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A 2 m long wire of resistance 4 Omega an...

A 2 m long wire of resistance `4 Omega` and diameter 0.64 mm is coated with plastic insulation of thickness 0.66 mm. A current of 5A flows through the wire. Find the temperature difference across the insulation in the steady state. Thermal conductivity of plastic is `0.16 xx (10^-2) cal//s cm^@-C`.

Text Solution

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The correct Answer is:
B, C
Thermal resistance of plastic coating`(R_t = l/(KA))`
`R_t = t/(K(pid)l)` (d = diameter)
`(0.6xx10^(3))/((0.16xx10^(-2)xx4.18xx10^(2))(pi)(0.64xx10^(-3))(2))`
`=0.0223^@C-s//J`
Now, `i^2R_e = (TD)/R_t`
`:. TD = i^2R_eR_t`
`=(5)^2(4)(0.0223)`
`=2.23^@C`
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