A wire has resistance of `10Omega`. If it is stretched by 1/10th of its length, then its resistance is nearly

Explanation

Volume of material of wire remains constant.
Volume V = (Area) `xx` (Length) = AL
Now `R=(rhoL)/(A)=(rhoL^(2))/(V)=((rho)/(V))L^(2)" "(because A=(V)/(L))`
`therefore (R_(1))/(R_(2))=((L_(1))/(L_(2)))^(2)=((L)/(1.1L))^(2)=(1)/(1.21)`
or `R_(2)=R_(1)xx1.21=10xx1.21=12.1Omega`
`therefore` final resistance `=12.1Omega`.