
The coefficient of linear expansion of iron is 11/180 of volume coefficient of expansion of mercury which is $18\times {{10}^{-5}}{{/}^{0}}C$. An iron rod is 10m long at 27$^{0}C$. The length of the rod will be decreased by 1.1mm when the temperature of the rod changes by
(A) 0$^{0}C$
(B) 10$^{0}C$
(C) 20$^{0}C$
(D) 170$^{0}C$
Answer
502.2k+ views
Hint:The coefficient of linear expansion for a solid is given as ${{\alpha }_{L}}=\dfrac{1}{L}\left( \dfrac{\Delta L}{\Delta T} \right)$, where l is the length of the solid,$\Delta T$ is change in the temperature and$\Delta L$ is the change in the length. By substituting the value of the coefficient of linear expansion of iron in the equation we get the value of changing the temperature.
Complete step by step answer:
Given, the volume coefficient of expansion of mercury$=18\times {{10}^{-5}}{{/}^{0}}C$
The coefficient of linear expansion of iron is $\dfrac{11}{180}$ of volume coefficient of expansion of mercury.Therefore,
coefficient of linear expansion of iron=$\dfrac{11}{180}\times 18\times {{10}^{-5}}=11\times {{10}^{-6}}/$℃.
Given temperature, ${{t}_{i}}=27$℃.
At 27℃ the length of the iron rod is, $L=10m$.
Decrease in the length of the rod (change in the length of the rod)is, $\Delta L=1.1mm=1.1\times {{10}^{-3}}m$.
The coefficient linear expansion is defined as
${{\alpha }_{L}}=\dfrac{1}{L}{{\left( \dfrac{\delta L}{\delta T} \right)}_{P}}$ .
where L is the length and T is temperature. δL is the change in length due to change in temperature at constant pressure.
For solids, we can neglect the pressure dependence, and also assume that if the temperature was increased linearly, $\Delta L\approx dL$. Therefore:
${{\alpha }_{L}}=\dfrac{1}{L}\left( \dfrac{\Delta L}{\Delta T} \right)$
So, the change in temperature can be written as$\Delta T=\dfrac{\Delta L}{{{\alpha }_{L}}L}$.
Substituting the given values in this equation we get,
$\therefore\Delta T=\dfrac{1.1\times {{10}^{-3}}}{10\times 11\times {{10}^{-6}}}=10\,℃$.
Therefore the temperature should be changed by 10℃ to decrease the height of the rod by 1.1mm.
So, the correct option is B.
Note:These coefficients are unique for every material. For copper and iron under the same conditions of temperature and pressure, their values will be different. Here, \[\Delta T\]is the change in temperature that is the difference between final value and initial value of temperatures. We need to keep in mind although we have to take the values of temperature always in kelvin but when we talk about difference then it does not matter what units are.
Complete step by step answer:
Given, the volume coefficient of expansion of mercury$=18\times {{10}^{-5}}{{/}^{0}}C$
The coefficient of linear expansion of iron is $\dfrac{11}{180}$ of volume coefficient of expansion of mercury.Therefore,
coefficient of linear expansion of iron=$\dfrac{11}{180}\times 18\times {{10}^{-5}}=11\times {{10}^{-6}}/$℃.
Given temperature, ${{t}_{i}}=27$℃.
At 27℃ the length of the iron rod is, $L=10m$.
Decrease in the length of the rod (change in the length of the rod)is, $\Delta L=1.1mm=1.1\times {{10}^{-3}}m$.
The coefficient linear expansion is defined as
${{\alpha }_{L}}=\dfrac{1}{L}{{\left( \dfrac{\delta L}{\delta T} \right)}_{P}}$ .
where L is the length and T is temperature. δL is the change in length due to change in temperature at constant pressure.
For solids, we can neglect the pressure dependence, and also assume that if the temperature was increased linearly, $\Delta L\approx dL$. Therefore:
${{\alpha }_{L}}=\dfrac{1}{L}\left( \dfrac{\Delta L}{\Delta T} \right)$
So, the change in temperature can be written as$\Delta T=\dfrac{\Delta L}{{{\alpha }_{L}}L}$.
Substituting the given values in this equation we get,
$\therefore\Delta T=\dfrac{1.1\times {{10}^{-3}}}{10\times 11\times {{10}^{-6}}}=10\,℃$.
Therefore the temperature should be changed by 10℃ to decrease the height of the rod by 1.1mm.
So, the correct option is B.
Note:These coefficients are unique for every material. For copper and iron under the same conditions of temperature and pressure, their values will be different. Here, \[\Delta T\]is the change in temperature that is the difference between final value and initial value of temperatures. We need to keep in mind although we have to take the values of temperature always in kelvin but when we talk about difference then it does not matter what units are.
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