Question

A metal rod 'A' of length $l_0$ expands by $\Delta l$ when its temperature is raised by ${100}^{0}C$. Another rod 'B' of different metal of length $2l_0$ expands by $\Delta l/2$ for same rise in temperature.A third rod 'C' of length $3l_0$ is made up of pieces of rods 'A' and 'B' placed end to end expands by $\Delta$ on heating through 100k . Find the length of each portion of the composite rod.

Answer 1

Case 1

Metal rod A, length $=l_0$

expansion$Al$

Temperature rise in $100°C$

Case 2

Another rod B of different metal of length $=2l_0$

Expansion$=\frac{\Delta l}{2}$

Temperature raised by $100°C$

Case-3

Third rod C of length$=3l_0$

Third rod is made up of a piece of rods A and B heating through $=100K$

Expansion $=\Delta =273-100=173°C$

Solution

Let, ${\alpha }_1$ and ${\alpha }_2$be the temperature coefficients of linear expansion of two rods A and B respectively.

$\Delta l=l_0{\alpha }_1\times 100\Delta l=100{\alpha }_1{l}_0{\alpha }_1=\frac{\Delta l}{100l_0}\to \left(1\right)$

and

$\frac{\Delta l}{2}=2l_0\times {\alpha }_2\times 100=200l_0{\alpha }_2\Delta l=400l_0{\alpha }_2{\alpha }_2=\frac{\Delta l}{400l_0}\to \left(2\right)$

Let, the composite rod be the made of $L_1^1$ length of A and $L_2$ length of B$

$\therefore L=L_1^1=L_2^1=3l_0\to \left(3\right)$

When, this rod is heated through $100K$

Then increase in the length is

$\Delta =[L_1^1{\alpha }_1+L_2^1{\alpha }_2]\times (273-100)°C=[L_1^1\frac{\Delta l}{100l_0}+L_2^1\frac{\Delta l}{400l_0}]\left(173°\right)C=\frac{\Delta l}{100l_0}(L_1^1+\frac{L_2^1}{4})\left(173°\right)°$