Question

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

α_{i r o n} = 1.2 \times 10^{- 5} / K

and

α_{b r a s s} = 1.8 \times 10^{- 5} / K

, what should we take as length of each strip?

Accepted Answer

Required scale can be made with the help of iron and brass rod so that their one end coincides at $A$ .

The other end of iron rod is at $α_{i r o n} = 1.2 \times 10^{- 5} / K$ and brass rod is at $P$ . Let the initial lengths of the iron and the brass rods be $L_{I 1}$ and $L_{B 1}$ and their final lengths after expansion be $L_{I 2} and L_{B 2}$ respectively. The difference between the two rods, $PB = 10 cm$ at any temperature $(T)$ .

Let the coefficient of linear expansion of iron and brass rods at any temperáture $T$ be $α_{i r o n} = 1.2 \times 10^{- 5} / K$ and $α_{b r a s s} = 1 .8 \times 10^{- 5} / K$ , respectively.

Coefficient of linear expansion $α = \frac{Δ L}{L_{0} Δ T}$ or $α = \frac{L_{2} - L_{1}}{L_{1} Δ T}$

Therefore $L_{2} = L_{1} + L_{1} α Δ T$

$\Rightarrow L_{2} = L_{1} (1 + α Δ T)$ ...............(i)

The given initial condition: $L_{1 I} - L_{1 B} = 10 cm$ ....(ii)

If combination of rods are heated through temperature $Δ T,$ according to the question the length of the scale remains constant i.e, $L_{2 I} - L_{2 B} = 10 cm$ .

Using (i) and (ii), $L_{1 I} (1 + α_{1} Δ T) - L_{1 B} (1 + α_{B} Δ T) = 10$

$\Rightarrow L_{1 I} + α_{I} L_{1 I} Δ T - L_{1 B} - L_{1 B} α_{B} Δ T = 10$

$\Rightarrow L_{1 I} - L_{1 B} + (α_{1} L_{1 I} - α_{B} L_{1 B}) Δ T = 10$

Since $L_{1 I} - L_{1 B} = 10$

$α_{1} L_{1 I} - α_{B} L_{1 B} = 0$

Hence:

$\frac{L_{1 I}}{L_{1 B}} = \frac{α_{B}}{α_{I}} = \frac{1.8 \times 10^{- 5}}{1.2 \times 10^{- 5}} = \frac{18}{12} = \frac{3}{2}$

$∴ \frac{L_{1 I}}{L_{1 B}} = \frac{3}{2}$

Then let $L_{1 I} = 3 x$ and $L_{1 B} = 2 x$

$L_{1 I} - L_{1 B} = 10$

$3 x - 2 x = 10$

$x = 10$

$∴$ Length of iron rod $= 3 \times 10 = 30 cm$

Length of brass rod $= 2 \times 10 = 20 cm$

Thus, the difference between the two ends remain $10 cm.$

B M Sharma Solutions for Chapter: Thermometry, Thermal Expansion and Calorimetry, Exercise 1: ILLUSTRATION

B M Sharma Physics Solutions for Exercise - B M Sharma Solutions for Chapter: Thermometry, Thermal Expansion and Calorimetry, Exercise 1: ILLUSTRATION

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