If the value of atmospheric pressure is `10^(6)` dyne `cm^(-2)`. Its value in SI units is:

Options

A. `10^(4)` N `m^(-2)`
B. `10^(6)`N `m^(-2)`
C. `10^(5)` N `m^(-2)`
D. `10^(3)` N `m^(-2)`

Explanation

To convert the atmospheric pressure from dyne/cm² to SI units (Pascals), we can follow these steps:

### Step 1: Understand the given value
The atmospheric pressure is given as:
\[ P = 10^6 \, \text{dyne/cm}^2 \]

### Step 2: Convert dyne to Newton
We know that:
\[ 1 \, \text{Newton} = 10^5 \, \text{dyne} \]
Thus, we can express dyne in terms of Newton:
\[ 1 \, \text{dyne} = 10^{-5} \, \text{Newton} \]

### Step 3: Convert cm² to m²
We also know that:
\[ 1 \, \text{cm} = 10^{-2} \, \text{m} \]
Therefore, for area:
\[ 1 \, \text{cm}^2 = (10^{-2} \, \text{m})^2 = 10^{-4} \, \text{m}^2 \]

### Step 4: Substitute the conversions into the pressure formula
Now we can convert the pressure:
\[ P = 10^6 \, \text{dyne/cm}^2 = 10^6 \, \text{dyne} \times \frac{1 \, \text{Newton}}{10^5 \, \text{dyne}} \times \frac{1 \, \text{cm}^2}{10^{-4} \, \text{m}^2} \]

### Step 5: Simplify the expression
Substituting the conversions:
\[ P = 10^6 \times \frac{1}{10^5} \times \frac{1}{10^{-4}} \, \text{N/m}^2 \]
\[ P = 10^6 \times 10^{-5} \times 10^{4} \, \text{N/m}^2 \]
\[ P = 10^{6 - 5 + 4} \, \text{N/m}^2 = 10^{5} \, \text{N/m}^2 \]

### Step 6: Final result
Thus, the value of atmospheric pressure in SI units (Pascals) is:
\[ P = 10^5 \, \text{Pa} \]

---