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Dimensional Analysis

A 6 volt ba...

Question

A $6 v o l t$ battery is connected to the terminals of a $t h r e e m e t r e$ long wire of uniform thickness and resistance of $100 o h m$ . The difference of potential between two points on the wire separated by a distance of $50 c m$ will be :-

3 v o l t

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1.5 v o l t

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2 v o l t

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1 v o l t

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Solution

The correct option is B $1.5 v o l t$
Resistance $R = \frac{ρ l}{A}$ where $ρ =$ resistivity, $l =$ length of wire and A, cross section of wire.
Thus, $R \propto l$
here, $\frac{R_{1}}{R_{2}} = \frac{l_{1}}{l_{2}} = \frac{300}{50} = 6$
since, $R_{1} = 100 Ω$ so, $R_{2} = R_{1} / 6 = 100 / 6 = 50 / 3 Ω$
The current through wire is $I = V / R_{1} = 6 / 100 = 3 / 50$
Voltage across $R_{2}$ is $V_{2} = I R_{2} = (3 / 50) (50 / 3) = 1 V$

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