We would like to prepare a scale whose length does not change with temperature. It is proposed to prepare a unit scale of this type whose length remains, say 10 cm. We can use a bimetallic strip made of brass and iron each of different length whose length (both components) would change in such a way that difference between their lengths remain constant. If alpha_("iron")=1.2xx10^(-5)//K and alpha_("brass")=1.8xx10^(-5)//K what should we take as length of each strip ?

We would like to prepare a scale whose length does not change with tem

1.	We would like to prepare a scale whose length does not change with temperature. It is proposed to prepare a unit scale of this type whose length remains, say 10 cm. We can use a bimetallic strip made of brass and iron each of different length whose length (both components) would change in such a way that difference between their lengths remain constant. If alpha_("iron")=1.2xx10^(-5)//K and alpha_("brass")=1.8xx10^(-5)//K what should we take as length of each strip ?
Answer» Solution :ACCORDING to question, `l_("iron")-l_("brass")=10cm` . . . (1) Constant at all temperature. Suppose LENGTH at `0^(@)C` is `l_(0)` and it becomes l due to INCREASE in temperature `DeltaT` `:.l_("iron")=l_("0iron")(1+alpha_("iron")DeltaT)` and `l_("brass")=l_("0brass")(1+alpha_("brass")DeltaT)` `:.` From equation (1), `l_("0iron")(1+alpha_("iron")DeltaT)-l_("0brass")(1+alpha_("brass")DeltaT)=10cm` But the difference of their LENGTHS remains constant, `:. L_("0iron")alpha_("iron")=l_("0brass")alpha_("brass")` `:.(l_("0iron"))/(l_("0brass"))=(alpha_("brass"))/(alpha_("iron"))=(1.8xx10^(-5))/(1.2xx10^(-5))=3/2` `:.l_("0iron")=3/2l_("0brass")` Now, from equation (1), `3/2l_("0brass")-l_("0brass")=10` `:.1/2l_("0brass")=10` `:.l_("0 brass")=20cm` `:.` Original length of brass scale is 10 cm. Now, again from equation (1), `l_("0iron")-l_("0 brass")=10` `l_("0iron")-20=10` `:.l_("0iron")=30cm` `:.` Original length of iron scale is 30 cm.