2}{V}$.
Since $\rho$ and $V$ are constant,$R \propto \ell^2$.
Given the new length $\ell_2 = 3 \ell_1$,the new resistance $R_2$ is related to the original resistance $R_1$ by the ratio $\frac{R_2}{R_1} = \left(\frac{\ell_2}{\ell_1}\right)^2$.
Substituting the values: $\frac{R_2}{5 \Omega} = (3)^2 = 9$.
Therefore,$R_2 = 5 \Omega \times 9 = 45 \Omega$.

(D) Electrical conductivity depends on the number of free electrons available for charge transport. Among the given options,$Copper$ $(Cu)$ is a transition metal with a very high density of free electrons and low resistivity,making it an excellent conductor of electricity. While $Gold$ and $Platinum$ are also good conductors,$Copper$ is widely used in electrical wiring due to its superior conductivity-to-cost ratio. $Silicon$ is a semiconductor,not a conductor.

(A) The resistance of a conductor is given by the formula $R = \rho \frac{l}{A}$,where $\rho$ is the resistivity,$l$ is the length of the conductor in the direction of current flow,and $A$ is the cross-sectional area perpendicular to the current flow.
To maximize the resistance $R$,we need to maximize the length $l$ and minimize the cross-sectional area $A$.
When the battery is connected across the $1 \text{ cm} \times 0.5 \text{ cm}$ faces,the current flows along the length of $10 \text{ cm}$. Thus,$l = 10 \text{ cm}$ and $A = 1 \text{ cm} \times 0.5 \text{ cm} = 0.5 \text{ cm}^2$.
The resistance is $R = \rho \frac{10}{0.5} = 20\rho$.
For the other two orientations,the length $l$ will be either $1 \text{ cm}$ or $0.5 \text{ cm}$,and the area $A$ will be larger,which results in a smaller resistance value.
Therefore,the resistance is maximum when the battery is connected across the $1 \text{ cm} \times 0.5 \text{ cm}$ faces.