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  • A wire of resistance is stretched till its length is double of the original wire. Then, the resistance of the stretched wire is:<di

A wire of resistance \mathrm{R} is stretched till its length is double of the original wire. Then, the resistance of the stretched wire is:

Option: 1

2\mathrm{R}


Option: 2

4\mathrm{R}


Option: 3

8\mathrm{R}


Option: 4

16\mathrm{R}


Answers (1)

Resistance \mathrm{R} of a wire of length l and radius r with specific resistance \mathrm{K} for its material is given by

\mathrm{ \mathrm{R}=\rho l / \pi \mathrm{r}^2 }            (i)

As the wire is stretched, there is no change in its mass \mathrm{M}.

But \mathrm{\mathrm{M}=\pi \mathrm{r}^2 l \mathrm{~d} \quad\: or\: \pi \mathrm{r}^2=\mathrm{M} / l \mathrm{~d}}.

Here, \mathrm{R}=\frac{\rho l}{\mathrm{M} / l \mathrm{~d}}=\left(\frac{\rho \mathrm{d}}{\mathrm{M}}\right) l^2    (ii)

Let the resistance of the stretched wire by \mathrm{R}^{\prime}; then

\mathrm{R}^{\prime}=\left(\frac{\rho \mathrm{d}}{\mathrm{M}}\right)(2 l)^2=4 \mathrm{R}

Hence option 2 is correct.

Posted by

Ramraj Saini

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