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 `3alpha`. 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 step by step, we will analyze the thermal expansion of both the beaker and the liquid inside it.

### Step 1: Understand the coefficients of expansion
The beaker has a coefficient of linear thermal expansion of \(3\alpha\), and the liquid has a coefficient of volume expansion of \(3\alpha\).

### Step 2: Determine the area expansion of the beaker
The area expansion coefficient \(\beta\) is related to the linear expansion coefficient \(\alpha\) by the relation:
\[ \beta = 2\alpha \]
Since the linear expansion coefficient of the beaker is \(3\alpha\), the area expansion coefficient becomes:
\[ \beta = 2 \times 3\alpha = 6\alpha \]

### Step 3: Calculate the change in area of the beaker
The change in area \(A\) of the beaker when heated by a temperature \(T\) can be expressed as:
\[ A = A_0 (1 + \beta T) = A_0 (1 + 6\alpha T) \]

### Step 4: Determine the volume expansion of the liquid
The volume expansion coefficient \(\gamma\) is related to the coefficient of volume expansion of the liquid, which is given as \(3\alpha\). The change in volume \(V\) of the liquid can be expressed as:
\[ V = V_0 (1 + \gamma T) = V_0 (1 + 3\alpha T) \]

### Step 5: Relate volume to area and height
The volume of the liquid can also be expressed in terms of the area of the base \(A\) and the height \(H\):
\[ V = A \cdot H \]
Thus, the new volume after heating can be expressed as:
\[ V' = A' \cdot H' \]
Where \(A'\) is the new area and \(H'\) is the new height.

### Step 6: Set up the equation for the height change
Equating the initial and final volumes gives:
\[ A_0 H = A_0 (1 + 6\alpha T) H' \]
From this, we can express \(H'\):
\[ H' = H \cdot \frac{1 + 3\alpha T}{1 + 6\alpha T} \]

### Step 7: Simplify using the small temperature approximation
Since we are given that \(\alpha T \ll 1\), we can use the binomial expansion to simplify:
\[ H' \approx H \cdot \left(1 + 3\alpha T - 6\alpha T\right) = H \cdot \left(1 - 3\alpha T\right) \]

### Step 8: Calculate the change in height
The change in height \(\Delta H\) can be expressed as:
\[ \Delta H = H' - H = H(1 - 3\alpha T) - H = -3H\alpha T \]
Thus, the reduction in the height of the liquid level is:
\[ \Delta H = -3H\alpha T \]

### Final Answer
The level of the liquid will reduce by \(3H\alpha T\).