Question

A rod of length $ L $ and radius $ r $ is held between two rigid walls so that it is not allowed to expand. If its temperature is increased then the force developed in it is proportional to
(A) $ L $
(B) $ 1/L $
(C) $ {r^2} $
(D) $ {r^{ - 2}} $

Answer 1

Hint: These types of problems are related to the topic of elastic properties of matter. There are three types of elastic moduli: Young’s modulus, Rigidity modulus, and Bulk modulus. To solve this question we need to use the formula of Young’s modulus.

Formula used:
 $ Y = \dfrac{{F/A}}{{\Delta L/L}} $
where, $ Y $ represents the young’s modulus,
 $ F $ is the force generated on the rod,
 $ A $ is the cross-sectional area of the rod,
 $ L $ is the length of the rod, and
 $ \Delta L $ is the increase in the length due to expansion caused by the increase in temperature.

Complete step by step answer:
We know that young’s modulus is defined as,
 $ Y = \dfrac{{stress}}{{strain}} $ ………. $ (1) $
Again, we know that,
 $ stress = \dfrac{F}{A} $
where, $ F $ is the force generated on the rod, and
 $ A $ is the cross-sectional area of the rod.
Also, $ strain = \dfrac{{\Delta L}}{L} $
where, $ \Delta L $ is the change is length, and
 $ L $ is the original length.
Substituting these values in equation $ (1) $ we get the young’s modulus as,
 $ Y = \dfrac{{F/A}}{{\Delta L/L}} $
Upon cross multiplying the above equation we get,
 $ F = \dfrac{{(Y\Delta L)A}}{L} $
Now, in the question, it is given that the rod is held between two rigid walls so that it is not allowed to expand. Hence, its length cannot be changed. Therefore from the above equation, we have the quantity $ \dfrac{{Y\Delta L}}{L} $ as constant.
Therefore we can say that the force is proportional to the area, or
 $ F \propto A $
The radius of the rod is given as $ r $ .
Hence the cross-sectional area of the rod, $ A = \pi {r^2} $ .
Therefore, $ F \propto \pi {r^2} $ .
Again, $ \pi $ is a constant.
Therefore we get, $ F \propto {r^2} $ .
Thus if its temperature is increased then the force developed in the rod, $ F $ is proportional to $ {r^2} $ .
Therefore the correct answer is option (3) $ {r^2} $ .

Note:
The three moduli of elasticity can be expressed in relation to each other as $ Y = \dfrac{{9KG}}{{G + 3K}} $ , where $ G $ is the rigidity modulus, and $ K $ is the bulk modulus. We can arrive at this equation by eliminating the Poisson’s ratio, $ \mu $ from the relations $ Y = 3K(1 - 2\mu ) $ and $ Y = 2G(1 + 2\mu ) $ .