MCQ
Two rods are joined between fixed supports as shown in the figure. Condition for no change in the lengths of individual rods with the increase of temperature will be

( ${\alpha _1},\,{\alpha _2},$ = linear expansion coefficient

$A_1, A_2$ = Area of rods

$Y_1, Y_2$ = Young modulus)

  • A
    $\frac{{{A_1}}}{{{A_2}}}\, = \,\frac{{{\alpha _1}{Y_1}}}{{{\alpha _2}{Y_2}}}$
  • B
    $\frac{{{A_1}}}{{{A_2}}}\, = \,\frac{{{L_1}{\alpha _1}{Y_1}}}{{{L_2}{\alpha _2}{Y_2}}}$
  • C
    $\frac{{{A_1}}}{{{A_2}}}\, = \,\frac{{{L_2}{\alpha _2}{Y_2}}}{{{L_1}{\alpha _1}{Y_1}}}$
  • $\frac{{{A_1}}}{{{A_2}}}\, = \,\frac{{{\alpha _2}{Y_2}}}{{{\alpha _1}{Y_1}}}$

Answer

Correct option: D.
$\frac{{{A_1}}}{{{A_2}}}\, = \,\frac{{{\alpha _2}{Y_2}}}{{{\alpha _1}{Y_1}}}$
d
$Y=\frac{\text { Stress }}{\text { Strain }}=\frac{T / A}{\Delta l / l}$

$\mathrm{T}=\frac{\mathrm{Y} \Delta l}{l}=\mathrm{A}=\mathrm{Y} \cdot \mathrm{A} \alpha \Delta \mathrm{T}$

In both the rods tension will be some so

$\mathrm{T}_{1}=\mathrm{T}_{2},$ Hence $\mathrm{Y}_{1} \mathrm{A}_{1} \alpha_{1} \Delta \mathrm{T}=\mathrm{Y}_{2} \mathrm{A}_{2} \alpha_{2} \Delta \mathrm{T}$

$\frac{A_{1}}{A_{2}}=\frac{Y_{2} \alpha_{2}}{Y_{1} \alpha_{1}}$

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