Question

The apparent coefficient of expansion of a liquid when heated in a brass vessel is $X$ and when heated in a tin vessel is $Y$. If $\alpha$ is the coefficient of linear expansion for brass, the coefficient of linear expansion of tin is, then:
$(A)\dfrac{{X + Y + 3\alpha }}{3}$
$(B)\dfrac{{Y - X + 3\alpha }}{3}$
$(C)\dfrac{{X + Y - 2\alpha }}{3}$
$(D)\dfrac{{X + Y - 2\alpha }}{2}$

Answer 1

On heating, the liquid as well as the vessels expand. In order to solve this particular question, we will be using the relationship between coefficient of linear expansion and coefficient of volume expansion. On applying the above concept, we will find the solution.

Complete step by step solution:
Expansion is basically defined as the change in volume and shape and when this change takes place due to the change in temperature, then it is known as thermal expansion. Thermal expansion causes an increase in the length of dimensions of the body.
When the liquid is heated in a brass or tin vessel, both the liquid and vessel expand. The apparent coefficient of volume expansion is defined as the difference between the volumetric expansion which takes place in the liquid and the volume expansion which takes place in the vessel.
According to the question,
Let us consider the volume expression of the liquid to be $\delta$ and the linear expansion of the tin be ${\alpha _t}$. Therefore,
$X = 3\alpha + \delta ........(1)$
$Y = 3{\alpha _t} + \delta ........(2)$
On rearranging equation (1) and (2), we get,
$\delta = X - 3\alpha ........(3)$
$\delta = Y - 3{\alpha _t}........(4)$
On comparing equation (3) and (4), we get,
$X - 3\alpha = Y - 3{\alpha _t}$
Now, we need to find ${\alpha _t}$
$3{\alpha _t} = Y - X + 3\alpha$
${\alpha _t} = \dfrac{{Y - X + 3\alpha }}{3}$
So, the coefficient of linear expansion of tin is ${\alpha _t} = \dfrac{{Y - X + 3\alpha }}{3}$.
Thus, the final answer is $(B)\dfrac{{Y - X + 3\alpha }}{3}$.

Note:
It is important to note that the value of temperature of one vessel is very often more than that of the other, on the basis of its material, in practical cases. But in this question, this fact has been ignored as it makes the process more complicated and the calculation part also increases.