Question

Two rods P and Q of different metals having the same area A and the same length L are placed between two rigid walls as shown in figure. The coefficients of linear expansion of P and Q are ${\alpha }_1$ and ${\alpha }_2$ respectively and their Youngs modulii are $Y_1$ and $Y_2$ . The temperature of both rods is now raised by T degrees. The new length of rod Q is

• A. $L_2=[1+{\alpha }_{2}T+\frac{F}{A{Y}_2}]L$
• B. $L_2=[1-{\alpha }_{2}T+\frac{F}{A{Y}_2}]L$
• C. $L_2=[1+{\alpha }_{2}T-\frac{F}{A{Y}_2}]L$
• D. $L_2=[1-{\alpha }_{2}T-\frac{F}{A{Y}_2}]L$

Answer 1

Answer: (C)

$L_2=[1+{\alpha }_{2}T-\frac{F}{A{Y}_2}]L$

Since both have same cross section area , hence stress will be same .

For P, $\Delta L=L{\alpha }_{1}T-\frac{\sigma L}{Y_1}$

For Q, $\Delta L=L{\alpha }_{2}T-\frac{\sigma L}{Y_2}$

Let the force be F so $\sigma =F/A$

For Q, $\Delta L=L{\alpha }_{2}T-\frac{FL}{A{Y}_2}$

Final length of Q =$L+\Delta L$ =$L[1+{\alpha }_{2}T-\frac{F}{A{Y}_2}]$