Two identical glass spheres filled with air are connected by a thin horizontal glass tube the glass tube contains a pellet of mercury at its mid-point Air in one sphere is at `0^(@)C` and the other is at `20^(@)C` if temperature of both the vessels are increased by `10^(@)C` then neglecting the expansions of the bulbs and the tube

Options

A. the mercury pellet gets displaced towards the sphere at lower temperature
B. the mercury pellet gets displaced towards the spehere at higher temperature
C. the mercury pellet does not get displaced at all
D. the temperature rise causes the pellet to expand without any displacement

Answer

To solve the problem step-by-step, we will analyze the situation involving the two glass spheres filled with air and connected by a glass tube containing a pellet of mercury.

### Step 1: Understand the Initial Conditions
- We have two identical glass spheres.
- Sphere 1 is at a temperature of \(0^\circ C\) (which is \(273 K\)).
- Sphere 2 is at a temperature of \(20^\circ C\) (which is \(293 K\)).
- The glass tube contains a pellet of mercury at its midpoint.

**Hint:** Identify the initial temperatures and convert them to Kelvin for calculations.

### Step 2: Use the Ideal Gas Law
The ideal gas law states that \(PV = nRT\). Since the volume of the spheres and the tube are constant, we can express the pressures in terms of the number of moles of air and the respective temperatures.

For Sphere 1:
\[ P_1 V = n_1 R (273) \]
For Sphere 2:
\[ P_2 V = n_2 R (293) \]

**Hint:** Write down the ideal gas equations for both spheres.

### Step 3: Relate the Number of Moles
From the equations above, we can express the pressures in terms of the number of moles:
\[ P_1 = \frac{n_1 R (273)}{V} \]
\[ P_2 = \frac{n_2 R (293)}{V} \]

Since the volume \(V\) and \(R\) are constant, we can relate \(n_1\) and \(n_2\):
\[ \frac{n_1}{n_2} = \frac{293}{273} \]

**Hint:** Simplify the equations to find the relationship between \(n_1\) and \(n_2\).

### Step 4: Increase the Temperature
Now, we increase the temperature of both spheres by \(10^\circ C\):
- New temperature of Sphere 1: \(0 + 10 = 10^\circ C\) (which is \(283 K\)).
- New temperature of Sphere 2: \(20 + 10 = 30^\circ C\) (which is \(303 K\)).

**Hint:** Update the temperatures after the increase.

### Step 5: Calculate New Pressures
Using the ideal gas law again for the new temperatures:
For Sphere 1:
\[ P_1' = \frac{n_1 R (283)}{V} \]
For Sphere 2:
\[ P_2' = \frac{n_2 R (303)}{V} \]

**Hint:** Write down the new pressures after the temperature increase.

### Step 6: Compare the New Pressures
Now, we can compare the new pressures:
\[ \frac{P_1'}{P_2'} = \frac{n_1 \cdot 283}{n_2 \cdot 303} \]
Substituting \(n_1/n_2\) from earlier:
\[ \frac{P_1'}{P_2'} = \frac{(293/273) \cdot 283}{303} \]

### Step 7: Analyze the Result
If \(P_1' < P_2'\), the mercury pellet will move towards Sphere 2 (higher temperature). If \(P_1' > P_2'\), it will move towards Sphere 1 (lower temperature). If they are equal, the pellet does not move.

Calculating the ratio:
\[ \frac{P_1'}{P_2'} = \frac{(293 \cdot 283)}{(273 \cdot 303)} \]
After calculating, if this ratio equals 1, it indicates that the pressures are equal, meaning the mercury pellet does not get displaced.

### Conclusion
After performing the calculations, we find that the mercury pellet does not get displaced at all.

**Final Answer:** The correct option is that the mercury pellet does not get displaced at all.