A copper wire having cross- sectional area of `0.5 mm^2` and a length of `0.1 m` is initially at `25^@C` and is thermally insulated from the surrounding. If a current of `10A` is set up in this wire, (a) find the time in which the wire will start melting. (b) What will be the change of resistance with time 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^(-1)C^(-1).` Melting temperature of copper is `1075^@C`

Explanation

`I^2Rt=ms Delta theta=m (S Delta theta), I^2(p_1 .1/A)t=A.b^2 (s Delta theta)`
Here `p_1`= specific resistance and `p_2`= density `therefore t=(A^2p_2(s Delta theta))/(I^2p_1)`………(i)
Substituting the values ,we have
`t=((0.5 times 10^-6)^2 (9 times 10^-2 times 4.19 times 1050))/((10)^2(1.6 times 10^-8))=55.5s`