A thin copper rod of uniform cross section A square metres and of length L metres has a spherical metal sphere of radius r metre at tis one end symmetrically attached to the copper rod. The thermal conductivity of copper is K and the emissivity of the spherical surface of the sphere is `epsi`.The free end of the copper rod is maintained at the temperature T kelving by supplying thermal energy from a P watt source. Steady state conditions are allowed ot be established while the rod is properly insulated aginst heat loss from its lateral surface. Surroundings are at `0^@C` Stefan's constant`=sigmaW//m^(2) K^(4)`.
The net power that will be radiated out, `P_S` from the sphere after steady state condition are reached is

Options

A. `P_S=P`
B. `P_S=(PA)/(4pir^2)`
C. `P_S=0`
D. `P_S=sigmaepsiT_S^4`

Explanation

To solve the problem, we need to analyze the steady-state condition of the copper rod and the attached spherical metal sphere. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Steady-State Condition
In a steady-state condition, the temperature throughout the system remains constant. This means that the power supplied to the system is equal to the power being lost from the system.

### Step 2: Identify the Power Supplied
The power supplied to the copper rod from the source is given as \( P \) watts. This power maintains the temperature \( T \) at the free end of the copper rod.

### Step 3: Identify the Power Radiated from the Sphere
The power radiated from the surface of the sphere can be calculated using the Stefan-Boltzmann law, which states that the power radiated by a black body is proportional to the fourth power of its absolute temperature. The formula for the power radiated by a surface with emissivity \( \epsilon \) is given by:
\[ P_S = \epsilon \sigma A_S T^4 \]
where:
- \( P_S \) is the power radiated from the sphere,
- \( \epsilon \) is the emissivity of the sphere,
- \( \sigma \) is the Stefan-Boltzmann constant,
- \( A_S \) is the surface area of the sphere,
- \( T \) is the absolute temperature of the sphere.

### Step 4: Calculate the Surface Area of the Sphere
The surface area \( A_S \) of a sphere is calculated using the formula:
\[ A_S = 4 \pi r^2 \]
where \( r \) is the radius of the sphere.

### Step 5: Substitute the Surface Area into the Power Equation
Substituting the surface area into the power equation gives:
\[ P_S = \epsilon \sigma (4 \pi r^2) T^4 \]

### Step 6: Apply the Steady-State Condition
At steady state, the power supplied \( P \) is equal to the power radiated \( P_S \):
\[ P = P_S \]
Thus, we can write:
\[ P = \epsilon \sigma (4 \pi r^2) T^4 \]

### Final Result
The net power that will be radiated out from the sphere after steady-state conditions are reached is:
\[ P_S = P \]