The length of a given cylindrical wire is increased by 100% Due to the consequent decrease in diameter the change in the resistance of the wire will be

  • Option 1)

    200%

  • Option 2)

    100%

  • Option 3)

    50%

  • Option 4)

    300%

 

Answers (2)

As we learnt in

Resistivity -

R=\rho\frac{l}{A}

- wherein

\rho\rightarrow  resistivity / specific resistance

 

 

 

Let the length of the wire be l, radius of the wire be r.

\dpi{100} \therefore\; Resistance\; R=\rho\frac{l}{\pi r^{2}} \: \: \: \: \rho \: \: = \: \: resistivity \; of\; the\; wire \;

Now l is increased by 100%             \therefore \: \: \: {l}'= l+\frac{100}{100}l= 2l

As length is increased, its radius is going to be decreased in such a way that the volume of the cylinder remains constant.

\pi r^{2}\times l= \pi r{}'^{2}\times l{}'\Rightarrow {r}'^{2}= \frac{r^{2}\times l}{{l}'}= \frac{r^{2}\times l}{2l}= \frac{r^{2}}{2}

\: \therefore\: \: The\: \: new\: resistance\: R{}'^{2}= \rho \frac{{l}'}{\pi r{}'^{2}}= \rho \frac{2l}{\pi \times \frac{r^{2}}{2}}= 4R

\therefore \: \: \: Change\: in \: resistance\: R{}'-R= 3R

 


Option 1)

200%

This option is incorrect.

Option 2)

100%

This option is incorrect.

Option 3)

50%

This option is incorrect.

Option 4)

300%

This option is correct.

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