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

Options

A. `12.5Omega`
B. `14.5Omega`
C. `15.6Omega`
D. `16.6Omega`

Answer

C

Explanation

Resistance of a wire, `R=rho(l)/(A)=10Omega`
where l is the length and A be area of cross-section of wire respectively.
New length `l.=l+(25)/(100)l=(5)/(4)l`
As the volume of wire remains unchanged on increaing length, hence `A.l.=Al`
`A.=A((l)/(l.))=(4)/(5)A`
New resistance, `R.=rho(l.)/(A.)=(rho((5)/(4)l))/(((4)/(5)A))=(25)/(16)rho(l)/(A)`
`=(25)/(16)xx10=15.6Omega`