For two gases, A and B with molecular weights `M_(A)` and `M_(B)`. It is observed that at a certain temperature. T, the mean velocity of A is equal to the root mean square velocity of B. thus the mean velocity of A can be made equal to the mean velocity of B, if:

Options

A. A is at temperature `T_1` and B at `T_2` and `T_1 gt T_2`
B. A is lowered to a temperature `T_1` and `T_1 lt T_2`
C. Both A & B are raised to a higher temperature
D. Both A & B are placed at a lower temperature

Answer

To solve the problem, we need to analyze the relationship between the mean velocity of gas A and the root mean square velocity of gas B, as well as the conditions under which their mean velocities can be made equal.

### Step-by-Step Solution:

1. **Understand the Definitions**:
- The mean velocity (v̄) of a gas is given by the formula:
\[ v̄ = \sqrt{\frac{8RT}{\pi M}} \]
- The root mean square velocity (v_rms) of a gas is given by the formula:
\[ v_{rms} = \sqrt{\frac{3RT}{M}} \]

2. **Set Up the Given Condition**:
- According to the problem, the mean velocity of gas A is equal to the root mean square velocity of gas B:
\[ v̄_A = v_{rms_B} \]
- Substituting the formulas:
\[ \sqrt{\frac{8RT}{\pi M_A}} = \sqrt{\frac{3RT}{M_B}} \]

3. **Square Both Sides**:
- Squaring both sides to eliminate the square roots:
\[ \frac{8RT}{\pi M_A} = \frac{3RT}{M_B} \]

4. **Cancel Common Terms**:
- Since \(RT\) is common on both sides, we can cancel it out (assuming \(T\) is constant and non-zero):
\[ \frac{8}{\pi M_A} = \frac{3}{M_B} \]

5. **Rearranging the Equation**:
- Cross-multiplying gives:
\[ 8M_B = 3\pi M_A \]
- From this, we can deduce:
\[ M_B = \frac{3\pi}{8} M_A \]

6. **Analyzing Molar Masses**:
- Since \(3\pi \approx 9.42\), it follows that:
\[ M_B > M_A \]
- This means gas B has a greater molar mass than gas A.

7. **Condition for Equal Mean Velocities**:
- To make the mean velocity of gas A equal to that of gas B, we need to consider the relationship of mean velocity with temperature. The mean velocity is inversely related to the square root of molar mass.
- Since gas A has a lower molar mass, it will have a higher mean velocity. To make the mean velocities equal, we can lower the temperature of gas A.

8. **Conclusion**:
- Therefore, the mean velocity of gas A can be made equal to the mean velocity of gas B if the temperature of gas A is reduced to a value \(T_1\) such that \(T_1 < T_2\) (where \(T_2\) is the original temperature).

### Final Answer:
The mean velocity of gas A can be made equal to the mean velocity of gas B if the temperature of gas A is lowered (option 2).