A rod of length $20cm$ is made of metal $A$ . It expands by $0.075cm$ when its temperature is raised from ${{0}^{0}}C$ to ${{100}^{0}}C$. Another rod of a different metal $B$ having the same length expands by $0.045cm$ for the same change in temperature. A third rod of the same length is composed of two parts, one of metal $A$ and the other of metal $B$. This rod expands by $0.060cm$ for the same change in temperature. The portion of metal $A$ has length
$A)\text{ }20cm$
$B)\text{ 1}0cm$
$C)\text{ 15}cm$
$D)\text{ 18}cm$
Answer
656.1k+ views
Hint: This problem can be solved by finding out the coefficients of thermal expansion for both the metals using the initial information given and the direct formula for the change in length of a body in terms of the temperature change and the thermal coefficient of expansion. For the composite rod, the total expansion will be the sum of the expansions for the two metals.
Formula used:
$\Delta L=L\alpha \Delta T$
Complete answer:
We will find out the thermal coefficients of expansion for both metals and this will be useful when we try to find out the expansion of each part in the composite rod.
The change in length $\Delta L$ of a rod when it is subjected to a temperature change $\Delta T$ is given by
$\Delta L=L\alpha \Delta T$ --(1)
Where $\alpha $ is the coefficient of thermal expansion of the material of the rod.
Now, let us analyze the question.
The length of the rod made of metal $A$ is ${{L}_{A}}=20cm=0.2m$ $\left( \because 1cm=0.01m \right)$
The temperature change in the rod is $\Delta {{T}_{A}}=373K-273K=100K$ $\left( \because {{100}^{0}}C=373K,{{0}^{0}}C=273K \right)$
The change in length of the rod is $\Delta {{L}_{A}}=0.075cm=0.00075m$ $\left( \because 1cm=0.01m \right)$
Let the coefficient of thermal expansion of metal $A$ be ${{\alpha }_{A}}$.
Now, using (1), we get
$0.00075=0.2{{\alpha }_{A}}\left( 100 \right)=20{{\alpha }_{A}}$
$\therefore {{\alpha }_{A}}=\dfrac{0.00075}{20}=3.75\times {{10}^{-5}}$ --(2)
Now, The length of the rod made of metal $B$ is ${{L}_{B}}=20cm=0.2m$ $\left( \because 1cm=0.01m \right)$
The temperature change in the rod is $\Delta {{T}_{B}}=373K-273K=100K$ $\left( \because {{100}^{0}}C=373K,{{0}^{0}}C=273K \right)$
The change in length of the rod is $\Delta {{L}_{B}}=0.045cm=0.00045m$ $\left( \because 1cm=0.01m \right)$
Let the coefficient of thermal expansion of metal $B$ be ${{\alpha }_{B}}$.
Now, using (1), we get
$0.00045=0.2{{\alpha }_{B}}\left( 100 \right)=20{{\alpha }_{B}}$
$\therefore {{\alpha }_{B}}=\dfrac{0.00045}{20}=2.25\times {{10}^{-5}}$ --(3)
Now, a composite rod of length $L=20cm=0.2m$ is made of the two metals.
Let the length of the part of the rod made of metal $A$ be $x$.
Therefore, the length of the part of the rod made of metal $B$ will be $0.2-x$.
The rod is subjected to the temperature change $\Delta T={{100}^{0}}C-{{0}^{0}}C=373K-273K=100K$ $\left( \because {{100}^{0}}C=373K,{{0}^{0}}C=273K \right)$
The change in length of the rod is $\Delta L=0.06cm=0.0006m$ $\left( \because 1cm=0.01m \right)$
Let the change in the length of the portion of metal $A$ in the rod be $\Delta {{L}_{A}}'$ and that of metal $B$ be $\Delta {{L}_{B}}'$.
Obviously, the sum of these changes will be the change in length of the entire rod.
$\therefore \Delta L=\Delta {{L}_{A}}'+\Delta {{L}_{B}}'$ --(4)
Now, using (1)
$\Delta {{L}_{A}}'=x{{\alpha }_{A}}\left( 100 \right)=100x{{\alpha }_{A}}$ --(5)
$\Delta {{L}_{B}}'=\left( 0.2-x \right){{\alpha }_{B}}\left( 100 \right)=100\left( 0.2-x \right){{\alpha }_{B}}$ --(6)
Putting (5) and (6) in (4), we get
$0.0006=100x{{\alpha }_{A}}+100\left( 0.2-x \right){{\alpha }_{B}}=100\left( x{{\alpha }_{A}}+\left( 0.2-x \right){{\alpha }_{B}} \right)$
Putting (2) and (3) in the above equation, we get
$0.0006=100\left( 3.75\times {{10}^{-5}}x+\left( 0.2-x \right)2.25\times {{10}^{-5}} \right)=100\times {{10}^{-5}}\left( 3.75x+\left( 0.2-x \right)2.25 \right)$
$\Rightarrow 0.0006={{10}^{-3}}\left( 3.75x+0.45-2.25x \right)={{10}^{-3}}\left( 1.5x+0.45 \right)$
$\Rightarrow \dfrac{0.0006}{{{10}^{-3}}}=1.5x+0.45$
$\Rightarrow 0.0006\times {{10}^{3}}=1.5x+0.45$
$\Rightarrow 0.6=1.5x+0.45$
$\Rightarrow 1.5x=0.6-0.45=0.15$
$\Rightarrow x=\dfrac{0.15}{1.5}=0.1m=10cm$ $\left( \because 1m=100cm \right)$
Therefore, the length of the rod made of metal $A$ is $10cm$.
Therefore, the correct option is $B)\text{ 1}0cm$.
Note:
Students must note that while writing the temperature change in thermodynamics problems or such thermal expansion problems, they must always change the units to Kelvin first and then proceed with the calculations. In this question, it did not make a difference since the change in temperature for the Celsius and Kelvin scale both are same but if the temperatures were given in Fahrenheit and we did not change in Kelvin, then we would have got a wrong value of the temperature change and hence, a wrong value for the coefficient of thermal expansion for the metals which would have ultimately led us to the wrong answer.
Formula used:
$\Delta L=L\alpha \Delta T$
Complete answer:
We will find out the thermal coefficients of expansion for both metals and this will be useful when we try to find out the expansion of each part in the composite rod.
The change in length $\Delta L$ of a rod when it is subjected to a temperature change $\Delta T$ is given by
$\Delta L=L\alpha \Delta T$ --(1)
Where $\alpha $ is the coefficient of thermal expansion of the material of the rod.
Now, let us analyze the question.
The length of the rod made of metal $A$ is ${{L}_{A}}=20cm=0.2m$ $\left( \because 1cm=0.01m \right)$
The temperature change in the rod is $\Delta {{T}_{A}}=373K-273K=100K$ $\left( \because {{100}^{0}}C=373K,{{0}^{0}}C=273K \right)$
The change in length of the rod is $\Delta {{L}_{A}}=0.075cm=0.00075m$ $\left( \because 1cm=0.01m \right)$
Let the coefficient of thermal expansion of metal $A$ be ${{\alpha }_{A}}$.
Now, using (1), we get
$0.00075=0.2{{\alpha }_{A}}\left( 100 \right)=20{{\alpha }_{A}}$
$\therefore {{\alpha }_{A}}=\dfrac{0.00075}{20}=3.75\times {{10}^{-5}}$ --(2)
Now, The length of the rod made of metal $B$ is ${{L}_{B}}=20cm=0.2m$ $\left( \because 1cm=0.01m \right)$
The temperature change in the rod is $\Delta {{T}_{B}}=373K-273K=100K$ $\left( \because {{100}^{0}}C=373K,{{0}^{0}}C=273K \right)$
The change in length of the rod is $\Delta {{L}_{B}}=0.045cm=0.00045m$ $\left( \because 1cm=0.01m \right)$
Let the coefficient of thermal expansion of metal $B$ be ${{\alpha }_{B}}$.
Now, using (1), we get
$0.00045=0.2{{\alpha }_{B}}\left( 100 \right)=20{{\alpha }_{B}}$
$\therefore {{\alpha }_{B}}=\dfrac{0.00045}{20}=2.25\times {{10}^{-5}}$ --(3)
Now, a composite rod of length $L=20cm=0.2m$ is made of the two metals.
Let the length of the part of the rod made of metal $A$ be $x$.
Therefore, the length of the part of the rod made of metal $B$ will be $0.2-x$.
The rod is subjected to the temperature change $\Delta T={{100}^{0}}C-{{0}^{0}}C=373K-273K=100K$ $\left( \because {{100}^{0}}C=373K,{{0}^{0}}C=273K \right)$
The change in length of the rod is $\Delta L=0.06cm=0.0006m$ $\left( \because 1cm=0.01m \right)$
Let the change in the length of the portion of metal $A$ in the rod be $\Delta {{L}_{A}}'$ and that of metal $B$ be $\Delta {{L}_{B}}'$.
Obviously, the sum of these changes will be the change in length of the entire rod.
$\therefore \Delta L=\Delta {{L}_{A}}'+\Delta {{L}_{B}}'$ --(4)
Now, using (1)
$\Delta {{L}_{A}}'=x{{\alpha }_{A}}\left( 100 \right)=100x{{\alpha }_{A}}$ --(5)
$\Delta {{L}_{B}}'=\left( 0.2-x \right){{\alpha }_{B}}\left( 100 \right)=100\left( 0.2-x \right){{\alpha }_{B}}$ --(6)
Putting (5) and (6) in (4), we get
$0.0006=100x{{\alpha }_{A}}+100\left( 0.2-x \right){{\alpha }_{B}}=100\left( x{{\alpha }_{A}}+\left( 0.2-x \right){{\alpha }_{B}} \right)$
Putting (2) and (3) in the above equation, we get
$0.0006=100\left( 3.75\times {{10}^{-5}}x+\left( 0.2-x \right)2.25\times {{10}^{-5}} \right)=100\times {{10}^{-5}}\left( 3.75x+\left( 0.2-x \right)2.25 \right)$
$\Rightarrow 0.0006={{10}^{-3}}\left( 3.75x+0.45-2.25x \right)={{10}^{-3}}\left( 1.5x+0.45 \right)$
$\Rightarrow \dfrac{0.0006}{{{10}^{-3}}}=1.5x+0.45$
$\Rightarrow 0.0006\times {{10}^{3}}=1.5x+0.45$
$\Rightarrow 0.6=1.5x+0.45$
$\Rightarrow 1.5x=0.6-0.45=0.15$
$\Rightarrow x=\dfrac{0.15}{1.5}=0.1m=10cm$ $\left( \because 1m=100cm \right)$
Therefore, the length of the rod made of metal $A$ is $10cm$.
Therefore, the correct option is $B)\text{ 1}0cm$.
Note:
Students must note that while writing the temperature change in thermodynamics problems or such thermal expansion problems, they must always change the units to Kelvin first and then proceed with the calculations. In this question, it did not make a difference since the change in temperature for the Celsius and Kelvin scale both are same but if the temperatures were given in Fahrenheit and we did not change in Kelvin, then we would have got a wrong value of the temperature change and hence, a wrong value for the coefficient of thermal expansion for the metals which would have ultimately led us to the wrong answer.
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