Question

A current of \(0.01\) A passes through the potentiometer wire of resistivity \(10^{6}\) \(\Omega\) cm and area of cross-section \(10^{-2}\) \(\text{cm}^{2}\). The potential gradient is (in V/m)

• A. \(10^{9}\)
• B. \(10^{11}\)
• C. \(10^{10}\)
• D. \(10^{8}\)

Hint:

Potential gradient = I times resistivity divided by area. Convert units to SI first.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
Potential gradient \(K=\dfrac{V}{l}=\dfrac{IR}{l}=\dfrac{I\rho}{A}\).

Step 2: Convert Units
\(\rho=10^{6}\ \Omega\text{ cm}=10^{4}\ \Omega\text{ m}\) and \(A=10^{-2}\ \text{cm}^{2}=10^{-6}\ \text{m}^{2}\).

Step 3: Substitute
\(K=\dfrac{0.01\times10^{4}}{10^{-6}}\).

Step 4: Calculate
\(K=\dfrac{10^{2}}{10^{-6}}=10^{8}\) V/m.

Final Answer:
The potential gradient is 10 to the power 8 V/m. \[ \boxed{\text{(D) }10^{8}} \]