The area of cross-section of a potentiometer wire is $6 \times 10^{-7}$ m$^2$. The potential difference per unit length of the potentiometer wire when it is connected to a cell of negligible internal resistance and a resistor in series is $0.15$ Vm$^{-1}$. If the current through potentiometer wire is $0.3$A, then the resistivity of the material of the potentiometer wire is
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The potential gradient ($k$) in a potentiometer is directly proportional to the current ($I$) and the resistivity ($\rho$), and inversely proportional to the cross-sectional area ($A$) of the wire ($k = I\rho/A$). This relationship is a consequence of Ohm's law and the definition of resistivity.
Step 1: State the relationship between potential gradient and current.
The potential difference across a length $L$ of a wire is $V = IR$, where $I$ is the current and $R$ is the resistance of that length.
The potential gradient ($k$) is the potential difference per unit length:
\[
k = \frac{V}{L} = \frac{IR}{L}.
\] Step 2: Express resistance in terms of resistivity.
The resistance of the wire is given by $R = \rho \frac{L}{A}$, where $\rho$ is the resistivity and $A$ is the area of cross-section.
Substitute this into the potential gradient equation:
\[
k = I \frac{\rho L/A}{L} = \frac{I\rho}{A}.
\] Step 3: Solve the equation for resistivity $\rho$.
Rearrange the equation to isolate $\rho$:
\[
\rho = \frac{kA}{I}.
\] Step 4: Substitute the given numerical values.
We are given:
Potential gradient: $k = 0.15$ V/m.
Area of cross-section: $A = 6 \times 10^{-7}$ m$^2$.
Current: $I = 0.3$ A.
\[
\rho = \frac{(0.15 \text{ V/m}) \times (6 \times 10^{-7} \text{ m}^2)}{0.3 \text{ A}}.
\] Step 5: Calculate the resistivity.
\[
\rho = \frac{0.15 \times 6}{0.3} \times 10^{-7} \frac{\Omega\cdot\text{m}}{\text{m}} \text{ m}^2 = \frac{0.9}{0.3} \times 10^{-7} \Omega\text{m}.
\]
\[
\rho = 3 \times 10^{-7} \Omega\text{m}.
\]
\[
\boxed{\rho = 3 \times 10^{-7} \Omega\text{m}}.
\]
