A barometer kept in an elevator accelerating upwards reads 76 cm of Hg. If the elevator is accelerating upwards at `4.9 "ms"^(-2)` , what will be the air pressure in the elevator?

Answer

To find the air pressure in the elevator when it is accelerating upwards, we can follow these steps:

### Step 1: Understand the situation
The barometer reads 76 cm of Hg when the elevator is at rest or moving at constant velocity. When the elevator accelerates upwards, the effective acceleration due to gravity increases.

### Step 2: Determine the effective acceleration
The effective acceleration \( g' \) when the elevator is accelerating upwards can be calculated as:
\[ g' = g + a \]
where:
- \( g \) is the acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \)),
- \( a \) is the upward acceleration of the elevator (\( 4.9 \, \text{m/s}^2 \)).

Substituting the values:
\[ g' = 9.8 \, \text{m/s}^2 + 4.9 \, \text{m/s}^2 = 14.7 \, \text{m/s}^2 \]

### Step 3: Relate the height of the mercury column to pressure
The pressure exerted by a column of mercury is given by:
\[ P = \rho g h \]
where:
- \( P \) is the pressure,
- \( \rho \) is the density of mercury (approximately \( 13,600 \, \text{kg/m}^3 \)),
- \( g \) is the effective acceleration due to gravity,
- \( h \) is the height of the mercury column.

### Step 4: Calculate the new height of the mercury column
Since the barometer reads 76 cm of Hg at normal conditions, we can find the new height \( h' \) of the mercury column under the new effective gravitational acceleration:
\[ P_0 = \rho g h \implies h' = \frac{P_0}{\rho g'} \]
Given that the original pressure \( P_0 \) corresponds to 76 cm of Hg, we can express this in meters:
\[ h' = \frac{0.76 \, \text{m} \cdot 9.8 \, \text{m/s}^2}{14.7 \, \text{m/s}^2} \]

### Step 5: Calculate the new height
Calculating \( h' \):
\[ h' = \frac{0.76 \cdot 9.8}{14.7} \approx 0.51 \, \text{m} \text{ (or 51 cm)} \]

### Step 6: Calculate the air pressure in the elevator
The air pressure in the elevator can be calculated using the formula:
\[ P = \rho g' h' \]
Substituting the values:
\[ P = 13,600 \, \text{kg/m}^3 \cdot 14.7 \, \text{m/s}^2 \cdot 0.51 \, \text{m} \]
Calculating this gives:
\[ P \approx 101,325 \, \text{Pa} \text{ (or 1 atm)} \]

### Final Result
The air pressure in the elevator when it is accelerating upwards at \( 4.9 \, \text{m/s}^2 \) is approximately \( 101,325 \, \text{Pa} \).