A copper wire having cross- sectional area of `0.5 mm^2` and a lentgth of 0.1 m is initially at `25^@C` and is thermally insulated form the surronding. If a current of 10A is set up in this wire, (a) find the time in which the wire will start melting. The change of resistance with time be, if the length of the wire is doubled? For copper, resistivity is `1.6xx10^(-8)Omega m, density is 9,000 kgm^(-3)` and specific heat capacity is `0.09 cal g^(-10)C^(-1).` Melting temperature fo copper is `1075^@C`

Explanation

`a=0.5mm^(2)=0.5xx10^(-6)m^(2),l=0.1m,T_(1)=25^(@)C,I=10A`
`T_(2)=1075^(@)C`
a. `I^(2)Rt=mstriangleT` or `I^(2)rho(l)/(a)xxt=mstriangleT`
or `t=(mstriangleTxxa)/(I^(2)rhol)=((d xx a xxI)striangleTxxa)/(I^(2)rhol)=(da^(2)triangleT)/(rhoI^(2))`
`9xx10^(3)xx(0.5xx10^(-6))^(2)`
`=(9xx10^(-2)xx10^(3)xx4.18xx1050)/(10xx10xx1.6xx10^(-8))`
`=555.5s=9min15s`
(b). Since length does not occur in the expression of time. the melting does ot depend on the length. So time taken will be the same.