A cubic vessel (with face horizontal + vetical ) contains an ideal gas at NTP. The vessel is being carried by a rocket which is moving at a speed of `500 ms^(-1)` in vertical direction. The pressure of the gas inside the vessel as observed by us on the ground.

Options

A. remains the same because `500 ms^(-1)` is very much smaller than `v_("rms")` of the gas
B. remains the same because motion of the vessel as a whole does not not affect the relative motion of the gas molecules and the walls
C. will increase by a factor equal to `(V_("rms")^(2)+ (500)^(2)) // V_("rms")^(2) " where " V_("rms")` was the original mean square velocity of the gas
D. will be different on the top wall and bottom wall of the vessel

Explanation

To solve the problem, we need to analyze the situation of the cubic vessel containing an ideal gas inside a rocket moving vertically at a speed of \(500 \, \text{m/s}\). We want to determine the pressure of the gas inside the vessel as observed from the ground.

### Step-by-Step Solution:

1. **Understanding the Situation**:
- The vessel is cubic and contains an ideal gas at Normal Temperature and Pressure (NTP).
- The rocket is moving vertically at a speed of \(500 \, \text{m/s}\).

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