Question

plz help......................... . ........ . . . .. ​

Answer 1

This is the correct answer...

Answer 2

Answer:

$\mathcal{\blue{answer = \frac{1}{4} of \: the \: resistance \: will \: change}}$

Explanation:

Given: Resistivity=1.6× 10-⁸ ohm/m ,Resistance =10 ohm,Diameter =0.5mm

$a = \frac{0.5}{2} \times {10}^{ - 3} m$

$a = \pi {r}^{2}$

$a = 3.14 \times (\frac{0.5 \times {10}^{ - 3} )}{2} \times \frac{5 \times {10}^{ - 4} }{2}$

$a = 19.6 \times {10}^{ - 8}$

$r = \frac{pl}{a}$

$l = \frac{ra}{p}$

$l = 10 \times \frac{1.96 \times {10}^{ - 8} }{1.6 \times {10}^{ - 8} }$

$l = 123m$

lenght =123m

$r = \frac{pl}{a}$

$r = \frac{pl}{\pi \frac{ {d}^{2} }{4} }$

$r = \frac{4pl}{\pi {d}^{2} }$

$r = \frac{4p {l}^{2} }{\pi {d}^{2} } \: \: \: \: \: \: \: (r \alpha \frac{1}{ {d}^{2} } )$

when double ,

${r}^{2} \alpha ( \frac{1}{4 {d}^{2} } )$

here the constant is 1

${r}^{2} = \frac{1}{4 {d}^{2} }$

$r = \frac{4pl}{\pi} \times \frac{1}{ {d}^{2} }$

when doubles,

$r = \frac{4pl}{\pi} \times \frac{1}{2 {d}^{2} }$

$r = \frac{4pl}{\pi} \times \frac{1}{4 {d}^{2} }$

$r = \frac{4pl}{\pi {d}^{2} } \alpha \frac{1}{4}$

So,

r is directly proportional

$r = \frac{1}{4}$

$<marquee>{Thank you}</marquee>$