Question 15: Properties of Matter

Question

An aluminium and steel rods having same lengths and cross-sections are joined to make total length of 120 cm at $30^{\circ} \mathrm{C}$. The coefficient of linear expansion of aluminium and steel are $24 	imes 10^{-6} /{ }^{\circ} \mathrm{C}$ and $1.2 	imes 10^{-5} /{ }^{\circ} \mathrm{C}$, respectively. The length of this composite rod when its temperature is raised to $100^{\circ} \mathrm{C}$, is $\_\_\_\_$ cm.

Accepted Answer

To find the final length of the composite rod, we need to calculate the expansion of both the aluminium and steel sections individually and then add them to the original length.
Total Initial Length $\left(\mathrm{L}_1ight)$ is 120 cm at $30^{\circ} \mathrm{C}$.
Since both rods have the same length, the initial length of each rod is, $\mathrm{L}_{\mathrm{Al}}=\frac{120}{2}=60 \mathrm{~cm}$ and $\mathrm{L}_{\mathrm{St}}=\frac{120}{2}=$ 60 cm .
The change in temperature is,
$$ \Delta \mathrm{T}=\mathrm{T}_{	ext {final }}-\mathrm{T}_{	ext {initial }}=100^{\circ} \mathrm{C}-30^{\circ} \mathrm{C}=70^{\circ} \mathrm{C} $$
Coefficients of linear expansion ( $\alpha$ ) for aluminium is $\alpha_{\mathrm{Al}}=24 	imes 10^{-6} /{ }^{\circ} \mathrm{C}$ and that for steel is $\alpha_{\mathrm{St}}=1.2 	imes 10^{-5} /{ }^{\circ} \mathrm{C}=12 	imes 10^{-6} /{ }^{\circ} \mathrm{C}$.
The linear expansion is $\Delta \mathrm{L}=\mathrm{L}_{	ext {initial }} \cdot \alpha \cdot \Delta \mathrm{T}$
For Aluminium :
$$ \Delta \mathrm{L}_{\mathrm{Al}}=60 	imes\left(24 	imes 10^{-6}ight) 	imes 70 $$
$$\Rightarrow $$ $\Delta \mathrm{L}_{\mathrm{Al}}=60 	imes 70 	imes 24 	imes 10^{-6}$
$$\Rightarrow $$ $$ \Delta \mathrm{L}_{\mathrm{Al}}=4200 	imes 24 	imes 10^{-6}=100800 	imes 10^{-6}=0.1008 \mathrm{~cm} $$
For Steel :
$$ \Delta \mathrm{L}_{\mathrm{St}}=60 	imes\left(12 	imes 10^{-6}ight) 	imes 70 $$
$$\Rightarrow $$ $$ \Delta \mathrm{L}_{\mathrm{St}}=4200 	imes 12 	imes 10^{-6}=50400 	imes 10^{-6}=0.0504 \mathrm{~cm} $$
The total change in length ( $\Delta \mathrm{L}_{	ext {total }}$ ) is :
$$ \Delta \mathrm{L}_{	ext {total }}=\Delta \mathrm{L}_{\mathrm{Al}}+\Delta \mathrm{L}_{\mathrm{St}} $$
$$\Rightarrow $$ $$ \Delta \mathrm{L}_{	ext {total }}=0.1008+0.0504=0.1512 \mathrm{~cm} $$
So, the final length of the rod is :
$$ \mathrm{L}_{\mathrm{final}}=120+0.1512=120.1512 \mathrm{~cm} $$
$$\Rightarrow $$ $$ \mathrm{L}_{\mathrm{final}} \approx 120.15 \mathrm{~cm} $$
Therefore, the final length of the combination is 120.15 cm .
Hence, the correct option is (C).

Answer

120.20

Answer

120.06

Answer

120.15

Answer

120.03

Properties of Matter

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