If length of a conductor is doubled by keeping volume constant, then what is its new resistance if initial were `4Omega` ?

Options

A. `16Omega`
B. `8Omega`
C. `4Omega`
D. `2Omega`

Answer

A

Explanation

`R_(1)=4 Omega, R_(2)=? ` if `l_(2)=2l_(1)` by keeping volume constant.
`(R_(2))/(R_(1))=((l_(2))/(l_(1)))^(2)`
`R_(2)=((2l_(1))/(l_(1)))^(2) R_(1)=4xx4=16 Omega`.