A wire of resistance `5 Omega` is stretched 20%. IF the volume remains constant , find the new resistance.

Explanation

Volume of the wire=V=constant , Let the initial length of the wire=l
`therefore` Initial cross sectional area , `A_1=V/l`
If the length of the wire is increased by `20%` ,
final length =`l times120/100=1.2l`
`therefore` Final cross sectional area , `A_2=V/(1.2l)`
Now, `R_1=rho.l_1/A_1`and `R_2=rho.l_2/A_2`
`thereforeR_2/R_1=l_2/l_1timesA_1/A_2`
or, `R_2=R_1timesl_2/l_1timesA_1/A_2=5times(1.2l)/l times (v/l)/(v/(1.2l))`
`=5times1.2times1.2=7.2Omega`