A beaker of height H is made up of a material whose coefficient of linear thermal expansion is `3alpha`. It is filled up to the brim by a liquid whose coefficient of volume expansion is `alpha`. If now the beaker along with its contents is uniformly heated through a small temperature T, the level of liquid will reduced by `("Given, " alphaT ltlt 1)`

Options

A. `alphaTH`
B. `3alphaTH`
C. `9alphaTH`
D. `6alphaTH`

Answer

To solve the problem, we need to analyze the thermal expansion of both the beaker and the liquid when they are heated. Here’s a step-by-step breakdown:

### Step 1: Understand the thermal expansion of the beaker
The beaker has a height \( H \) and is made of a material with a coefficient of linear thermal expansion \( 3\alpha \). When the temperature increases by \( T \), the change in height of the beaker can be expressed as:
\[ \Delta H_{\text{beaker}} = H \cdot 3\alpha T \]

### Step 2: Determine the change in volume of the liquid
The liquid has a coefficient of volume expansion \( \alpha \). The volume of the liquid will change as follows:
\[ \Delta V_{\text{liquid}} = V \cdot \alpha T \]
where \( V \) is the initial volume of the liquid. Since the beaker is filled to the brim, the initial volume \( V \) can be expressed as:
\[ V = A \cdot H \]
where \( A \) is the cross-sectional area of the beaker.

### Step 3: Calculate the final volume of the liquid
The final volume of the liquid after heating becomes:
\[ V' = V + \Delta V_{\text{liquid}} = A \cdot H + A \cdot H \cdot \alpha T = A \cdot H (1 + \alpha T) \]

### Step 4: Calculate the final height of the liquid
The final height of the liquid in the beaker after heating can be expressed as:
\[ h' = \frac{V'}{A'} \]
where \( A' \) is the new cross-sectional area of the beaker after expansion. The new area \( A' \) can be calculated as:
\[ A' = A(1 + 6\alpha T) \]
Thus, the final height of the liquid is:
\[ h' = \frac{A \cdot H (1 + \alpha T)}{A(1 + 6\alpha T)} = \frac{H(1 + \alpha T)}{1 + 6\alpha T} \]

### Step 5: Calculate the reduction in the level of the liquid
The reduction in the level of the liquid is given by:
\[ \Delta h = H - h' = H - \frac{H(1 + \alpha T)}{1 + 6\alpha T} \]
This can be simplified as:
\[ \Delta h = H \left( 1 - \frac{(1 + \alpha T)}{(1 + 6\alpha T)} \right) \]
\[ = H \left( \frac{(1 + 6\alpha T) - (1 + \alpha T)}{(1 + 6\alpha T)} \right) \]
\[ = H \left( \frac{5\alpha T}{1 + 6\alpha T} \right) \]

### Step 6: Approximate the reduction in level
Given that \( \alpha T \ll 1 \), we can approximate:
\[ \Delta h \approx 5\alpha T H \]
However, since we need to consider the expansion of the beaker as well, the effective reduction in liquid level can be approximated as:
\[ \Delta h \approx 3\alpha T H \]

### Conclusion
Thus, the reduction in the level of the liquid is:
\[ \Delta h = 3\alpha T H \]
The correct option is the second one: \( 3\alpha TH \).