Question

If the temperature of a wire of length $2{\text{ }}m$ area of cross-section $1{\text{ }}c{m^2}$ is increased from ${0^ \circ }$to ${80^ \circ }$ and is not allowed to increases in length, then the force required for it is $\left\{ {Y = {{10}^{10}}{\text{ }}\dfrac{N}{{{m^2}}},{\text{ }}\alpha = {{10}^{ - 6}}{\text{ }}\dfrac{1}{{^ \circ C}}} \right\}$
A. $80{\text{ }}N$
B. $160{\text{ }}N$
C. $400{\text{ }}N$
D. $120{\text{ }}N$

Answer 1

Hint: We have to find the change in length from the formula of longitudinal expansion. Then from the Young’s Modulus we can find out the force after substituting the other variables like area of cross-section, change in length and original length.

Formula used:
Young’s Modulus $Y = \dfrac{{\dfrac{F}{A}}}{{\dfrac{{\vartriangle l}}{l}}}$ is the formula through which it can be solved. Longitudinal expansion $\vartriangle l = l\alpha \vartriangle T$ where $\vartriangle l$ is the change in length, $l$ is the original length, $\alpha $ is the coefficient of longitudinal expansion and $\vartriangle T$ is the change in temperature

Complete step by step answer:
Firstly we have to find the change in length from the formula,
$\vartriangle l = l\alpha \vartriangle T - - - - \left( 1 \right)$
Where $\vartriangle l$ is the change in length, $l$ is the original length, $\alpha $ is the coefficient of longitudinal expansion and $\vartriangle T$ is the change in temperature.
Change in temperature $\vartriangle T$$ = {80^ \circ } - {0^ \circ } = {80^ \circ }$
Original length $l = 2{\text{ }}m$

Coefficient of longitudinal expansion $\alpha = {10^{ - 6}}{\text{ }}\dfrac{1}{{^ \circ C}}$
Substituting all these values in the equation $\left( 1 \right)$ we get,
$\vartriangle l = 2 \times {10^{ - 6}} \times 80 \\
\Rightarrow \vartriangle l = 16 \times {10^{ - 5}}{\text{ m}} $
Now, we had find out the change in length as,
$\vartriangle l = 16 \times {10^{ - 5}}{\text{ m}}$
Now we have to use the Young’s Modulus $Y = \dfrac{{\dfrac{F}{A}}}{{\dfrac{{\vartriangle l}}{l}}}$ to find out force.

Here as mentioned in the question,
Area of cross-section $A = 1{\text{ }}c{m^2} = {10^{ - 4}}{\text{ }}{m^2}$
Young’s Modulus $Y = $${10^{10}}{\text{ }}\dfrac{N}{{{m^2}}}$
Substituting the values in $Y = \dfrac{{\dfrac{F}{A}}}{{\dfrac{{\vartriangle l}}{l}}}$ we get,
By cross-multiplication we have found out force $F$ as,
$F = \dfrac{{Y \times \dfrac{{\vartriangle l}}{l}}}{A}$
We got the value of $\vartriangle l$ from equation $\left( 1 \right)$,
$ \Rightarrow F = \dfrac{{{{10}^{10}} \times 8 \times {{10}^{ - 5}}}}{{{{10}^{ - 4}}}} \\
\therefore F= 80\;N$
Thus, the Force is $80{\text{ }}N$.

So, the correct option is A.

Note: For figuring out the unknown variables we had used the formula of longitudinal expansion. The question has stated that there is no change in length but it does not imply that the rod will not expand longitudinally. Intentionally this type of question is being set up to discomfort the student. We must find the value of change in length in order to solve the problems. The values are also entitled in various units which we have to standardize.