Question

If specific resistance of a potentiometer wire is $10^{-7}\, \Omega m$ and current flow through it is 0.1 amp., cross-sectional area of wire is $10^{-6}\, m^2$ then potential gradient will be

• A. $10^{-2}\, \, volt/ m$
• B. $10^{-4}\, \, volt/ m$
• C. $10^{-6}\, \, volt/ m$
• D. $10^{-8}\, \, volt/ m$

Answer 1

The Correct Option is A

To find the potential gradient of the potentiometer wire, we need to first understand the relationship between resistance, resistivity, current, and potential difference across the wire.

Step 1: Calculate the resistance of the wire.

The resistance R of a wire is given by the formula:

R = \frac{\rho \cdot L}{A}

where:

• \rho = 10^{-7}\, \Omega \, m is the specific resistance (resistivity) of the wire.
• A = 10^{-6}\, m^2 is the cross-sectional area of the wire.
• L is the length of the wire, which is not directly given but will cancel out in our calculations, as we will see.

Step 2: Calculate the potential gradient.

The potential gradient K across the wire is given by:

K = \frac{V}{L}

where V is the potential difference across the wire. Using Ohm's law, we have:

V = I \cdot R

Substituting R from Step 1 into the equation:

V = I \cdot \frac{\rho \cdot L}{A}

Then, the potential gradient is:

K = \frac{V}{L} = \frac{I \cdot \rho}{A}

Substitute the given values:

I = 0.1 \, \text{A}, \rho = 10^{-7}\, \Omega \, m, and A = 10^{-6}\, m^2.

K = \frac{0.1 \times 10^{-7}}{10^{-6}} = \frac{0.1 \times 10^{-7}}{10^{-6}} = 10^{-2} \, \text{volt/m}

Thus, the potential gradient is 10^{-2} \, \text{volt/m}.

Conclusion: The correct answer is 10^{-2} \, \text{volt/m}, which matches with option A.