Saved Bookmarks
| 1. |
A copper wire of resistance `R_(0)` is strerched till its length is increased to n times of its original length. What will be its new resistance? |
|
Answer» Original resistnace, `R_(0) = rho l_(0)//A_(0)` New resistance, `R = rho l//A`. As the volume of wire remains constant, so `l_(0)A_(0)=lA= (n l_(0)) A or A= A_(0)//n` Now `R=(rho l)/(A)= rho (n l_(0))/(A_(0)//n)=n^(2)((rho l)/(A))=n^(2) R_(0)` |
|