First slide

Ohm's law

Question

A wire of length L is drawn such that its diameter is reduced to half of its original diameter. If the initial resistance of the wire were $10 Ω,$ its new resistance would be

Moderate

A

$40 Ω$
B

$80 Ω$
C

$160 Ω$
D

$120 Ω$

Solution

For a wire of length l, area A, specific resistance $ρ$ , the resistance is given by
$R = ρ \frac{l}{A} . ... (1)$
If original diameter of wire be d, then new diameter is d/2
Original area of cross-section is $π d^{2} / 4$ and final area of cross-section is $π d^{2} / 16$
Since volume remains constant, on pulling the wire we have
$\begin{array}{l} \frac{{πd}^{2}}{4} \times l = \frac{{πd}^{2}}{16} \times l^{'} \\ \Rightarrow l^{'} = \frac{16}{4} l = 4 l \end{array}$
Also $R^{'} = ρ \frac{l^{'}}{A^{'}} . ... (2)$
$∴ R^{'} = ρ . \frac{4 l}{A / 4} = 16 R$
Hence, new resistance $16 \times 10 = 160 Ω$

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