Question

There is some change in length when a 33000N tensile force is applied on a steel rod of area of cross-section ${{10}^{-3}}{{m}^{2}}$. The change of temperature required to produce the same elongation, if the steel rod is heated is (The modulus of elasticity is $3\times {{10}^{11}}N/{{m}^{2}}$and the coefficient of linear equation if steel is $1.1\times {{10}^{-5}}{{/}^{0}}C$
A. $20^{0}C$
B. ${{15}^{0}}C$
C. ${{10}^{0}}C$
D. ${{0}^{0}}C$

Answer 1

We are given that a force is being exerted, we know force can produce some extension in length and that can be found out using Young’s Modulus. We also know that change in temperature can also result in extension. So, we can find out the extension in the two cases and since the change is the same, we can equate both to find out the change in temperature.

Complete step by step answer:
We know Young’s modulus is given by $Y=\dfrac{Fl}{A\Delta l}$, where F is the force applied, l is the original length and A is the area. So, elongation produced can be given as $\Delta l=\dfrac{Fl}{AY}$--(1)
Also, elongation in length due to change of temperature is given by $\Delta l=l\alpha \Delta T$, where $\Delta l$is the change in the length and $\alpha$ is the coefficient of linear expansion.
Since, the change in length is equal in both the case, equating this with the equation (1) we get,
$\dfrac{Fl}{AY}=l\alpha \Delta T$
$\Rightarrow\dfrac{F}{AY}=\alpha \Delta T$
$\Rightarrow\dfrac{F}{AY\alpha }=\Delta T$
Putting the values, we get,
$\Delta T=\dfrac{33000}{{{10}^{-3}}\times 3\times {{10}^{11}}\times 1.1\times {{10}^{-5}}}$
$\therefore\Delta T={{10}^{0}}C$

So, the correct option is C.
Note: Young’s modulus is dependent on the nature of the material and it does not depend on other parameters such as length or area of the cross-section or the temperature.Here the area was uniform and thus we did not have to take into account deformation of shape. If we are given a problem in which two wires of different materials, then we have to use two sets of young’s moduli for each of them.