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Question: A iron rod of length 50 *cm* is joined at an end to an aluminium rod of length 100 *cm*. All measure...

A iron rod of length 50 cm is joined at an end to an aluminium rod of length 100 cm. All measurements refer to 200C. The coefficients of linear expansion of iron and aluminium are 12×10−6⥂/∘C12 \times 10^{- 6} ⥂ /{^\circ}C and 24×10−6⥂/∘C24 \times 10^{- 6} ⥂ /{^\circ}C

respectively. The average coefficient of composite system is

A

36×10−6⥂/∘C36 \times 10^{- 6} ⥂ /{^\circ}C

B

12×10−6⥂/∘C12 \times 10^{- 6} ⥂ /{^\circ}C

C

20×10−6⥂/∘C20 \times 10^{- 6} ⥂ /{^\circ}C

D

48×10−6⥂/∘C48 \times 10^{- 6} ⥂ /{^\circ}C

Answer

20×10−6⥂/∘C20 \times 10^{- 6} ⥂ /{^\circ}C

Explanation

Solution

Initially (at 200C) length of composite system L = 50 + 100 =

150 cm

Length of iron rod at 1000C

=50[1+12×10−6×(100−20)]=50.048cm= 50\lbrack 1 + 12 \times 10^{- 6} \times (100 - 20)\rbrack = 50.048cmLength of

aluminum rod at 1000C

=100[1+24×10−6×(100−20)]=100.192cm= 100\lbrack 1 + 24 \times 10^{- 6} \times (100 - 20)\rbrack = 100.192cm

Finally (at 1000C) length of composite system L' =

50.048+100.192=150.24cm50.048 + 100.192 = 150.24cm

Change in length of the composite system ∆L = L' – L =

150.24 –150 = 0.24 cm

∴ Average coefficient of expansion at 1000C α=ΔLL×ΔT\alpha = \frac{\Delta L}{L \times \Delta T}= 0.24150×(100−20)\frac{0.24}{150 \times (100 - 20)} = 20×10−6/∘C20 \times 10^{- 6}/{^\circ}C