Question

A unit scale is to be prepared whose length does not change with temperature and remains $20\, cm$, using a bimetallic strip made of brass and iron each of different length The length of both components would change in such a way that difference between their lengths remains constant If length of brass is $40 \,cm$ and length of iron will be ________$cm$\((\alpha_{iron}=1.2\times10^{-5}K^{-1} \,\,\,and \,\,\,\alpha_{brass}=1.8\times10^{-5}K^{-1}).\)

Answer 1

Correct Answer: 60

To solve this problem, we need to ensure that the length of the bimetallic strip remains constant at \(20\, \text{cm}\) despite temperature changes. The strip is made of brass and iron, and the key is that the difference in their lengths due to thermal expansion must remain constant.

The thermal expansion of a material is given by the formula: \[ \Delta L = L_0 \alpha \Delta T \] where \(L_0\) is the initial length, \(\alpha\) is the coefficient of linear expansion, and \(\Delta T\) is the change in temperature.

Let's denote:

• \(L_{\text{brass}} = 40\, \text{cm}\)
• \(L_{\text{iron}}\) (unknown)

The total length of the strip must remain at \(20\, \text{cm}\), so:

\(L_{\text{total}} = L_{\text{brass}} - L_{\text{iron}} = 20\, \text{cm}\)

The condition for constant difference is:

• \(L_{\text{brass}}\alpha_{\text{brass}}\Delta T - L_{\text{iron}}\alpha_{\text{iron}}\Delta T = \text{constant}\)

This simplifies to:

\(40 \cdot 1.8 \times 10^{-5} - L_{\text{iron}} \cdot 1.2 \times 10^{-5} = \text{constant}\)

This constant reflects the requirement that initially, \(L_{\text{brass}} - L_{\text{iron}} = 20\, \text{cm}\).

Simplifying:

\(72 \times 10^{-5} = 1.2 \times 10^{-5} L_{\text{iron}}\)

Solving for \(L_{\text{iron}}\):

\(L_{\text{iron}} = \frac{72 \times 10^{-5}}{1.2 \times 10^{-5}} = 60\, \text{cm}\)

The calculated length of the iron strip is \(60\, \text{cm}\), which is within the given range of 60, 60. Thus, the solution is verified as correct.