Question

A metal rod of length 2m has its cross-sectional area \[1c{m^2}\]. By raising the temperature from ${0^ \circ }$ to ${100^ \circ }$ and holding it so that it is not permitted to expand or bend, the force developed is? (given, $Y = {10^{12}}\dfrac{{dyne}}{{c{m^2}}}$, $\alpha = \dfrac{{{{10}^{ - 5}}}}{{{C^ \circ }}}$).
a. ${10^3}N$
b. $2 \times {10^3}N$
c. ${10^4}N$
d. ${10^6}N$

Answer 1

Hint: Stress is defined as the ratio of force to the area of cross section and strain is defined as the ratio of change in length to the original length. The linear thermal expansion is directly proportional to the coefficient of thermal expansion and the coefficient of thermal expansion is different for different materials.

Formula used:
The formula of the young’s modulus in the stress-strain relationship is defined by,
$ \Rightarrow Y = \dfrac{{Fl}}{{A\Delta l}}$
Where force is F, the young’s modulus is Y, the area of cross section is A, the change in length is $\Delta l$ and the length is l.
The formula of the linear thermal expansion is given by,
$ \Rightarrow \Delta l = \alpha l\Delta T$
Where $\Delta T$=change in temperature, $\alpha $= coefficient of thermal expansion, the change in length is $\Delta l$ and original length is l.

Complete step by step answer:
Step1: writing what is given is and what is need to calculate-
Given that, $l = 2m = 2 \times 100cm$
$l = 200cm$,$A = 1c{m^2}$, $Y = {10^{12}}\dfrac{{dyne}}{{c{m^2}}}$ ,$\alpha = \dfrac{{{{10}^{ - 5}}}}{{{C^ \circ }}}$ and we need to calculate the force.
Step2: Use the equation of Young’s modulus of the material and coefficient of thermal linear expansion.
Therefore,
The formula of the young’s modulus in the stress-strain relationship is defined by,
$ \Rightarrow Y = \dfrac{{Fl}}{{A\Delta l}}$
Where force is F, the young’s modulus is Y, the area of cross section is A, the change in length is $\Delta l$ and the length is l.
$ \Rightarrow Y = \dfrac{{Fl}}{{A\Delta l}}$
$ \Rightarrow F = \dfrac{{YA\Delta l}}{l}$………(1)
Now from the Linear thermal expansion equation we have,
$ \Rightarrow \Delta l = \alpha l\Delta T$ ……….(2)
Where $\Delta T$=change in temperature, $\alpha $= coefficient of thermal expansion.
From eq. (1) and (2) we have,
$ \Rightarrow F = \dfrac{{YA\alpha l\Delta T}}{l}$
Step3: Now substitute all the values in above equation and simplifying,
$ \Rightarrow F = \dfrac{{{{10}^{12}} \times 200 \times {{10}^{ - 5}} \times (100 - 0)}}{{200}}$
$ \Rightarrow F = {10^9}Dynes$
$ \Rightarrow F = {10^4}N$

Hence, the correct answer is option (C).

Note: Always remember the unit conversions, Either convert all the units in CGS system or in SI system before doing the final calculations also students are advised to remember the relation between the stress and strain.