
A metallic wire with tension T and at temperature \[30^\circ {\text{C}}\] vibrates with its fundamental frequency of 1 KHz. The same wire with the same tension but at \[10^\circ {\text{C}}\] temperature vibrates with a fundamental frequency of 1.001 KHz. The coefficient of linear expansion of the wire is:
A. \[2 \times {10^{ - 4}}/^\circ {\text{C}}\]
B. \[1.5 \times {10^{ - 4}}/^\circ {\text{C}}\]
C. \[1 \times {10^{ - 4}}/^\circ {\text{C}}\]
D. \[0.5 \times {10^{ - 4}}/^\circ {\text{C}}\]
Answer
569.4k+ views
Hint: Use the expression for frequency of vibrations to express the fundamental frequencies at the respective temperatures and take the ratio. Use the formula for linear expansion of the material to express the ratio of the final length to the initial length of the metallic wire. Consider \[30^\circ {\text{C}}\] as the initial temperature and \[10^\circ {\text{C}}\] as the final temperature.
Formula used:
The frequency of the vibration is given as,
\[f = \dfrac{1}{{2\pi }}\sqrt {\dfrac{g}{l}} \]
Here, g is the acceleration due to gravity and l is the length of the wire.
\[l = {l_0}\left( {1 + \alpha \,\Delta T} \right)\]
Here, l is the final length, \[{l_0}\] is the initial length, \[\alpha \] is the linear expansion coefficient and \[\Delta T\] is the change in temperature.
Complete step by step answer:
We have given the initial temperature of the wire is \[{T_1} = 30^\circ {\text{C}}\] and the final temperature of the wire is \[{T_2} = 10^\circ {\text{C}}\]. At \[{T_1} = 30^\circ {\text{C}}\], the wire vibrates at the fundamental frequency\[{f_1} = 1\,{\text{KHz}}\] and at \[{T_2} = 10^\circ {\text{C}}\], the wire vibrates at the fundamental frequency \[{f_2} = 1.001\,{\text{KHz}}\].
We know the expression for the frequency of the vibration is given as,
\[f = \dfrac{1}{{2\pi }}\sqrt {\dfrac{g}{l}} \]
Here, g is the acceleration due to gravity and l is the length of the wire.
Let’s express the fundamental frequency of the vibration at \[{T_1} = 30^\circ {\text{C}}\] as follows,
\[{f_1} = \dfrac{1}{{2\pi }}\sqrt {\dfrac{g}{{{l_1}}}} \] …… (1)
Let’s express the fundamental frequency of the vibration at \[{T_2} = 10^\circ {\text{C}}\] as follows,
\[{f_2} = \dfrac{1}{{2\pi }}\sqrt {\dfrac{g}{{{l_2}}}} \] …… (2)
Dividing equation (2) by equation (1), we get,
\[\dfrac{{{f_1}}}{{{f_2}}} = \dfrac{{\dfrac{1}{{2\pi }}\sqrt {\dfrac{g}{{{l_1}}}} }}{{\dfrac{1}{{2\pi }}\sqrt {\dfrac{g}{{{l_2}}}} }}\]
\[ \Rightarrow \dfrac{{{f_1}}}{{{f_2}}} = \sqrt {\dfrac{{{l_2}}}{{{l_1}}}} \]
\[ \Rightarrow \dfrac{{{l_2}}}{{{l_1}}} = {\left( {\dfrac{{{f_1}}}{{{f_2}}}} \right)^2}\] …… (3)
We can express the linear expansion of the wire due to change in temperature as,
\[{l_2} = {l_1}\left( {1 + \alpha \,\Delta T} \right)\]
\[ \Rightarrow \dfrac{{{l_2}}}{{{l_1}}} = \left( {1 + \alpha \,\Delta T} \right)\]
Here, \[\alpha \] is the linear expansion coefficient and \[\Delta T\] is the change in temperature.
Using equation (3), we can rewrite the above equation as,
\[\dfrac{{f_1^2}}{{f_2^2}} = 1 + \alpha \,\Delta T\]
Substituting 1 KHz for \[{f_1}\], 1.001 KHz for \[{f_2}\] and \[ - 20^\circ {\text{C}}\] for \[\Delta T\] in the above equation, we get,
\[\dfrac{1}{{{{\left( {1.001} \right)}^2}}} = 1 + \alpha \left( { - 20} \right)\]
\[ \Rightarrow 0.998 = 1 - 20\alpha \]
\[ \Rightarrow \alpha = \dfrac{{2 \times {{10}^{ - 3}}}}{{20}}\]
\[ \therefore \alpha = 1 \times {10^{ - 4}}/^\circ {\text{C}}\]
So, the correct answer is option C.
Note:We have assumed the final temperature as \[10^\circ {\text{C}}\] and the initial temperature as \[30^\circ {\text{C}}\]. Therefore, the change in temperature is \[ - 20^\circ {\text{C}}\]. You don’t need to convert the temperature in Kelvin since the coefficient of linear expansion is in \[/^\circ C\].
Formula used:
The frequency of the vibration is given as,
\[f = \dfrac{1}{{2\pi }}\sqrt {\dfrac{g}{l}} \]
Here, g is the acceleration due to gravity and l is the length of the wire.
\[l = {l_0}\left( {1 + \alpha \,\Delta T} \right)\]
Here, l is the final length, \[{l_0}\] is the initial length, \[\alpha \] is the linear expansion coefficient and \[\Delta T\] is the change in temperature.
Complete step by step answer:
We have given the initial temperature of the wire is \[{T_1} = 30^\circ {\text{C}}\] and the final temperature of the wire is \[{T_2} = 10^\circ {\text{C}}\]. At \[{T_1} = 30^\circ {\text{C}}\], the wire vibrates at the fundamental frequency\[{f_1} = 1\,{\text{KHz}}\] and at \[{T_2} = 10^\circ {\text{C}}\], the wire vibrates at the fundamental frequency \[{f_2} = 1.001\,{\text{KHz}}\].
We know the expression for the frequency of the vibration is given as,
\[f = \dfrac{1}{{2\pi }}\sqrt {\dfrac{g}{l}} \]
Here, g is the acceleration due to gravity and l is the length of the wire.
Let’s express the fundamental frequency of the vibration at \[{T_1} = 30^\circ {\text{C}}\] as follows,
\[{f_1} = \dfrac{1}{{2\pi }}\sqrt {\dfrac{g}{{{l_1}}}} \] …… (1)
Let’s express the fundamental frequency of the vibration at \[{T_2} = 10^\circ {\text{C}}\] as follows,
\[{f_2} = \dfrac{1}{{2\pi }}\sqrt {\dfrac{g}{{{l_2}}}} \] …… (2)
Dividing equation (2) by equation (1), we get,
\[\dfrac{{{f_1}}}{{{f_2}}} = \dfrac{{\dfrac{1}{{2\pi }}\sqrt {\dfrac{g}{{{l_1}}}} }}{{\dfrac{1}{{2\pi }}\sqrt {\dfrac{g}{{{l_2}}}} }}\]
\[ \Rightarrow \dfrac{{{f_1}}}{{{f_2}}} = \sqrt {\dfrac{{{l_2}}}{{{l_1}}}} \]
\[ \Rightarrow \dfrac{{{l_2}}}{{{l_1}}} = {\left( {\dfrac{{{f_1}}}{{{f_2}}}} \right)^2}\] …… (3)
We can express the linear expansion of the wire due to change in temperature as,
\[{l_2} = {l_1}\left( {1 + \alpha \,\Delta T} \right)\]
\[ \Rightarrow \dfrac{{{l_2}}}{{{l_1}}} = \left( {1 + \alpha \,\Delta T} \right)\]
Here, \[\alpha \] is the linear expansion coefficient and \[\Delta T\] is the change in temperature.
Using equation (3), we can rewrite the above equation as,
\[\dfrac{{f_1^2}}{{f_2^2}} = 1 + \alpha \,\Delta T\]
Substituting 1 KHz for \[{f_1}\], 1.001 KHz for \[{f_2}\] and \[ - 20^\circ {\text{C}}\] for \[\Delta T\] in the above equation, we get,
\[\dfrac{1}{{{{\left( {1.001} \right)}^2}}} = 1 + \alpha \left( { - 20} \right)\]
\[ \Rightarrow 0.998 = 1 - 20\alpha \]
\[ \Rightarrow \alpha = \dfrac{{2 \times {{10}^{ - 3}}}}{{20}}\]
\[ \therefore \alpha = 1 \times {10^{ - 4}}/^\circ {\text{C}}\]
So, the correct answer is option C.
Note:We have assumed the final temperature as \[10^\circ {\text{C}}\] and the initial temperature as \[30^\circ {\text{C}}\]. Therefore, the change in temperature is \[ - 20^\circ {\text{C}}\]. You don’t need to convert the temperature in Kelvin since the coefficient of linear expansion is in \[/^\circ C\].
Recently Updated Pages
Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

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

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

Discuss the various forms of bacteria class 11 biology CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

