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The variation of length of two metal rod...


The variation of length of two metal rods A and B with change in temperature is shown in Fig. the coefficient of linear expansion `alpha_A` for the metal A and the temperature T will be

A

`alpha_(A)=3 xx 10^(-6)//^(@)C,500^(@)C`

B

`alpha_(A)=3 xx 10^(-6)//^(@)C,222.22^(@)C`

C

`alpha_(A)=27 xx 10^(-6)//^(@)C,500^(@)C`

D

`alpha_(A)=27xx 10^(-6)//^(@)C,222.22^(@)C`

Text Solution

Verified by Experts

The correct Answer is:
D

Slope of line `A=((1006-1000)mm)/(T^@C)=(DeltaL)/(DeltaT)=Lalpha_A`
i.e., `(6)/(T)mm//^(@)C=(1000mm)alpha_A`
similary for line B
`(2)/(T)mm//^(@)C=(1002mm)alpha_B`
Dividing Eq. (i) by Eq. (ii)
`3=(1000 alpha_A)/(1002alpha_B)approx=3alpha=3alpha_B`
From Eq. (iii) `alpha_A=3xx9xx10^-6=27xx10^(-6)//^(@)C`
from eq. (i) `T=(6)/(1000alpha_A)=(6xx10^6)/(1000xx27)`
`=222.22^@C`
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