A barometer kept in an elevator accelerating upward reads 76 cm. The air pressure in the elevator is

Options

A. 76cm
B. lt76cm
C. gt76cm
D. zero

Answer

To find the air pressure in an elevator that is accelerating upward and where a barometer reads 76 cm of mercury, we can follow these steps:

### Step 1: Understand the basic principle of a barometer
A barometer measures atmospheric pressure by balancing the weight of a column of mercury against the atmospheric pressure. The height of the mercury column (H) is related to the pressure (P) by the equation:
\[ P = H \cdot \rho \cdot g \]
where:
- \( P \) is the atmospheric pressure,
- \( H \) is the height of the mercury column,
- \( \rho \) is the density of mercury,
- \( g \) is the acceleration due to gravity.

### Step 2: Consider the effect of the elevator's acceleration
When the elevator accelerates upward with an acceleration \( a \), the effective gravitational acceleration becomes:
\[ g' = g + a \]
This means that the pressure exerted by the mercury column in the barometer will also change because the effective weight of the mercury column increases.

### Step 3: Write the modified pressure equation
In the accelerating elevator, the pressure at the bottom of the mercury column can be expressed as:
\[ P' = H \cdot \rho \cdot (g + a) \]
where \( P' \) is the pressure in the elevator.

### Step 4: Relate the pressures
Since the barometer reads 76 cm of mercury, we can substitute \( H = 76 \, \text{cm} \) into the equation:
\[ P' = 76 \, \text{cm} \cdot \rho \cdot (g + a) \]

### Step 5: Compare the pressures
The pressure at sea level (normal atmospheric pressure) is given by:
\[ P = 76 \, \text{cm} \cdot \rho \cdot g \]
Since \( g + a > g \), it follows that:
\[ P' = 76 \, \text{cm} \cdot \rho \cdot (g + a) > 76 \, \text{cm} \cdot \rho \cdot g = P \]

### Conclusion
Thus, the air pressure in the elevator is greater than the normal atmospheric pressure of 76 cm of mercury.

### Final Answer
The air pressure in the elevator is greater than 76 cm of mercury.

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