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 αiron = 1.2 x 10-5 / K and αbrass = 1.8 × 10-5 / K, what should we take as length of each strip?
Answer» From question. l_iron − l_brass = 10 cm = constant at all temperatures. Let l₀ be length at temperature 0ºC and length l after change in temperature. \(l^o_{iron}\)− \(l^o_{brass}\) = 10 cm at all temperature ∴ \(l^o_{iron}\) (1 + α_iron∆t) − \(l^o_{brass}\)(1 + α_brass∆t) = 10 cm \(l^o_{iron}\)\(\alpha^o_{iron}\) = \(l^0_{brass} \)α_brass ∴ \(\frac{l^o_{iron}}{l^o_{brass}}=\frac{1.8\times10^{-5}}{1.2\times10^{-5}}=\frac{1.8}{1.2}=\frac{3}{2}\) Then, \(l^o_{iron}\)- \(l^0_{brass} \) = 10 3x - 2x = 10 x = 10 Length of iron rod = 3 × 10 = 30 cm Length of brass rod = 2 × 10 = 20 cm The difference between is 10cm.

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 αiron = 1.2 x 10-5 / K and αbrass = 1.8 × 10-5 / K, what should we take as length of each strip?

Answer»

From question. l_iron − l_brass = 10 cm = constant at all temperatures. Let l₀ be length at temperature 0ºC and length l after change in temperature.

\(l^o_{iron}\)− \(l^o_{brass}\) = 10 cm at all temperature

∴ \(l^o_{iron}\) (1 + α_iron∆t) − \(l^o_{brass}\)(1 + α_brass∆t) = 10 cm

\(l^o_{iron}\)\(\alpha^o_{iron}\) = \(l^0_{brass} \)α_brass

∴ \(\frac{l^o_{iron}}{l^o_{brass}}=\frac{1.8\times10^{-5}}{1.2\times10^{-5}}=\frac{1.8}{1.2}=\frac{3}{2}\)

Then,

\(l^o_{iron}\)- \(l^0_{brass} \) = 10

3x - 2x = 10

x = 10

Length of iron rod = 3 × 10 = 30 cm

Length of brass rod = 2 × 10 = 20 cm

The difference between is 10cm.