When a wire of resistance R …
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Naveen Sharma 8 years, 7 months ago
Ans. Let the Initial Length of Wire = l
Final length of Wire = L
Intial Radius of wire = r
FInal Radius of Wire r1 = {tex}r\over 2{/tex}
So, Initial Volume of Wire = Final Volume of Wire
{tex}=> \pi r^2 l = \pi r_1^2L{/tex}
{tex}=> r^2 l = ({r\over 2})^2L{/tex}
{tex}=> r^2 l = {r^2\over 4}L{/tex}
{tex}=> L = 4l{/tex}
Initial Area of Cross Section of wire (A) = {tex}\pi r^2{/tex}
Final area of Cross section of wire (A') = {tex}\pi r_1^2 = \pi{r^2\over 4}{/tex}
Initial Resistance of Wire (R)= {tex}\rho {l\over A} = \rho {l\over \pi r^2}{/tex}
Final Resistance of Wire (R') = {tex}\rho {L\over A'} = \rho {4l\over {\pi r^2\over 4}} ={ \rho 4\times 4l\over\pi r^2}{/tex}
= {tex}16\rho{l\over \pi r^2} = 16R{/tex}
So new Resistance ll be 16 times of initial Resistance.
1Thank You