Light is incident normally on a completely absorbing surface with an energy flux of `25 Wcm^(-2)` If the surface has an area of `25 cm^(2)` the momentum transferred to the surface in 40 min time duration will be:

Options

A. `6.3 xx 10^(-4) Ns`
B. `5.0 xx 10^(-3) Ns`
C. `1.4 xx 10^(-6) Ns`
D. `3.5 xx 10^(-6) Ns`

Explanation

To solve the problem, we need to calculate the momentum transferred to a completely absorbing surface when light is incident on it. We will follow these steps:

### Step 1: Understand the given parameters
- Energy flux (intensity) \( I = 25 \, W/cm^2 \)
- Area of the surface \( A = 25 \, cm^2 \)
- Time duration \( t = 40 \, minutes \)

### Step 2: Convert units
1. Convert the energy flux from \( W/cm^2 \) to \( W/m^2 \):
\[ I = 25 \, W/cm^2 = 25 \times 10^4 \, W/m^2 \]
2. Convert the area from \( cm^2 \) to \( m^2 \):
\[ A = 25 \, cm^2 = 25 \times 10^{-4} \, m^2 \]
3. Convert time from minutes to seconds:
\[ t = 40 \, minutes = 40 \times 60 = 2400 \, seconds \]

### Step 3: Calculate the radiation pressure
The radiation pressure \( P \) exerted by the light on the surface can be calculated using the formula:
\[ P = \frac{I}{c} \]
where \( c \) is the speed of light (\( c \approx 3 \times 10^8 \, m/s \)).

Substituting the values:
\[ P = \frac{25 \times 10^4}{3 \times 10^8} \, N/m^2 \]

### Step 4: Calculate the force
The force \( F \) exerted on the surface can be calculated using the formula:
\[ F = P \times A \]
Substituting the values:
\[ F = \left(\frac{25 \times 10^4}{3 \times 10^8}\right) \times (25 \times 10^{-4}) \, N \]

### Step 5: Calculate the momentum transferred
The momentum transferred \( \Delta P \) can be calculated using the formula:
\[ \Delta P = F \times t \]
Substituting the values:
\[ \Delta P = \left(\frac{25 \times 10^4}{3 \times 10^8} \times 25 \times 10^{-4}\right) \times 2400 \]

### Step 6: Simplify the expression
1. Calculate \( F \):
\[ F = \frac{25 \times 25 \times 10^4 \times 10^{-4}}{3 \times 10^8} = \frac{625}{3 \times 10^8} \, N \]
2. Now calculate \( \Delta P \):
\[ \Delta P = \left(\frac{625}{3 \times 10^8}\right) \times 2400 \]
\[ \Delta P = \frac{625 \times 2400}{3 \times 10^8} = \frac{1500000}{3 \times 10^8} \approx 5 \times 10^{-3} \, kg \cdot m/s \]

### Final Answer
The momentum transferred to the surface in 40 minutes is approximately:
\[ \Delta P \approx 5 \times 10^{-3} \, kg \cdot m/s \]