Figure. Shows the graphs of elongation versus temperature for two different metals. If these metals are employed to form a straight bimetallic strip of thickness 6 cm and heated, it bends in the form of an arc, the radius of cuvature chaging with temperature approximately as shown in the figure. The linear expansivities of the two metals are

Options

A. `24xx10^(-6)//^@C` and `12xx10^(-6)//^@C`
B. `20xx10^(-6)//^@C` and `10xx10^(-6)//^@C`
C. `18xx10^(-6)//^@C` and 9xx10^(-6)//^@C`
D. `16xx10^(-6)//^@C` and `8xx10^(-6)//^@C`

Answer

D

Explanation

The slope of lines are
`((DeltaL)/(DeltaT))_1=Lalpha_1=(4-0)/(4-0)=1` (For the first metal)
and `((DeltaL)/(DeltaT))_2=Lalpha_2=(2-0)/(4-0)=(1)/(2)` (For the second metal)
and `L_0(1+alpha_2T)=(R-1.5)theta`
Dividing we get `(1+alpha_1T)/(1+alpha_2T)=(R+1.5)/(R-1.5)`
or `(alpha_1-alpha_2)T=(3)/(R )` and `alpha_1-alpha_2=(3)/(RT)` from the graph shown in fig. (b)
`alpha_1-alpha_2=(3)/(7500xx50)=8xx10^(-6)//^@C`
But `(alpha_1)/(alpha_2)=2`
`:. alpha_1=16xx10^(-6)//^@C` and `a_1=8xx10^(-6)//^(@)C`