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Physics-

General

Easy

Question

The coefficient of thermal conductivity of copper is nine times that of steel. In the composite cylindrical bar show in figure, what will be the temperature at the junction of copper ad steel?

75 $℃$
67 $℃$
33 $℃$
25 $℃$

The correct answer is: 75 $℃$

Let the temperature of junction be $theta.$
$open parentheses fraction numerator increment Q over denominator d subscript 1 end subscript end fraction close parentheses subscript c o p p e r end subscript blank equals open parentheses fraction numerator increment Q over denominator increment T end fraction close parentheses subscript s t e e l end subscript$
$K subscript 1 end subscript A equals blank fraction numerator open parentheses 100 minus theta close parentheses over denominator 18 end fraction equals fraction numerator K subscript 2 end subscript A open parentheses theta minus 0 close parentheses over denominator 6 end fraction$
$9 K subscript 2 end subscript fraction numerator open parentheses 100 minus theta close parentheses over denominator 3 end fraction equals K subscript 2 end subscript theta$
3 $theta equals 900 minus 9 theta$
12 $theta equals 900$
$theta equals 75 ℃$

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Five rods of same dimensions are arranged as shown in figure. They have thermal conductivities $K subscript 1 end subscript comma blank K subscript 2 end subscript comma blank K subscript 3 end subscript comma blank K subscript 4 end subscript a n d K subscript 5 end subscript.$ When points $A blank a n d blank B$ are maintained at different temperature, no heat would flow through central rod, if

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fraction numerator P over denominator Q end fraction equals fraction numerator R over denominator S end fraction

(in terms of resistances)

fraction numerator 1 divided by K subscript 1 end subscript over denominator 1. K subscript 2 end subscript end fraction equals fraction numerator 1 divided by K subscript 3 end subscript over denominator 1 divided by K subscript 4 end subscript end fraction

fraction numerator K subscript 2 end subscript over denominator K subscript 1 end subscript end fraction equals fraction numerator K subscript 4 end subscript over denominator K subscript 3 end subscript end fraction

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physics-General

The equivalent electrical circuit, figure in these cases is of Wheatstone bridge. No current would flow through central rod

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fraction numerator P over denominator Q end fraction equals fraction numerator R over denominator S end fraction

(in terms of resistances)

fraction numerator 1 divided by K subscript 1 end subscript over denominator 1. K subscript 2 end subscript end fraction equals fraction numerator 1 divided by K subscript 3 end subscript over denominator 1 divided by K subscript 4 end subscript end fraction

fraction numerator K subscript 2 end subscript over denominator K subscript 1 end subscript end fraction equals fraction numerator K subscript 4 end subscript over denominator K subscript 3 end subscript end fraction

K subscript 1 end subscript K subscript 4 end subscript equals K subscript 2 end subscript K subscript 3 end subscript

If two metallic plates of equal thicknesses and thermal conductivities $K subscript 1 end subscript$ and $K subscript 2 end subscript$ are put together face to face and a common plate is constructed, then the equivalent thermal conductivity of this plate will be

R subscript e q end subscript equals R subscript 1 end subscript plus R subscript 2 end subscript rightwards double arrow fraction numerator 2 l over denominator K subscript e q end subscript A end fraction equals fraction numerator l over denominator K subscript 1 end subscript A end fraction plus fraction numerator l over denominator K subscript 2 end subscript A end fraction

rightwards double arrow fraction numerator 2 over denominator K subscript e q end subscript end fraction equals fraction numerator 1 over denominator K subscript 1 end subscript end fraction plus fraction numerator 1 over denominator K subscript 2 end subscript end fraction rightwards double arrow K subscript e q end subscript equals fraction numerator 2 K subscript 1 end subscript K subscript 2 end subscript over denominator K subscript 1 end subscript plus K subscript 2 end subscript end fraction

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therefore e subscript x end subscript greater than e subscript y end subscript

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left parenthesis negative fraction numerator d T over denominator d t end fraction right parenthesis proportional to

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From the graph,

open parentheses negative fraction numerator d T over denominator d t end fraction close parentheses subscript x end subscript greater than open parentheses negative fraction numerator d T over denominator d t end fraction close parentheses subscript y end subscript

therefore e subscript x end subscript greater than e subscript y end subscript

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e

proportional to

absorptive power (

a

) (good absorbers are good emitters also)

therefore a subscript x end subscript greater than a subscript y end subscript

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