Length of a wire at room temperature is \[4.55\;{\rm{m}}\], when the temperature increases upto \[100\;^\circ {\rm{C}}\] then its length becomes \[4.57\;{\rm{m}}\]. The coefficient of linear expansion \[\left( \alpha \right)\] of the given wire is:
A. \[5.021 \times {10^{ - 5}}\;{{\rm{K}}^{{\rm{ - 1}}}}\]
B. \[6.021 \times {10^{ - 5}}\;{{\rm{K}}^{{\rm{ - 1}}}}\]
C. \[7.021 \times {10^{ - 5}}\;{{\rm{K}}^{{\rm{ - 1}}}}\]
D. \[8.021 \times {10^{ - 5}}\;{{\rm{K}}^{{\rm{ - 1}}}}\]
Answer
654.6k+ views
Hint:The above problem can be resolved using the concepts and the fundamentals involved in thermal expansion. When any linear objects are maintained by increasing the object's temperature, then we can observe that some variation in the length of the linear dimension is observed. Then, this phenomenon is known as thermal expansion, and the value of the coefficient involved in this process is known as the coefficient of thermal expansion.
Complete step by step answer:
Given:
The length of wire at room temperature is, \[{L_1} = 4.55\;{\rm{m}}\].
The room temperature is, \[{T_1} = 27\;^\circ {\rm{C}}\].
The increase in the temperature is, \[{T_2} = 100\;{\rm{^\circ }}\]C.
The final length is, \[{L_2} = 4.57\;{\rm{m}}\].
The expression for the final length due to expansion is given as,
\[{L_2} = {L_1}\left[ {1 + \alpha \left( {{T_2} - {T_2}} \right)} \right]\]
Solve by substituting the value in above expression as,
\[\begin{array}{l}
4.57\;{\rm{m}} = 4.55\;{\rm{m}}\left[ {1 + \alpha \left( {\left( {100^\circ {\rm{C}} + 273} \right)\;{\rm{K}} - \left( {27^\circ {\rm{C}} + 273} \right)\;{\rm{K}}} \right)} \right]\\
\left[ {1 + \alpha \left( {373\;{\rm{K}} - 300\;{\rm{K}}} \right)} \right] = \dfrac{{4.57\;{\rm{m}}}}{{4.55\;{\rm{m}}}}\\
\left[ {\alpha \left( {373\;{\rm{K}} - 300\;{\rm{K}}} \right)} \right] = \dfrac{{4.57\;{\rm{m}}}}{{4.55\;{\rm{m}}}} - 1\\
\alpha = 6.021 \times {10^{ - 5}}\;{{\rm{K}}^{{\rm{ - 1}}}}
\end{array}\]
Therefore, the coefficient of linear expansion \[\left( \alpha \right)\] of the given wire is \[6.021 \times {10^{ - 5}}\;{{\rm{K}}^{{\rm{ - 1}}}}\] and option (B) is correct.
Note:To resolve the given problem, one must understand the meaning of thermal expansion and how this phenomenon works. There occurs some linear change in the object's dimension taken into consideration when that object is heated or supplied with a definite amount of thermal energy. Moreover, the mathematical relation for this change is given in terms of length, width, and height of the object. The thermal expansion has significant applications in various fields of applied physics like in thermal power plants and laboratories.
Complete step by step answer:
Given:
The length of wire at room temperature is, \[{L_1} = 4.55\;{\rm{m}}\].
The room temperature is, \[{T_1} = 27\;^\circ {\rm{C}}\].
The increase in the temperature is, \[{T_2} = 100\;{\rm{^\circ }}\]C.
The final length is, \[{L_2} = 4.57\;{\rm{m}}\].
The expression for the final length due to expansion is given as,
\[{L_2} = {L_1}\left[ {1 + \alpha \left( {{T_2} - {T_2}} \right)} \right]\]
Solve by substituting the value in above expression as,
\[\begin{array}{l}
4.57\;{\rm{m}} = 4.55\;{\rm{m}}\left[ {1 + \alpha \left( {\left( {100^\circ {\rm{C}} + 273} \right)\;{\rm{K}} - \left( {27^\circ {\rm{C}} + 273} \right)\;{\rm{K}}} \right)} \right]\\
\left[ {1 + \alpha \left( {373\;{\rm{K}} - 300\;{\rm{K}}} \right)} \right] = \dfrac{{4.57\;{\rm{m}}}}{{4.55\;{\rm{m}}}}\\
\left[ {\alpha \left( {373\;{\rm{K}} - 300\;{\rm{K}}} \right)} \right] = \dfrac{{4.57\;{\rm{m}}}}{{4.55\;{\rm{m}}}} - 1\\
\alpha = 6.021 \times {10^{ - 5}}\;{{\rm{K}}^{{\rm{ - 1}}}}
\end{array}\]
Therefore, the coefficient of linear expansion \[\left( \alpha \right)\] of the given wire is \[6.021 \times {10^{ - 5}}\;{{\rm{K}}^{{\rm{ - 1}}}}\] and option (B) is correct.
Note:To resolve the given problem, one must understand the meaning of thermal expansion and how this phenomenon works. There occurs some linear change in the object's dimension taken into consideration when that object is heated or supplied with a definite amount of thermal energy. Moreover, the mathematical relation for this change is given in terms of length, width, and height of the object. The thermal expansion has significant applications in various fields of applied physics like in thermal power plants and laboratories.
Recently Updated Pages
If x a + bt + ct2 where x is in meters and t is in class 11 physics CBSE

A car covers the first half distance between two places class 11 physics CBSE

The resultant of two vectors overrightarrow P and overrightarrow class 11 physics CBSE

Find the value of cos 135 class 11 maths CBSE

A mass M is held in place by an applied force F and class 11 physics CBSE

A solution of glucose in water is labelled as 10 dfracwv class 11 chemistry CBSE

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Find the value of the expression given below sin 30circ class 11 maths CBSE

What do you mean by retardation What is its SI uni class 11 physics CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

10 examples of friction in our daily life

Difference between physical and chemical change class 11 chemistry CBSE

