Physics-
General
Easy

Question

A bimetallic strip consists of metals X and Y. It is mounted rigidly at the base as shown. The metal X has a higher coefficient of expansion compared to that for metal Y. when bimetallic strip is placed in a cold bath

  1. It will bend towards the right    
  2. It will bend towards the left    
  3. It will not bend but shrink    
  4. It will neither bend nor shrink    

The correct answer is: It will bend towards the left


    The metal X has a higher coefficient of expansion compared to that for metal Y so, on placing bimetallic strip in a cold bath, X will shrink more than Y. Hence, the strip will bend towards the left.

    Related Questions to study

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    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 T subscript 2 end subscript and T subscript 1 end subscript left parenthesis T subscript 2 end subscript greater than T subscript 1 end subscript right parenthesis. The rate of heat transfer through the slab, in a steady state is open parentheses fraction numerator A open parentheses T subscript 2 end subscript minus T subscript 1 end subscript close parentheses K over denominator x end fraction close parentheses f, with f equals to

    Let the temperature of common interface be T ℃. Rate of heat flow
    H equals fraction numerator Q over denominator t end fraction equals blank fraction numerator K A increment T over denominator l end fraction
    therefore H subscript 1 end subscript equals open parentheses fraction numerator Q over denominator t end fraction close parentheses subscript 1 end subscript=fraction numerator 2 K A left parenthesis T minus T subscript 1 right parenthesis end subscript over denominator 4 x end fraction
    And H subscript 2 end subscript equals open parentheses fraction numerator Q over denominator t end fraction close parentheses subscript 2 end subscript=fraction numerator K A left parenthesis T subscript 2 end subscript minus T right parenthesis over denominator x end fraction
    In steady state, the rate of heat flow should be same in whole system i e comma
    H subscript 1 end subscript equals blank H subscript 2 end subscript
    rightwards double arrow fraction numerator 2 K A left parenthesis T minus T subscript 1 end subscript right parenthesis over denominator 4 x end fraction = fraction numerator K A left parenthesis T subscript 2 end subscript minus T right parenthesis over denominator x end fraction
    rightwards double arrow fraction numerator T minus T subscript 1 end subscript over denominator 2 end fraction equals blank T subscript 2 end subscript minus T
    rightwards double arrow T minus T subscript 1 end subscript equals blank 2 T subscript 2 end subscript minus 2 T
    rightwards double arrow T equals fraction numerator 2 T subscript 2 end subscript plus T subscript 1 end subscript over denominator 3 end fraction (i)
    Hence, heat flow from composite slab is
    H equals blank fraction numerator K A left parenthesis T subscript 2 end subscript minus T right parenthesis over denominator x end fraction
    = fraction numerator K A over denominator x end fraction open parentheses T subscript 2 end subscript minus fraction numerator 2 T subscript 2 end subscript plus T subscript 1 end subscript over denominator 3 end fraction close parentheses equals blank fraction numerator K A over denominator 3 x end fraction blank left parenthesis T subscript 2 end subscript minus T subscript 1 end subscript right parenthesis (ii)
    Accordingly, H equals open square brackets fraction numerator A open parentheses T subscript 2 end subscript minus T subscript 1 end subscript close parentheses K over denominator x end fraction close square brackets f (iii)
    By comparing Eqs. (ii) and (iii), we get
    rightwards double arrow f equals blank fraction numerator 1 over denominator 3 end fraction

    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 T subscript 2 end subscript and T subscript 1 end subscript left parenthesis T subscript 2 end subscript greater than T subscript 1 end subscript right parenthesis. The rate of heat transfer through the slab, in a steady state is open parentheses fraction numerator A open parentheses T subscript 2 end subscript minus T subscript 1 end subscript close parentheses K over denominator x end fraction close parentheses f, with f equals to

    physics-General
    Let the temperature of common interface be T ℃. Rate of heat flow
    H equals fraction numerator Q over denominator t end fraction equals blank fraction numerator K A increment T over denominator l end fraction
    therefore H subscript 1 end subscript equals open parentheses fraction numerator Q over denominator t end fraction close parentheses subscript 1 end subscript=fraction numerator 2 K A left parenthesis T minus T subscript 1 right parenthesis end subscript over denominator 4 x end fraction
    And H subscript 2 end subscript equals open parentheses fraction numerator Q over denominator t end fraction close parentheses subscript 2 end subscript=fraction numerator K A left parenthesis T subscript 2 end subscript minus T right parenthesis over denominator x end fraction
    In steady state, the rate of heat flow should be same in whole system i e comma
    H subscript 1 end subscript equals blank H subscript 2 end subscript
    rightwards double arrow fraction numerator 2 K A left parenthesis T minus T subscript 1 end subscript right parenthesis over denominator 4 x end fraction = fraction numerator K A left parenthesis T subscript 2 end subscript minus T right parenthesis over denominator x end fraction
    rightwards double arrow fraction numerator T minus T subscript 1 end subscript over denominator 2 end fraction equals blank T subscript 2 end subscript minus T
    rightwards double arrow T minus T subscript 1 end subscript equals blank 2 T subscript 2 end subscript minus 2 T
    rightwards double arrow T equals fraction numerator 2 T subscript 2 end subscript plus T subscript 1 end subscript over denominator 3 end fraction (i)
    Hence, heat flow from composite slab is
    H equals blank fraction numerator K A left parenthesis T subscript 2 end subscript minus T right parenthesis over denominator x end fraction
    = fraction numerator K A over denominator x end fraction open parentheses T subscript 2 end subscript minus fraction numerator 2 T subscript 2 end subscript plus T subscript 1 end subscript over denominator 3 end fraction close parentheses equals blank fraction numerator K A over denominator 3 x end fraction blank left parenthesis T subscript 2 end subscript minus T subscript 1 end subscript right parenthesis (ii)
    Accordingly, H equals open square brackets fraction numerator A open parentheses T subscript 2 end subscript minus T subscript 1 end subscript close parentheses K over denominator x end fraction close square brackets f (iii)
    By comparing Eqs. (ii) and (iii), we get
    rightwards double arrow f equals blank fraction numerator 1 over denominator 3 end fraction
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    A solid material is supplied with heat at constant rate and the temperature of the material changes as shown. From the graph, the false conclusion drawn is

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    A metal rod of length 2 m has cross sectional areas 2 A and A as shown in figure. The ends are maintained at temperatures 100 ℃ and 70 ℃. The temperature at middle point C is

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    rightwards double arrow K open parentheses 2 A close parentheses open parentheses 100 minus theta close parentheses equals K A left parenthesis theta minus 70 right parenthesis
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    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?

    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
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    12theta equals 900
    theta equals 75 ℃

    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?

    physics-General
    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
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    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 or 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
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    maths-General
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    parallel
    General
    maths-

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