InterviewSolution
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InterviewSolution
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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 form 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` |
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Answer» `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 axxI)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. |
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