Question

The coefficient of linear expansion of copper is one and half times that of iron. Identical rods of copper and iron are heated through the same temperature range. The ratio of forces developed in them will be (The Young's modulus for copper and iron is the same)
(A) $ \dfrac{2}{3} $
(B) $ \dfrac{4}{9} $
(C) $ \dfrac{9}{4} $
(D) $ \dfrac{3}{2} $

Answer 1

Hint: We need to use concepts of the modulus of elasticity to find the force acting on each rod. Remember, linear expansion means change or increase in length.

Formula Used: The formulae used in the solution are given here.
The modulus of elasticity $ Y = \dfrac{{{F \mathord{\left/
 {\vphantom {F A}} \right.} A}}}{{{{\Delta l} \mathord{\left/
 {\vphantom {{\Delta l} l}} \right.} l}}} $ .
Where $ Y $ is the Young’s modulus, $ F $ is the force, $ A $ is the area of cross-section, $ \Delta L $ is the increase in length and $ L $ is the length.

Complete step by step answer:
Expansion means, change or increase in length. If the change in length is along one dimension (length) over the volume, then it is called linear expansion. Here the reason behind the expansion is the change in temperature Thus, it is implied that the change in temperature will reflect in the rate of expansion. How much material can withstand its original shape and size under the influence of heat radiation is well explained using this concept.
It has been given that the coefficient of linear expansion of copper is one and half times that of iron. Identical rods of copper and iron are heated through the same temperature range. The Young's modulus for copper and iron is the same.
If $ \alpha $ be coefficient of linear expansion of iron, coefficient of linear expansion of copper is $ \alpha + \dfrac{\alpha }{2} = \dfrac{{3\alpha }}{2} $ .
Let us assume that there are two rods, one of copper and the other of iron of cross-sectional area $ A $ , and length $ l $ . Let the force on the copper rod be $ {F_{Cu}} $ and that on the iron rod be $ {F_{Fe}} $ .
The modulus of elasticity $ Y = \dfrac{{{F \mathord{\left/
 {\vphantom {F A}} \right.} A}}}{{{{\Delta l} \mathord{\left/
 {\vphantom {{\Delta l} l}} \right.} l}}} $ .
Thus, $ F = AY{\alpha _L}dT $ where $ {\alpha _L} $ is the change in length/linear expansion coefficient and $ dT $ is the change in temperature.
The thermal force on copper is given by, $ {F_{Cu}} = AY\left( {\dfrac{{3\alpha }}{2}} \right)dT $ where
The thermal force on iron is given by, $ {F_{Fe}} = AY\alpha dT $ .
To find the ratio between the two,
 $ \dfrac{{{F_{Cu}}}}{{{F_{Fe}}}} = \dfrac{{AY\left( {\dfrac{{3\alpha }}{2}} \right)dT}}{{AY\alpha dT}} = \dfrac{3}{2} $ .
Hence, the correct answer is Option D.

Note:
Assuming that the effect of pressure is negligible, the coefficient of Linear Expansion is the rate of change of unit length per unit degree change in temperature. The coefficient of linear expansion can be mathematically written as: $ {\alpha _L} = \dfrac{{dL}}{{dT}} $ where, $ {\alpha _L} $ is the coefficient of linear expansion, $ dL $ is the unit change in length and $ dT $ is the unit change in temperature.