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A wire of resistance $4 Ω$ is redrawn by pulling it doubled. Its new resistance is [[1]] $Ω$ .

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Detailed Solution

A wire of resistance

4 Ω

is redrawn by pulling it doubled. Its new resistance is

16 Ω

.
Let

L and L'

be the length of wire before and after stretching respectively.

A and A'

be the area of cross section before and after stretching and

R and R'

be the resistance of the wire before and after stretching respectively.
As volume of wire does not change,

AL = A' L'

Now, it is given that the wire is pulled to double, so,

L^{'} = 2 L

So,

AL = A^{'} \times 2 L

\Rightarrow A = 2 A'

Now, resistance,

R = ρ \frac{L}{A}

And final resistance,

R' = ρ \frac{L'}{A'}

So,

\frac{R'}{R} = \frac{ρL'}{A'} \times \frac{A}{ρL}

\Rightarrow \frac{R'}{R} = \frac{ρ \times 2 L \times A}{\frac{A}{2} ρL}

\Rightarrow \frac{R'}{R} = 4

{\Rightarrow R}^{'} = 4 R

\Rightarrow R^{'} = 4 \times 4

\Rightarrow R^{'} = 16 Ω

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