First slide

Ohm's law

Question

A wire of resistance $4 Ω$ is stretched to twice its original length. The resistance of stretched wire would be

Moderate

A

$8 Ω$
B

$16 Ω$
C

$2 Ω$
D

$4 Ω$

Solution

Resistance of a wire,
$R = ρ \frac{1}{A} = 4 Ω - - - (i)$
When wire is stretched twice, its new length be ‘l’. Then $l^{'} = 2 l$
On stretching volume of the wire remains constant.
$∴ l A = l^{'} A^{'} w h e r e A^{'} i s t h e n e w c r o s s - s e c t i o n a l a r e a O r A^{'} = \frac{1}{l^{'}} A = \frac{1}{2 l} A = \frac{A}{2}$
$∴ R e s i s \tan c e o f a s t r e t c h e d w i r e i s R^{'} = ρ \frac{l^{'}}{A^{'}} = ρ \frac{2 l}{(A / 2)} = 4 ρ \frac{l}{A}$
$= 4 (4 Ω) = 16 Ω (U \sin g (i))$

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