Physics-
General
Easy

Question

The temperature of the two outer surfaces of a composite slab, consisting of two materials having coefficients of thermal conductivity K and 2K and thickness x and 4x, respectively are T2 and T1 (T2 > T(a) . The rate of heat transfer through the slab, in a steady state is open parentheses fraction numerator A left parenthesis T subscript 2 end subscript minus T subscript 1 end subscript right parenthesis K over denominator x end fraction close parentheses f, with ¦ which equal to

  1. 1    
  2. fraction numerator 1 over denominator 2 end fraction    
  3. fraction numerator 2 over denominator 3 end fraction    
  4. fraction numerator 1 over denominator 3 end fraction    

The correct answer is: fraction numerator 1 over denominator 3 end fraction


    Equation of thermal conductivity of the given combination K subscript e q end subscript equals fraction numerator l subscript 1 end subscript plus l subscript 2 end subscript over denominator fraction numerator l subscript 1 end subscript over denominator K subscript 1 end subscript end fraction plus fraction numerator l subscript 2 end subscript over denominator K subscript 2 end subscript end fraction end fraction equals fraction numerator x plus 4 x over denominator fraction numerator x over denominator K end fraction plus fraction numerator 4 x over denominator 2 K end fraction end fraction equals fraction numerator 5 over denominator 3 end fraction K. Hence rate of flow of heat through the given combination is fraction numerator Q over denominator t end fraction equals fraction numerator K subscript e q end subscript. A left parenthesis T subscript 2 end subscript minus T subscript 1 end subscript right parenthesis over denominator left parenthesis x plus 4 x right parenthesis end fraction equals fraction numerator fraction numerator 5 over denominator 3 end fraction K A left parenthesis T subscript 2 end subscript minus T subscript 1 end subscript right parenthesis over denominator 5 x end fraction=fraction numerator fraction numerator 1 over denominator 3 end fraction K A left parenthesis T subscript 2 end subscript minus T subscript 1 end subscript right parenthesis over denominator x end fraction
    On comparing it with given equation we get f equals fraction numerator 1 over denominator 3 end fraction

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