Question

A brass rod of length 50cm and diameter 3cm is joined to a steel rod of the same length and diameter. What is the change in length of the combined rod at \[{250^{\text{o}}}\,{\text{C}}\], if the original lengths are at \[{40^{\text{o}}}\,{\text{C}}\]? (Coefficient of linear expansion of brass \[ = \] \[2.0 \times {10^{ - 5}}{{\text{K}}^{ - 1}}\], steel \[ = \] \[1.2 \times {10^{ - 5}}{{\text{K}}^{ - 1}}\])
A. \[0.27\,{\text{cm}}\]
B. \[{\text{0}}{\text{.34}}\,{\text{cm}}\]
C. \[{\text{0}}{\text{.21}}\,{\text{cm}}\]
D. \[{\text{0}}{\text{.18 cm}}\]

Answer 1

Hint:We use the given information about brass rod and steel rod, and also use the coefficient of linear expansion of both brass and steel. We will find the expansions separately and then add them to find the desired result.

Complete step by step solution:
Given,
Initial temperature, \[{T_1} = {40^{\text{o}}}\,{\text{C}}\]
Final temperature, \[{T_2} = {250^{\text{o}}}\,{\text{C}}\]
So, the temperature difference is,
$\Delta T = {T_2} - {T_1} \\
\Delta T = {210^{\text{o}}}\,{\text{C}} \\$
The brass rod length at \[{T_1},\,{l_1} = 50\,{\text{cm}}\]
The brass rod diameter at \[{T_1},{d_1} = 3.0\,{\text{cm}}\]
The steel rod length at \[{T_2},\,{l_2} = 50\,{\text{cm}}\]
The steel rod diameter at \[{T_2},\,{d_2} = 3.0\,{\text{mm}}\]
Coefficient of linear expansion of brass is, \[{\alpha _1} = 2.0 \times {10^{ - 5}}{{\text{K}}^{ - 1}}\]
Coefficient of linear expansion of steel is, \[{\alpha _2} = 1.2 \times {10^{ - 5}}{{\text{K}}^{ - 1}}\]
The expansion in the brass rod is
\[\dfrac{{\Delta {l_1}}}{{{l_1}}} = {\alpha _1}\Delta T\]
Where,
\[\Delta {l_1}\] indicates change in length.
\[{l_1}\] indicates the original length.
Therefore,
\Delta {l_1} = 50 \times \left( {2.1 \times {{10}^{ - 5}}} \right) \times 210 \\
\Delta {l_1} = 0.2205\,{\text{cm}} \\
The expansion in the steel rod is
\[\dfrac{{\Delta {l_2}}}{{{l_2}}} = {\alpha _2}\Delta T\]
Where,
\[\Delta {l_2}\] indicates change in length.
\[{l_2}\] indicates the original length.
Therefore,
$\Delta {l_2} = 50 \times \left( {1.2 \times {{10}^{ - 5}}} \right) \times 210 \\
\Rightarrow\Delta {l_2} = 0.126\,{\text{cm}} \\$
So, the change in the lengths of steel and brass rod is,
$\Delta l = \Delta {l_1} + \Delta {l_2} \\
\Rightarrow\Delta l = 0.2205 + 0.126 \\
\therefore\Delta l = 0.346\,{\text{cm}}$

Hence, the required answer is \[0.34\,{\text{cm}}\] therefore option B is the right answer.

Additional information:
Coefficient of linear expansion: Expansion means length change or length rise. If the length shift is over a volume along a single dimension (length), then the linear expansion is called. Therefore, it is implied that the change in temperature represents the rate of expansion. This is why the expansion is due. The degree to which the material under the influence of heat radiation can withstand its original form and size is clarified by this definition.

Note: Any material is defined by its linear expansion coefficient. It also differs from material to material. The rate of expansion of a substance depends solely on the cohesive strength between atoms. The force that connects two or more atoms is a unified force.