Question

Two meters scale, one of steel and the other of aluminum, agree at $20^\circ C$ . Calculate the ratio of length of aluminum and steel at,
(a) $0^\circ C$
(b) $40^\circ C$ and
(c) $100^\circ C$
($\alpha $ for steel =$1.1 \times {10^{ - 5}}^\circ {C^{ - 1}}$ and for aluminum = $2.3 \times {10^{ - 5}}^\circ {C^{ - 1}}$)

Answer 1

Hint: We can use the formula ${L_f} = {L_i}(1 + \alpha \Delta T)$, where ${L_f}$ and ${L_i}$ is the final length and the initial length of the meter scale respectively, the formula for linear thermal expansion of a material.

Complete step by step solution:
Let the length of the meter scale of steel at $20^\circ C$ be ${L_s}$ and for that of aluminum be ${L_a}$.
As given, the coefficient of linear expansion for steel and aluminum are ${\alpha _s} = 1.1 \times {10^{ - 5}}^\circ {C^{ - 1}}$ and ${\alpha _a} = 2.3 \times {10^{ - 5}}^\circ {C^{ - 1}}$ respectively.
(a) The ratio of aluminum-centimeter / steel-centimeter at $0^\circ C$ will be:
As given in the question:
${L_{s(20)}} = {L_{a(20)}}$
Let the length of the meter scales of steel and aluminum be ${L_{s(0)}}$ and ${L_{a(0)}}$ respectively, the length of both the rods at $20^\circ C$ will be:
$\Rightarrow {L_{s(20)}} = {L_{s(0)}}(1 - \alpha \Delta T)$
$ \Rightarrow {L_{s(0)}}(1 - \alpha \times 20)$
$ \Rightarrow {L_{s(0)}}(1 - (1.1 \times {10^{ - 5}} \times 20))$
$\Rightarrow {L_{a(20)}} = {L_{a(0)}}(1 - \alpha \Delta T)$
$ \Rightarrow {L_{a(0)}}(1 - (\alpha \times 20))$
$ \Rightarrow {L_{a(0)}}(1 - (2.3 \times {10^{ - 5}} \times 20))$
Since ${L_{s(20)}} = {L_{a(20)}}$
Therefore, ${L_{a(0)}}(1 - (2.3 \times {10^{ - 5}} \times 20)) = {L_{s(0)}}(1 - (1.1 \times {10^{ - 5}} \times 20))$
$\Rightarrow \dfrac{{{L_{a(0)}}}}{{{L_{s(0)}}}} = \dfrac{{1 - (1.1 \times {{10}^{ - 5}} \times 20)}}{{1 - (2.3 \times {{10}^{ - 5}} \times 20)}}$
$\Rightarrow \dfrac{{{L_{a(0)}}}}{{{L_{s(0)}}}} = 0.000240011$
Therefore, the ratio of aluminum-centimeter/steel-centimeter at $0^\circ C$ is $0.000240011$.

(b) The ratio of aluminum-centimeter / steel-centimeter at $40^\circ C$ will be:
Similarly, we can find out the ratio at $40^\circ C$ as we found out at $0^\circ C$.
The length of the meter scales of steel and aluminum be ${L_{s(40)}}$ and ${L_{a(40)}}$ respectively, the length of both the rods at $20^\circ C$ will be:
$\Rightarrow {L_{s(20)}} = {L_{s(40)}}(1 + \alpha \Delta T)$
$ \Rightarrow {L_{s(40)}}(1 + \alpha \times 20)$
$ \Rightarrow {L_{s(40)}}(1 + (1.1 \times {10^{ - 5}} \times 20))$
$\Rightarrow {L_{a(20)}} = {L_{a(40)}}(1 + \alpha \Delta T)$
$ \Rightarrow {L_{a(40)}}(1 + (\alpha \times 20))$
$ \Rightarrow {L_{a(40)}}(1 + (2.3 \times {10^{ - 5}} \times 20))$
Since ${L_{s(20)}} = {L_{a(20)}}$
Therefore, ${L_{a(40)}}(1 + (2.3 \times {10^{ - 5}} \times 20)) = {L_{s(40)}}(1 + (1.1 \times {10^{ - 5}} \times 20))$
$\Rightarrow \dfrac{{{L_{a(40)}}}}{{{L_{s(40)}}}} = \dfrac{{1 + (1.1 \times {{10}^{ - 5}} \times 20)}}{{1 + (2.3 \times {{10}^{ - 5}} \times 20)}}$
$\Rightarrow \dfrac{{{L_{a(40)}}}}{{{L_{s(40)}}}} = 1.00023995$
Therefore, the ratio of aluminum-centimeter/steel-centimeter at $40^\circ C$ is $1.00023995$.

(c) The ratio of aluminum-centimeter / steel-centimeter at $100^\circ C$ will be:
The length of the meter scales of steel and aluminum be ${L_{s(100)}}$ and ${L_{a(100)}}$ respectively, the length of both the rods at $20^\circ C$ will be:
$\Rightarrow {L_{s(20)}} = {L_{s(100)}}(1 + \alpha \Delta T)$
$ \Rightarrow {L_{s(100)}}(1 + \alpha \times 80)$
$ \Rightarrow {L_{s(100)}}(1 + (1.1 \times {10^{ - 5}} \times 80))$
$\Rightarrow {L_{a(20)}} = {L_{a(100)}}(1 + \alpha \Delta T)$
$ \Rightarrow {L_{a(100)}}(1 + (\alpha \times 80))$
$ \Rightarrow {L_{a(100)}}(1 + (2.3 \times {10^{ - 5}} \times 80))$
Since ${L_{s(20)}} = {L_{a(20)}}$
Therefore, ${L_{a(100)}}(1 + (2.3 \times {10^{ - 5}} \times 80)) = {L_{s(100)}}(1 + (1.1 \times {10^{ - 5}} \times 80))$
$\Rightarrow \dfrac{{{L_{a(100)}}}}{{{L_{s(100)}}}} = \dfrac{{1 + (1.1 \times {{10}^{ - 5}} \times 80)}}{{1 + (2.3 \times {{10}^{ - 5}} \times 80)}}$
$\Rightarrow \dfrac{{{L_{a(100)}}}}{{{L_{s(100)}}}} = 1.00095916$
Therefore, the ratio of aluminum-centimeter/steel-centimeter at $100^\circ C$ is $1.00095916$.

Note: The condition that both the aluminum meter scale and steel scale agree at $20^\circ C$ should be used in each case. $\Delta T$ is the difference between the final and the initial temperature of the body and therefore affects the sign of the equation: the value would be negative if the body goes from higher temperature to lower one and positive for the other case.