A plate of mass 20mg is in equilibrium in air due to force exerted by light beam on the plate as shown is the figure. Calculate the power of the beam (in mega-watt) if the plate is perfectly absorbing.
'(##CEN_KSR_PHY_JEE_CO28_E01_018_Q01##)'

Answer

To solve the problem, we need to calculate the power of a light beam that exerts enough force to keep a plate of mass 20 mg in equilibrium. Here are the steps to arrive at the solution:

### Step 1: Convert mass to kilograms
The mass of the plate is given as 20 mg. We need to convert this to kilograms for our calculations.

\[ \text{Mass} = 20 \, \text{mg} = 20 \times 10^{-3} \, \text{g} = 20 \times 10^{-6} \, \text{kg} = 2 \times 10^{-5} \, \text{kg} \]

### Step 2: Calculate the weight of the plate
The weight (force due to gravity) of the plate can be calculated using the formula:

\[ \text{Weight} = m \cdot g \]

Where \( g \) (acceleration due to gravity) is approximately \( 9.81 \, \text{m/s}^2 \).

\[ \text{Weight} = 2 \times 10^{-5} \, \text{kg} \cdot 9.81 \, \text{m/s}^2 = 1.962 \times 10^{-4} \, \text{N} \]

### Step 3: Relate the force exerted by the light beam to the power of the beam
For a perfectly absorbing plate, the force exerted by the light beam can be expressed in terms of the power \( P \) of the beam and the speed of light \( c \):

\[ F = \frac{P}{c} \]

Where \( c \) (speed of light) is approximately \( 3 \times 10^8 \, \text{m/s} \).

### Step 4: Set the forces equal for equilibrium
Since the plate is in equilibrium, the force exerted by the light beam is equal to the weight of the plate:

\[ \frac{P}{c} = \text{Weight} \]

Substituting the values we have:

\[ \frac{P}{3 \times 10^8} = 1.962 \times 10^{-4} \]

### Step 5: Solve for power \( P \)
Now we can solve for \( P \):

\[ P = 1.962 \times 10^{-4} \times 3 \times 10^8 \]

Calculating this gives:

\[ P = 5.886 \times 10^4 \, \text{W} \]

### Step 6: Convert power to megawatts
To convert watts to megawatts, we divide by \( 10^6 \):

\[ P = \frac{5.886 \times 10^4}{10^6} = 0.05886 \, \text{MW} \]

### Final Answer
Thus, the power of the beam is approximately:

\[ P \approx 0.05886 \, \text{MW} \]