Q. A current of 0.01mA passes through the potentiometer wire of a resistivity of $10^{9}$ $\Omega$-cm and area of cross-section $10^{- 2}cm^{2}$. The potential gradient is
A) $10^{9}Vm^{- 1}$
B) $10^{11}Vm^{- 1}$
C) $10^{10}Vm^{- 1}$
D) $10^{8}Vm^{- 1}$
Explanation
(d) Potential gradient is given by
$k = \frac{V}{l} = \frac{IR}{l}\ \ \ \ \ \ \ \ \ \ \ (\because V = I\ R)$
$\ \ \ \ = \frac{I \times \rho l/A}{l}\ \ \ \ \ \ \ \ \ \left( \because R = \frac{\rho l}{A} \right)$
$\ \ \ \ = \frac{I\rho}{A}$
$\therefore\ \ k = \frac{0.01 \times 10^{- 3} \times 10^{9} \times 10^{- 2}}{10^{- 2} \times 10^{- 4}} = 10^{8}Vm^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.