Q. A battery of 6 volts is connected to the terminals of a three metre long wire of uniform thickness and resistance of the order of $100\Omega.$ The difference of potential between two points separated by $50\ cm$ on the wire will be
A) $1\ V$
B) $1.5\ V$
C) $2\ V$
D) $3\ V$
Explanation
(a) Here same current is passing throughout the length of the wire, hence $V \propto R \propto l$
$\Rightarrow \frac{V_{1}}{V_{2}} = \frac{l_{1}}{l_{2}} \Rightarrow \frac{6}{V_{2}} = \frac{300}{50} \Rightarrow V_{2} = 1\ V$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.