Question

The resistance of a 10 m long wire is 10 `Omega`. Its length is increased by 25% by stretching the wire uniformly . The resistance of wire will change to
A. `12.5Omega`
B. `14.5Omega`
C. `15.6Omega`
D. `16.6Omega`

Answer 1

Correct Answer - C
Given, `l_(1)=l+(25)/(100)l=(5l)/(4)`.
Since, volume of wire remains unchanged on increasing length, hence
`A_(1)l_(1)=Al`
`A_(1)xx(5l)/(4)=Al` or `A_(1)=4A//5`
Given, `R= rho l//A= 10 Omega`
and `R_(1)=(rho l_(1))/(A_(1))=(rho5l //4)/(4A//5)=(25 rho l)/(16 A)`
`therefore " " R_(1)=(25)/(16)xx10=(250)/(16)=15.6 Omega`