Light with an energy flux` 20 W//cm^2` falls on a non-reflecting surface at normal incidence. If the surface has an area of `30 cm^2`. the total momentum delivered ( for complete absorption)during 30 minutes is

Options

A. `36 xx 10^(-5) "kg m/s"`
B. `36xx10^(-4) "kg m/s"`
C. `108 xx 10^(4) "kg m/s"`
D. `1.08 xx 10^(7) "kg m/s"`

Explanation

To find the total momentum delivered to a non-reflecting surface by light over a given time period, we can follow these steps:

### Step 1: Understand the given values
- Energy flux (intensity) \( I = 20 \, \text{W/cm}^2 \)
- Area of the surface \( A = 30 \, \text{cm}^2 \)
- Time \( t = 30 \, \text{minutes} \)

### Step 2: Convert units
- Convert the area from \( \text{cm}^2 \) to \( \text{m}^2 \):
\[ A = 30 \, \text{cm}^2 = 30 \times 10^{-4} \, \text{m}^2 = 0.003 \, \text{m}^2 \]
- Convert time from minutes to seconds:
\[ t = 30 \, \text{minutes} = 30 \times 60 \, \text{seconds} = 1800 \, \text{s} \]

### Step 3: Calculate the energy delivered
The energy \( E \) delivered by the light can be calculated using the formula:
\[ E = I \times A \times t \]
Substituting the values:
\[ E = 20 \, \text{W/cm}^2 \times 0.003 \, \text{m}^2 \times 1800 \, \text{s} \]
First, convert the intensity from \( \text{W/cm}^2 \) to \( \text{W/m}^2 \):
\[ I = 20 \, \text{W/cm}^2 = 20 \times 10^4 \, \text{W/m}^2 = 200000 \, \text{W/m}^2 \]
Now calculate the energy:
\[ E = 200000 \, \text{W/m}^2 \times 0.003 \, \text{m}^2 \times 1800 \, \text{s} = 1080 \, \text{J} \]

### Step 4: Calculate the momentum delivered
The momentum \( p \) delivered by the light can be calculated using the relation between energy and momentum:
\[ p = \frac{E}{c} \]
Where \( c \) is the speed of light, approximately \( 3 \times 10^8 \, \text{m/s} \).

Substituting the values:
\[ p = \frac{1080 \, \text{J}}{3 \times 10^8 \, \text{m/s}} = 3.6 \times 10^{-6} \, \text{kg m/s} \]

### Final Result
The total momentum delivered during 30 minutes is:
\[ p = 3.6 \times 10^{-6} \, \text{kg m/s} \]