Question

Two rods having length $l_1$ and $l_2$ made of materials with the linear expansion coefficient $a_1$ and $a_2$ were soldered together. The equivalent coefficients of linear expansion for the obtained rod:

• A. $\frac{l_1{a}_2+l_2{a}_1}{a_1+a_2}$
• B. $\frac{l_1{a}_1+l_2{a}_2}{l_1+l_2}$
• C. $\frac{l_2{a}_1+l_1{a}_2}{a_1+a_2}$
• D. $\frac{l_1{a}_2+l_2{a}_1}{l_1+l_2}$

Answer 1

Answer: (B)

STEP 1: The coefficient of the linear expansion of first rod of length $L_1$ is:

$\alpha 1=\frac{\Delta I1}{I_1\times \Delta t}$

$\Delta I1=\alpha 1I_1\Delta t.....\left(1\right)$

STEP 2: The coefficient of the linear expansion of second rod of length $I_2$ is:

$\alpha 2=\frac{\Delta I2}{I_2\times \Delta t}$

$\Delta I2=\alpha 2I_2\Delta t.......\left(2\right)$

STEP 3: Now, if a single rod of linear expansion $\alpha$is taken instead of two rods.

$\alpha =\frac{\Delta I1+\Delta I2}{(I1+I2)\times \Delta t}......\left(3\right)$

STEP 4: Putting the values of $\Delta I1and\Delta I2$in equation 3, we get;

$\alpha =\frac{\alpha 1I1\Delta t+\alpha 2I2\Delta t}{(I1+I2)\Delta t}$

$\alpha =\frac{I1\alpha 1+I2\alpha 2}{I1+I2}$

Hence, the correct option is (c) $\frac{I1\alpha 1+I2\alpha 2}{I1+I2}$