A collimated beam of light of flux density `30kWm^(-2)` is incident normally on `100mm^(2)` completely absorbing screen. If P is the pressure exerted on the screen and `trianglep` is the momentum transferred to the screen during a 1000 s interval, then

Options

A. `P=10^(-3)Nm^(-2)`
B. `P=10^(-4)Nm^(-2)`
C. `trianglep=10^(-4)kgms^(-1)`
D. `trianglep=10^(-5)kgms^(-1)`

Explanation

To solve the problem, we need to calculate the pressure exerted on the screen and the momentum transferred to the screen during a specified time interval.

### Step-by-Step Solution:

1. **Identify the given values:**
- Flux density (Intensity, I) = 30 kW/m² = 30 × 10³ W/m²
- Area of the screen (A) = 100 mm² = 100 × 10⁻⁶ m²
- Time interval (Δt) = 1000 s
- Speed of light (c) = 3 × 10⁸ m/s

2. **Calculate the pressure (P) exerted on the screen:**
- The formula for pressure due to light is given by:
\[ P = \frac{I}{c} \]
- Substituting the values:
\[ P = \frac{30 \times 10^3 \text{ W/m}^2}{3 \times 10^8 \text{ m/s}} \]
- Simplifying this:
\[ P = \frac{30}{3} \times 10^{3 - 8} = 10 \times 10^{-5} = 10^{-4} \text{ N/m}^2 \]

3. **Calculate the momentum transferred (Δp) to the screen:**
- The formula for momentum transferred is given by:
\[ \Delta p = P \times A \times \Delta t \]
- Substituting the values:
\[ \Delta p = (10^{-4} \text{ N/m}^2) \times (100 \times 10^{-6} \text{ m}^2) \times (1000 \text{ s}) \]
- Simplifying this:
\[ \Delta p = 10^{-4} \times 100 \times 10^{-6} \times 1000 = 10^{-4} \times 10^{-4} \times 10^{3} = 10^{-5} \text{ kg m/s} \]

4. **Final Results:**
- Pressure (P) = \(10^{-4} \text{ N/m}^2\)
- Momentum transferred (Δp) = \(10^{-5} \text{ kg m/s}\)

### Summary of Results:
- Pressure (P) = \(10^{-4} \text{ N/m}^2\)
- Momentum transferred (Δp) = \(10^{-5} \text{ kg m/s}\)