Question

The coefficient of apparent expansion of mercury in a glass vessel is \[153\times {{10}^{-6}}/{}^\circ C\]and in a steel vessel is\[144\times {{10}^{-6}}/{}^\circ C\]. If \[\alpha \]for steel is\[12\times {{10}^{-6}}/{}^\circ C\], then that of glass is:
\[\begin{align}
  & A.\,\,6\times {{10}^{-6}}/{}^\circ C \\
 & B.\,\,9\times {{10}^{-6}}/{}^\circ C \\
 & C.\,\,36\times {{10}^{-6}}/{}^\circ C \\
 & D.\,\,27\times {{10}^{-6}}/{}^\circ C \\
\end{align}\]

Answer 1

Hint: The formula that we will be using to solve this problem is, the coefficient of the real expansion is equal to the sum of the coefficient of the apparent expansion and the coefficient of the expansion of the vessel. Considering the 2 situations, that is, using the glass and the steel vessel, we will compute the coefficient of the linear expansion for the glass.
Formula used:
\[{{\gamma }_{real}}={{\gamma }_{apparent}}+{{\gamma }_{vessel}}\]

Complete answer:
From the given information, we have the data as follows.
The coefficient of apparent expansion of mercury in a glass vessel is \[153\times {{10}^{-6}}/{}^\circ C\]and in a steel vessel is\[144\times {{10}^{-6}}/{}^\circ C\]. \[\alpha \]for steel is\[12\times {{10}^{-6}}/{}^\circ C\].
The formula that we will be using to solve this problem is given as follows.
\[{{\gamma }_{real}}={{\gamma }_{apparent}}+{{\gamma }_{vessel}}\]
Where \[{{\gamma }_{real}}\]is the coefficient of the real expansion, \[{{\gamma }_{apparent}}\]is the coefficient of the apparent expansion and \[{{\gamma }_{vessel}}\]is the coefficient of expansion of vessel.
Now, consider the vessel steel.
Consider the formula.
\[{{\gamma }_{real}}={{\gamma }_{apparent}}+{{\gamma }_{vessel}}\]
Substitute the values in the above formula.
\[{{\gamma }_{real}}=144\times {{10}^{-6}}+{{\gamma }_{steel}}\]…… (1)
The coefficient of expansion of vessel steel is computed as follows.
\[\begin{align}
  & {{\gamma }_{steel}}=3\alpha \\
 & \Rightarrow {{\gamma }_{steel}}=3\times 12\times {{10}^{-6}} \\
 & \therefore {{\gamma }_{steel}}=36\times {{10}^{-6}}/{}^\circ C \\
\end{align}\]
Substitute this value in equation (1).
\[\begin{align}
  & {{\gamma }_{real}}=144\times {{10}^{-6}}+36\times {{10}^{-6}} \\
 & \therefore {{\gamma }_{real}}=180\times {{10}^{-6}} \\
\end{align}\]
Now, consider the vessel glass.
Consider the formula.
\[{{\gamma }_{real}}={{\gamma }_{apparent}}+{{\gamma }_{vessel}}\]
Substitute the values in the above formula.
\[\begin{align}
  & 180\times {{10}^{-6}}=153\times {{10}^{-6}}+{{\gamma }_{glass}} \\
 & \Rightarrow {{\gamma }_{glass}}=180\times {{10}^{-6}}-153\times {{10}^{-6}} \\
 & \therefore {{\gamma }_{glass}}=27\times {{10}^{-6}}/{}^\circ C \\
\end{align}\]
We have computed the value of the volume expansion of the glass vessel. Using this value, we have to find the linear expansion of the glass.
Consider the formula.
\[\begin{align}
  & {{\gamma }_{glass}}=3\alpha \\
 & \Rightarrow 27\times {{10}^{-6}}=3\times \alpha \\
 & \Rightarrow \alpha =\dfrac{27\times {{10}^{-6}}}{3} \\
 & \therefore \alpha =9\times {{10}^{-6}}/{}^\circ C \\
\end{align}\]
\[\therefore \] The coefficient of the linear expansion of the vessel glass is, \[9\times {{10}^{-6}}/{}^\circ C\].

Thus, option (B) is correct.

Note:
The coefficient of the volume expansion is equal to thrice the coefficient of the linear expansion. The values of the coefficient of the volume and the linear expansion are different for different materials.