Answer
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Hint:A metal expands when the temperature of the metal is increased and vice-versa. The coefficient of expansion tells us how many meters of the length will increase for 1 degrees Celsius of increase in temperature.
Formula used:The formula of expanded length due to increase in the temperature is given by ${L_2} = {L_1}\left( {1 + \alpha \Delta T} \right)$ where ${L_2}$ is the length of the metal after the expansion ${L_1}$ is the original length $\alpha $ is the coefficient of expansion and $\Delta T$ is the change in the temperature.
Step by step solution:
It is given that the metal tape is 2m long and the initial temperature is $10^\circ C$ also the coefficient of thermal expansion is given as$\left( {\alpha = 1 \times {{10}^{ - 4}}\dfrac{m}{{^\circ C}}} \right)$.
The new length of the metal tape after the increase in temperature is given by,
$ \Rightarrow {L_2} = {L_1}\left( {1 + \alpha \Delta T} \right)$
Replace the values of original length of tape the value of coefficient of thermal expansion and the value of change in temperature.
$\Rightarrow {L_2} = 2\left[ {1 + {{10}^{ - 4}} \cdot \left( {30^\circ - 10^\circ } \right)} \right]$
Solving this relation we get,
$ \Rightarrow {L_2} = 2\left[ {1 + {{10}^{ - 4}} \cdot \left( {20} \right)} \right]$
$ \Rightarrow {L_2} = 2\left[ {1 + 0 \cdot 002} \right]$
$ \Rightarrow {L_2} = 2\left( {1 \cdot 002} \right)$
$ \Rightarrow {L_2} = 2 \cdot 004m$
The new length of the metal tape is equal to${L_2} = 2 \cdot 004m$.
Let us calculate the value of change of length due to the increase in the temperature.
Change of length is given by,
$ \Rightarrow \Delta L = {L_2} - {L_1}$
Replace the value of original length and new length in the above relation.
$ \Rightarrow \Delta L = {L_2} - {L_1}$
$ \Rightarrow \Delta L = 2 \cdot 004 - 2$
$ \Rightarrow \Delta L = 0 \cdot 004m$
The change in the length of the metal tape is equal to$\Delta L = 0 \cdot 004m$.
The correct answer for this problem is option D.
Note:The formula of the change in length due to the change in temperature should be understood and learned as it can be used to solve these types of problems also if you understand it you can use it whenever necessary.
Formula used:The formula of expanded length due to increase in the temperature is given by ${L_2} = {L_1}\left( {1 + \alpha \Delta T} \right)$ where ${L_2}$ is the length of the metal after the expansion ${L_1}$ is the original length $\alpha $ is the coefficient of expansion and $\Delta T$ is the change in the temperature.
Step by step solution:
It is given that the metal tape is 2m long and the initial temperature is $10^\circ C$ also the coefficient of thermal expansion is given as$\left( {\alpha = 1 \times {{10}^{ - 4}}\dfrac{m}{{^\circ C}}} \right)$.
The new length of the metal tape after the increase in temperature is given by,
$ \Rightarrow {L_2} = {L_1}\left( {1 + \alpha \Delta T} \right)$
Replace the values of original length of tape the value of coefficient of thermal expansion and the value of change in temperature.
$\Rightarrow {L_2} = 2\left[ {1 + {{10}^{ - 4}} \cdot \left( {30^\circ - 10^\circ } \right)} \right]$
Solving this relation we get,
$ \Rightarrow {L_2} = 2\left[ {1 + {{10}^{ - 4}} \cdot \left( {20} \right)} \right]$
$ \Rightarrow {L_2} = 2\left[ {1 + 0 \cdot 002} \right]$
$ \Rightarrow {L_2} = 2\left( {1 \cdot 002} \right)$
$ \Rightarrow {L_2} = 2 \cdot 004m$
The new length of the metal tape is equal to${L_2} = 2 \cdot 004m$.
Let us calculate the value of change of length due to the increase in the temperature.
Change of length is given by,
$ \Rightarrow \Delta L = {L_2} - {L_1}$
Replace the value of original length and new length in the above relation.
$ \Rightarrow \Delta L = {L_2} - {L_1}$
$ \Rightarrow \Delta L = 2 \cdot 004 - 2$
$ \Rightarrow \Delta L = 0 \cdot 004m$
The change in the length of the metal tape is equal to$\Delta L = 0 \cdot 004m$.
The correct answer for this problem is option D.
Note:The formula of the change in length due to the change in temperature should be understood and learned as it can be used to solve these types of problems also if you understand it you can use it whenever necessary.
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