
Consider the following statements.
(A) The coefficient of linear expansion has dimension $ {{K}^{-1}} $
(B) The coefficient of volume expansion has dimension $ {{K}^{-1}} $
A. both A and B are correct
B. A is correct but B is wrong
C. B is correct but A is wrong
D. both A and B are wrong
Answer
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Hint: The unit for coefficient of linear expansion is to be found first. And also the unit of coefficient of volume of expansion should be found then. Check whether the units are matched to $ {{K}^{-1}} $ . These all will help us to find out the correct answer for this question.
Complete step-by-step answer:
First of all the coefficient of linear expansion is given by the equation,
$ \alpha =\dfrac{\Delta l}{l\Delta t} $
Where $ l $ the initial length of the rod is, $ \Delta l $ is the change in length of the rod and $ \Delta t $ is the difference in the temperature.
Let us take the initial length of the rod as $ 1m $ .
That is,
$ l=1m $
Change in length of the rod is can be taken as,
$ \Delta l=1m $
And also change in temperature can be taken as,
$ \Delta t=1K $
Substituting this values in the equation will give,
$ \alpha =\dfrac{1m}{1m\times 1K}=1{{K}^{-1}} $
Therefore the first statement is correct.
Now let us check the second statement.
The coefficient of volume expansion can be given by the equation,
$ \gamma =\dfrac{\Delta V}{V\Delta t} $
Where $ \gamma $ the coefficient of volume expansion is, $ \Delta V $ is the change in volume, $ V $ is the initial volume and $ \Delta t $ is the change in temperature.
Let us take all these parameters as unity,
$ \Delta t=1K $
$ \Delta V=1{{m}^{3}} $
$ V=1{{m}^{3}} $
Therefore the coefficient of volume expansion is given as,
$ \gamma =\dfrac{1{{m}^{3}}}{1{{m}^{3}}\times 1K} $
Simplifying the equation will give,
$ \gamma ={{K}^{-1}} $
Therefore the second statement is also correct. Hence we can conclude that the both statements are correct.
So, the correct answer is “Option A”.
Note: The ratio of the increase in length to the original length for 1 degree rise in the temperature is known as the coefficient of linear expansion. And also the ratio of increase in volume to the initial volume for 1 degree rise in temperature is known as the coefficient of volume expansion. The larger is the coefficient for a material, the more is the expansion per degree temperature rise.
Complete step-by-step answer:
First of all the coefficient of linear expansion is given by the equation,
$ \alpha =\dfrac{\Delta l}{l\Delta t} $
Where $ l $ the initial length of the rod is, $ \Delta l $ is the change in length of the rod and $ \Delta t $ is the difference in the temperature.
Let us take the initial length of the rod as $ 1m $ .
That is,
$ l=1m $
Change in length of the rod is can be taken as,
$ \Delta l=1m $
And also change in temperature can be taken as,
$ \Delta t=1K $
Substituting this values in the equation will give,
$ \alpha =\dfrac{1m}{1m\times 1K}=1{{K}^{-1}} $
Therefore the first statement is correct.
Now let us check the second statement.
The coefficient of volume expansion can be given by the equation,
$ \gamma =\dfrac{\Delta V}{V\Delta t} $
Where $ \gamma $ the coefficient of volume expansion is, $ \Delta V $ is the change in volume, $ V $ is the initial volume and $ \Delta t $ is the change in temperature.
Let us take all these parameters as unity,
$ \Delta t=1K $
$ \Delta V=1{{m}^{3}} $
$ V=1{{m}^{3}} $
Therefore the coefficient of volume expansion is given as,
$ \gamma =\dfrac{1{{m}^{3}}}{1{{m}^{3}}\times 1K} $
Simplifying the equation will give,
$ \gamma ={{K}^{-1}} $
Therefore the second statement is also correct. Hence we can conclude that the both statements are correct.
So, the correct answer is “Option A”.
Note: The ratio of the increase in length to the original length for 1 degree rise in the temperature is known as the coefficient of linear expansion. And also the ratio of increase in volume to the initial volume for 1 degree rise in temperature is known as the coefficient of volume expansion. The larger is the coefficient for a material, the more is the expansion per degree temperature rise.
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