Question

The ratio of the lengths of two rods is 4:3. The ratio of their coefficients of cubical expansion is 2:3. Then the ratio of their linear expansions when they are heated through same temperature difference is:
A. 2:1
B. 1:2
C. 8:9
D. 9:8

Answer 1

When the temperature of a material increases, the material expands. This expansion happens in three forms: linear, superficial (or areal) and volumetric expansion. Each material has different values of coefficients for each different type of these expansions.

Complete step-by-step answer:
There are three types of expansion: Linear, Superficial and Volumetric.
Linear expansion is the change in length of the material due to change in the temperature. If $\Delta l$ is the change in length of a material from the original length $l$ due to change in temperature $\Delta T$, the coefficient of linear expansion $\left( \alpha \right)$ is given by,
$\dfrac{{\Delta l}}{l} = \alpha \Delta T$
Superficial expansion is the change in area of the material due to change in the temperature. If $\Delta A$ is the change in area of a material from the original length A due to change in temperature $\Delta T$, the coefficient of linear expansion $\left( \beta \right)$ is given by,
$\dfrac{{\Delta A}}{A} = \beta \Delta T$
Volumetric expansion is the change in volume of the material due to change in the temperature. If $\Delta V$ is the change in volume of a material from the original length V due to change in temperature $\Delta T$, the coefficient of linear expansion $\left( \gamma \right)$ is given by,
$\dfrac{{\Delta V}}{V} = \gamma \Delta T$
The relationship between the three coefficients of expansion is given by this formula:
$\dfrac{\alpha }{1} = \dfrac{\beta }{2} = \dfrac{\gamma }{3}$
$\gamma = 3\alpha$
Given in the question:
$\dfrac{{{\gamma _2}}}{{{\gamma _1}}} = \dfrac{2}{3}$
Given the relation of cubic coefficient and linear coefficient
$\gamma = 3\alpha \\\ \to \dfrac{{{\gamma _2}}}{{{\gamma _1}}} = \dfrac{{3{\alpha _2}}}{{3{\alpha _1}}} \\\$
So, the ratio of linear coefficient is –
$\dfrac{{{\alpha _2}}}{{{\alpha _1}}} = \dfrac{2}{3}$
We have the equation for linear expansion as:
$\dfrac{{\Delta l}}{l} = \alpha \Delta T$
Comparing for two different rods, we have –
$\dfrac{{\dfrac{{\Delta {l_2}}}{{{l_2}}}}}{{\dfrac{{\Delta {l_1}}}{{{l_1}}}}} = \dfrac{{{\alpha _2}\Delta T}}{{{\alpha _1}\Delta T}} \\\ \to \dfrac{{\Delta {l_2}}}{{\Delta {l_1}}} = \dfrac{{{\alpha _2}}}{{{\alpha _1}}} \times \dfrac{{{l_2}}}{{{l_1}}} \\\$
Give ratio of the lengths of the rods, $\dfrac{{{l_2}}}{{{l_1}}} = \dfrac{4}{3}$
$\to \dfrac{{\Delta {l_2}}}{{\Delta {l_1}}} = \dfrac{{{\alpha _2}}}{{{\alpha _1}}} \times \dfrac{{{l_2}}}{{{l_1}}}$
$\dfrac{{\Delta {l_2}}}{{\Delta {l_1}}} = \dfrac{2}{3} \times \dfrac{4}{3} = \dfrac{8}{9}$

Hence, the correct option is Option C.

Note: In this problem, we got to understand that the ratios of the coefficients of expansions are equal. Hence, when you get this kind of question next time, do not waste time by deducing the relations between them because you know that primarily,
Ratio of coefficient of linear expansion = Ratio of coefficient of superficial expansion = Ratio of coefficient of volumetric expansion
$\dfrac{{{\alpha _1}}}{{{\alpha _2}}} = \dfrac{{{\beta _1}}}{{{\beta _2}}} = \dfrac{{{\gamma _1}}}{{{\gamma _2}}}$