A wire of resistance 4- is redrawn by pulling it doubled- what is its new resistance -

The correct option is A 16 We know R _ 1 =ρ l _ 1 A Given R _ 1 = 4Ω l _ 1 =2 l _ 2 and R _ 2 =ρ 2 l _ 2 A Since the ...

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Byju's Answer

Standard XII

Physics

Conductivity and Resistivity

A wire of res...

Question

A wire of resistance $4 Ω$ is redrawn by pulling it doubled, what is its new resistance ?

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Solution

The correct option is A 16
$W e k n o w R_{1} = ρ \frac{l_{1}}{A} G i v e n R_{1} = 4 Ω l_{1} = 2 l_{2} a n d R_{2} = ρ \frac{2 l_{2}}{A} S i n c e t h e v o l u m e r e m a i n s c o n s t a n t t h e n e w a r e a = A / 2 \frac{R_{2}}{R_{1}} = 4 R_{2} = 4 R_{1} R_{2} = 4 \times 4 = 16 Ω$

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A wire of resistance 4Ω is redrawn by pulling it doubled, what is its new resistance ?

The correct option is A 16WeknowR1=ρl1AGivenR1=4Ωl1=2l2andR2=ρ2l2ASincethevolumeremainsconstantthenewarea=A/2R2R1=4R2=4R1R2=4×4=16Ω

A wire of resistance $4 Ω$ is redrawn by pulling it doubled, what is its new resistance ?

The correct option is A 16
$W e k n o w R_{1} = ρ \frac{l_{1}}{A} G i v e n R_{1} = 4 Ω l_{1} = 2 l_{2} a n d R_{2} = ρ \frac{2 l_{2}}{A} S i n c e t h e v o l u m e r e m a i n s c o n s t a n t t h e n e w a r e a = A / 2 \frac{R_{2}}{R_{1}} = 4 R_{2} = 4 R_{1} R_{2} = 4 \times 4 = 16 Ω$