Assuming that the atmosphere has the same density anywhere as at sea level `(rho =1.3 kgm^(-3))` and g to be constant `(g=10 ms^(-2))`. What should be the approximate height of atmosphere ?
`(rho_(0)=1.01xx10^(5)Nm^(-2))`

Options

A. 6 km
B. 8 km
C. 12 km
D. 18 km

Explanation

To find the approximate height of the atmosphere, we can use the hydrostatic pressure equation, which relates pressure, density, gravitational acceleration, and height. The equation is given by:

\[ P = \rho g h \]

Where:
- \( P \) is the atmospheric pressure (given as \( \rho_0 = 1.01 \times 10^5 \, \text{N/m}^2 \)),
- \( \rho \) is the density of the atmosphere (given as \( \rho = 1.3 \, \text{kg/m}^3 \)),
- \( g \) is the acceleration due to gravity (given as \( g = 10 \, \text{m/s}^2 \)),
- \( h \) is the height of the atmosphere (which we need to find).

### Step-by-Step Solution:

1. **Write down the hydrostatic pressure equation:**
\[ P = \rho g h \]

2. **Rearrange the equation to solve for height \( h \):**
\[ h = \frac{P}{\rho g} \]

3. **Substitute the known values into the equation:**
- \( P = 1.01 \times 10^5 \, \text{N/m}^2 \)
- \( \rho = 1.3 \, \text{kg/m}^3 \)
- \( g = 10 \, \text{m/s}^2 \)

Thus,
\[ h = \frac{1.01 \times 10^5}{1.3 \times 10} \]

4. **Calculate the denominator:**
\[ 1.3 \times 10 = 13 \]

5. **Now, calculate the height \( h \):**
\[ h = \frac{1.01 \times 10^5}{13} \]

6. **Perform the division:**
\[ h \approx 7776.92 \, \text{m} \]

7. **Convert meters to kilometers:**
\[ h \approx 7.78 \, \text{km} \approx 8 \, \text{km} \]

### Final Answer:
The approximate height of the atmosphere is about **8 kilometers**.