Question

The height of mercury column measured with brass scale at temperature $t_1$ is $n_1$ units. Height of the mercury column measured by brass scale at t = $0^\circ C$ is $n_2$ units. The coefficient of linear expansion of brass is $\alpha$ and coefficient of volume expansion of mercury is $\gamma$. (Assume the expansion in area of vessel containing mercury is negligible).

• A. n_1 (1 - \alpha t_1)
• B. $\frac{n_1(1 + \alpha t_1)}{1 + (\gamma / 3)t_1}$
• C. $\frac{n_1(1 - \gamma t_1)}{(1 - \alpha t_1)}$
• D. $\frac{n_1(1 - \alpha t_1)}{(1 - \gamma t_1)}$

Answer 1

$\frac{n_1 (1 + \alpha t_1)}{1 + \gamma t_1}$

Answer 2

Let $h_{true}(T)$ be the true height of the mercury column at temperature $T$. Let $L_{scale}(T)$ be the length of one unit on the brass scale at temperature $T$. Let $L_{scale}(0)$ be the length of one unit on the brass scale at $0^\circ C$. Then, $L_{scale}(t_1) = L_{scale}(0) (1 + \alpha t_1)$.

At temperature $t_1$, the reading on the scale is $n_1$. This means the true height of the mercury column is: $h_{true}(t_1) = n_1 \times L_{scale}(t_1) = n_1 \times L_{scale}(0) (1 + \alpha t_1)$.

At temperature $0^\circ C$, the reading on the scale is $n_2$. This means the true height of the mercury column is: $h_{true}(0) = n_2 \times L_{scale}(0)$.

The true height of the mercury column itself expands with temperature. Thus, the true height at $t_1$ is related to the true height at $0^\circ C$ by: $h_{true}(t_1) = h_{true}(0) (1 + \gamma t_1)$.

Substituting the expressions for $h_{true}(t_1)$ and $h_{true}(0)$: $n_1 \times L_{scale}(0) (1 + \alpha t_1) = [n_2 \times L_{scale}(0)] (1 + \gamma t_1)$.

Canceling $L_{scale}(0)$ from both sides: $n_1 (1 + \alpha t_1) = n_2 (1 + \gamma t_1)$.

We need to find $n_2$. Rearranging the equation: $n_2 = \frac{n_1 (1 + \alpha t_1)}{1 + \gamma t_1}$.

Comparing this result with the given options, none of the options exactly match. However, if we assume there might be a typo in the options and consider the structure of similar problems, the derived answer is $\frac{n_1 (1 + \alpha t_1)}{1 + \gamma t_1}$. Given the provided options, it's highly probable that there is an error in the question or its options. The correct answer is $\frac{n_1(1 + \alpha t_1)}{1 + \gamma t_1}$.