Question

A conducting wire has length 'L₁' and diameter 'd₁'. After stretching the same wire length becomes 'L₂' and diameter 'd₂' The ratio of resistances before and after stretching is ______.

Options

• `"d"_2^4:"d"_1^4`

• `"d"_1^4:"d"_2^4`

• `"d"_2^2:"d"_1^2`

• `"d"_1^2:"d"_2^2`

Answer 1

A conducting wire has length 'L₁' and diameter 'd₁'. After stretching the same wire length becomes 'L₂' and diameter 'd₂' The ratio of resistances before and after stretching is `bbunderline("d"_2^2:"d"_1^2`.

Explanation:

Using, R = f`ℓ/"A"`

`"R"_1/"R"_2="L"_1/"L"_2xx"A"_2/"l"_2`

= `"L"_1/"L"_2xx((pi"d"_2^2)/4)/((pi"d"_1^2)/4)`

∴ Ratio = `"R"_1/"R"_2="d"_2^2/"d"_1^2`