Question

एक तार जिसकी लम्बाई ℓ और प्रितरोध R है तो खींचे जाने पर उसके अनुप्रस्थ काट की त्रिज्या आधी हो जाती है ।खींची गई इस तार का प्रितरोध क्या है?​

Answer 1

Given:

A wire of length l and resistance R is stretched to get the radius of cross-section halved

To find:

The new   resistance of the wire after stretching

Solution:

The initial measurements of the wire before stretching:

The initial length of the wire = l

The initial resistance of the wire = R

Let the initial radius of the cross-section be "r".

After stretching the wire, the measurements will be:

The new length of the wire = l '

The new resistance of the wire = R'

The new radius of the cross-section after stretching, r' = $\frac{r}{2}$

Since the wire is being stretched, so, we can say that the volume of the wire in both cases remains the same.

∴ $V = V'$

$\implies \pi r^2 l = \pi r'^2 l'$

substituting the value of r₂'

$\implies \pi r^2 l = \pi( \frac{r^2}{2}) l'$

$\implies \pi r^2 l = \frac{1}{4}\pi r^2 l'$

cancelling similar terms

$\implies \bold{ l' = 4l}$ ............ (i)

Let' assume,

"R" → represents the initial resistance of the before stretching.  

" R' " → represents the new resistance of the after stretching.  

We know the formula of the resistance of a wire is as follows:

$\boxed{\bold{R = \frac{ \rho \:l}{A}}}$ ...... (ii)

Now, using the above formula, we will find the value of the new resistance (R'),

$R' = \frac{ \rho \:l'}{A'}$

$\implies R' = \frac{ \rho \:l'}{\pi r'^2}$

substituting the value of l' from (i) and r' = $\frac{r}{2}$, we get

$\implies R' = \frac{ \rho \:(4l)}{\pi(\frac{r}{2}) ^2}$

$\implies R' = \frac{ \rho \:(4l)}{\pi(\frac{r^2}{4}) }$

$\implies R' = 16 \times \frac{ \rho \:l}{\pi r^2 }$

substituting from (ii), we get

$\implies \bold{R' = 16 \times R}$

Thus, the new   resistance of the wire after stretching becomes → 16 times the initial resistance of the wire.

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