Question

Two wires $A$ and $B$ of the same cross-sectional areas, Young's modulii $Y_1$ and $Y_2$ and coefficients of linear expansion ${\alpha }_1$ and ${\alpha }_2$ respectively, are joined together and fixed between rigid supports at either ends at $0{}^{\circ }\text{C}$. The tension in the compound wire when wire $A$ is heated to temperature $T_1$ and wire $B$ is cooled to temperature $T_2$ is equal to tension in the compund wire when only wire $A$ is cooled to temperature $T_2$. Find the correct option.

• A. $\frac{{\alpha }_1}{{\alpha }_2}>\frac{Y_2}{2Y_1}$
• B. $\frac{{\alpha }_1}{{\alpha }_2}<\frac{Y_2}{2Y_1}$
• C. $\frac{{\alpha }_1}{{\alpha }_2}>\frac{2Y_2}{Y_1}$
• D. $\frac{{\alpha }_1}{{\alpha }_2}<\frac{2Y_2}{Y_1}$

Answer 1

The correct option is B $\frac{{\alpha }_1}{{\alpha }_2}<\frac{Y_2}{2Y_1}$

Given:
For wire $A$,
Coefficient of thermal expansion $={\alpha }_1$
Young's modulus $=Y_1$
For wire $B$,
Coefficient of thermal expansion $={\alpha }_2$
Young's modulus $=Y_2$
Initial temperature, $T=0{}^{\circ }\text{C}$

For case-1,
Temperature upto which wire $A$ is heated $=T_1$
Temperature upto which wire $B$ is cooled $=T_2$
For case-2,
Temperature upto which only wire $A$ is cooled $=T_2$

F.B.D of wire $A$:


Here, $F_1$ is the force on wire $A$ due to its expansion on heating to temperature $T_1$ and $F_2$ is the force on wire $A$ due to contraction on cooling of wire $B$ to temperature $T_2$. Also $F_1^{{}^{\prime }}$ is the force on wire $A$ due to contraction on cooling of wire $A$ to temperature $T_2$.
So, in case 1, net tension is
$F_2-F_1=Y_2{A}_2{\alpha }_2{T}_2-Y_1{A}_1{\alpha }_1{T}_1$ .....(1)
[ from formula of linear expansion and Young's modulus ]
and in case 2, net tension is $F_1^{{}^{\prime }}=Y_1{A}_1{\alpha }_1{T}_2$ .......(2)
[taking rightwards as positive]
Given, tension in both cases are equal.
From (1) and (2),
$\Rightarrow Y_2{A}_2{\alpha }_2{T}_2-Y_1{A}_1{\alpha }_1{T}_1=Y_1{A}_1{\alpha }_1{T}_2$
$\Rightarrow Y_2{A}_2{\alpha }_2{T}_2=Y_1{A}_1{\alpha }_1(T_1+T_2)$
$\Rightarrow \frac{(T_1+T_2)}{T_2}=\frac{Y_2{A}_2{\alpha }_2}{Y_1{A}_1{\alpha }_1}$
On applying componendo and dividendo, if $\frac{a}{b}=\frac{c}{d}$ then $\frac{a-2b}{b}=\frac{c-2d}{d}$
$\Rightarrow \frac{T_1-T_2}{T_2}=\frac{Y_2{A}_2{\alpha }_2-2Y_1{A}_1{\alpha }_1}{Y_1{A}_1{\alpha }_1}$
$\Rightarrow \frac{\Delta T}{T_2}=\frac{Y_2{\alpha }_2-2Y_1{\alpha }_1}{Y_1{\alpha }_1}$
[ given cross sectional area are equal ]
As $T_1$ is positive and $T_2$ is negative, so, $\Delta T=T_1-(-T_2)=$ positive
Hence, $Y_2{\alpha }_2-2Y_1{\alpha }_1>0$
$\Rightarrow \frac{{\alpha }_1}{{\alpha }_2}<\frac{Y_2}{2Y_1}$