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Question: The length of metal rod at \[20^\circ {\text{C}}\] is \[1\,{\text{m}}\]. When the temperature is inc...

The length of metal rod at 20∘C20^\circ {\text{C}} is 1 m1\,{\text{m}}. When the temperature is increased to 50∘C50^\circ {\text{C}} the expansion is 1 mm1\,{\text{mm}}. But to reduce it by 1 mm1\,{\text{mm}} to what temperature it should be cooled?
A. 30∘C30^\circ {\text{C}}
B. −50∘C - 50^\circ {\text{C}}
C. −10∘C - 10^\circ {\text{C}}
D. −20∘C - 20^\circ {\text{C}}

Explanation

Solution

Use the formula for linear thermal expansion of a material. Using this formula determines the coefficient of linear thermal expansion of the metal rod. Then use this value of coefficient of linear thermal expansion and determine the value of the temperature to which the metal rod should be cooled.

Formula used:
The expression for linear thermal expansion of a material is given by
L=L0+αL0ΔTL = {L_0} + \alpha {L_0}\Delta T …… (1)
Here, L0{L_0} is the original length of the material, LL is the changed length of the material, α\alpha is the coefficient of linear thermal expansion and ΔT\Delta T is the change in temperature of the material.

Complete step by step answer:
We have given that the original length of the metal rod is 1 m1\,{\text{m}}.
L0=1 m{L_0} = 1\,{\text{m}}

The initial temperature of the metal rod is 20∘C20^\circ {\text{C}}.
T0=20∘C{T_0} = 20^\circ {\text{C}}

First the temperature of the metal rod is increased to the temperature 50∘C50^\circ {\text{C}}.
T1=50∘C{T_1} = 50^\circ {\text{C}}

Let us first determine the change in temperature for this thermal expansion of metal rod.
ΔT=T1−T0\Delta T = {T_1} - {T_0}

Substitute 50∘C50^\circ {\text{C}} for T1{T_1} and 20∘C20^\circ {\text{C}} for T0{T_0} in the above equation.
ΔT=(50∘C)−(20∘C)\Delta T = \left( {50^\circ {\text{C}}} \right) - \left( {20^\circ {\text{C}}} \right)
⇒ΔT=30∘C\Rightarrow \Delta T = 30^\circ {\text{C}}

The increase in length of the metal rod for this expansion is 1 mm1\,{\text{mm}}.
L1−L0=1 mm{L_1} - {L_0} = 1\,{\text{mm}}

Here, L1{L_1} is the changed length of the metal rod.

Convert the unit of above increase in length of the rod in the SI system of units.
L1−L0=(1 mm)(10−3 m1 mm){L_1} - {L_0} = \left( {1\,{\text{mm}}} \right)\left( {\dfrac{{{{10}^{ - 3}}\,{\text{m}}}}{{1\,{\text{mm}}}}} \right)
⇒L1−L0=10−3 m\Rightarrow {L_1} - {L_0} = {10^{ - 3}}\,{\text{m}}

Hence, the increase in the length of metal rod is 10−3 m{10^{ - 3}}\,{\text{m}}.

Let us determine the coefficient of linear expansion for the metal rod using equation (1).

Rewrite equation (1) for the linear expansion of metal.
L1=L0+αL0ΔT{L_1} = {L_0} + \alpha {L_0}\Delta T
⇒L1−L0=αL0ΔT\Rightarrow {L_1} - {L_0} = \alpha {L_0}\Delta T

Substitute 10−3 m{10^{ - 3}}\,{\text{m}} for L1−L0{L_1} - {L_0}, 1 m1\,{\text{m}} for L0{L_0} and 30∘C30^\circ {\text{C}} for ΔT\Delta T in the above equation.
⇒10−3 m=α(1 m)(30∘C)\Rightarrow {10^{ - 3}}\,{\text{m}} = \alpha \left( {1\,{\text{m}}} \right)\left( {30^\circ {\text{C}}} \right)
⇒α=10−330∘C\Rightarrow \alpha = \dfrac{{{{10}^{ - 3}}}}{{30^\circ {\text{C}}}}
⇒α=3.34×10−5/∘C\Rightarrow \alpha = 3.34 \times {10^{ - 5}}/^\circ {\text{C}}

Hence, the coefficient of linear expansion for the metal rod is 3.34×10−5/∘C3.34 \times {10^{ - 5}}/^\circ {\text{C}}.

Now let us determine the temperature TT to which the metal rod should be decreased to have the decrease in length of metal rod by 1 mm1\,{\text{mm}}.
L0−L2=1 mm{L_0} - {L_2} = 1\,{\text{mm}}
⇒L0−L2=10−3 m\Rightarrow {L_0} - {L_2} = {10^{ - 3}}\,{\text{m}}

Here, L2{L_2} is the decreased length of the metal rod.

Rewrite equation (1) for the linear compression of metal rod.
L2=L0+αL0(T−T0){L_2} = {L_0} + \alpha {L_0}\left( {T - {T_0}} \right)
⇒L2−L0=αL0(T−T0)\Rightarrow {L_2} - {L_0} = \alpha {L_0}\left( {T - {T_0}} \right)

Substitute −10−3 m - {10^{ - 3}}\,{\text{m}} for L2−L0{L_2} - {L_0}, 3.34×10−5/∘C3.34 \times {10^{ - 5}}/^\circ {\text{C}} for α\alpha , 1 m1\,{\text{m}} for L0{L_0} and 20∘C20^\circ {\text{C}} for T0{T_0} in the above equation.
⇒(−10−3 m)=(3.34×10−5/∘C)(1 m)(T−20∘C)\Rightarrow \left( { - {{10}^{ - 3}}\,{\text{m}}} \right) = \left( {3.34 \times {{10}^{ - 5}}/^\circ {\text{C}}} \right)\left( {1\,{\text{m}}} \right)\left( {T - 20^\circ {\text{C}}} \right)
⇒T=−10−3 m3.34×10−5/∘C+20∘C\Rightarrow T = - \dfrac{{{{10}^{ - 3}}\,{\text{m}}}}{{3.34 \times {{10}^{ - 5}}/^\circ {\text{C}}}} + 20^\circ {\text{C}}
⇒T=−30∘C+20∘C\Rightarrow T = - 30^\circ {\text{C}} + 20^\circ {\text{C}}
⇒T=−10∘C\Rightarrow T = - 10^\circ {\text{C}}

Therefore, the temperature to which the temperature to which the metal rod should be cooled is −10∘C - 10^\circ {\text{C}}.

So, the correct answer is “Option C”.

Note:
The students should not forget to use the negative value of the change in temperature for the decrease in length of the metal rod while substituting the values in the equation for linear thermal expansion of the rod.
Because the final value of the length of rod is less than the initial value of length of the rod.