A battery of $6\, volts$ is connected to the terminals of a three meter 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 .......... $V$
AIPMT 2004, Easy
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(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.$
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