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Potential Energy - External Viewpoint

Two wires of ...

Question

Two wires of the same material have lengths 3 cm and 5 cm and radii 1 mm and 3 mm respectively. They are connected in series across a battery of 16 V. The p.d. across the shorter wire is

A

2.5 V

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B

6.5 V

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C

1.5 V

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D

13.5 V

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Solution

The correct option is D $13.5 V$
Length $= 3 c m$ $(L_{1}) = 60 m m$
$5 c m (L_{2}) = 100 m m$
radii $= 1 m m (r_{1}) = 0.5 m m$
$= 3 m (r_{2}) = 1.5 m m$
The $E M F$ of the battery $= 16 V$
material is same so resistivity would also be same
$\frac{A i r 1}{L_{1}} = A$
$\frac{A_{1} R_{1}}{L_{1}} = \frac{A_{2} R_{2}}{L_{2}}$
$A_{1} = π (0.5) \times 0.5 = 0.25 P i$
$\frac{R_{1}}{R_{2}} = \frac{A_{2} L_{1}}{A_{1} L_{2}}$
$\frac{R_{1}}{R_{2}} = \frac{2.25 π \times 60}{0.25 π \times 100}$
or, $\frac{R_{1}}{R_{2}} = \frac{225 \times 60}{25 \times 100}$
or $\frac{R_{1}}{R_{2}} = 5.4 = \frac{54}{10} = \frac{27}{5}$
(Hence, $R_{1} = 27$ and $R_{2} = 5$ )
Since they are in series $R_{1} + R_{2} = 27 + 5 = 32$
$i = \frac{V}{R}$
$i = \frac{16}{32} = \frac{1}{2} = 0.5 A$
Now, $P d$ across $6 c m$ wire is
$V = i R$
$= 0.5 \times 27$
$= 13.5 V$

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