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 its 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 kelvin by supplying thermal energy from a P watt source. Steady state conditions are allowed to be established while the rod is properly insulated against heat loss from its lateral surface. Surroundings are at `0^@C` Stefan's constant`=sigma W//m^(2) K^(4)`.
After the steady state conditions are reached,the temperature of the spherical end of the rod, `T_S` is

Options

A. `T_(S)=T-(PL)/(KA)`
B. `T_(S)=^(@)C`
C. `T_(S)=(PL)/(KA)`
D. `T_(S)=T-(P(L+r))/(KA)`

Explanation

To solve for the temperature of the spherical end of the copper rod, \( T_S \), we will follow these steps:

### Step 1: Understand the system
We have a copper rod of length \( L \) and cross-sectional area \( A \) with a spherical metal sphere of radius \( r \) attached to one end. The free end of the rod is maintained at temperature \( T \) by a power source supplying \( P \) watts. The system is insulated, meaning no heat is lost from the sides of the rod.

### Step 2: Apply the heat conduction equation
In steady state conditions, the power supplied to the rod is equal to the power conducted through the rod to the sphere. The heat conduction through the rod can be expressed using Fourier's law:
\[ P = \frac{K \cdot A \cdot (T - T_S)}{L + r} \]
Where:
- \( K \) is the thermal conductivity of copper,
- \( A \) is the cross-sectional area,
- \( T \) is the temperature at the free end of the rod,
- \( T_S \) is the temperature at the spherical end,
- \( L \) is the length of the rod,
- \( r \) is the radius of the sphere.

### Step 3: Rearrange the equation to solve for \( T_S \)
Rearranging the equation to isolate \( T_S \):
\[ T - T_S = \frac{P \cdot (L + r)}{K \cdot A} \]
\[ T_S = T - \frac{P \cdot (L + r)}{K \cdot A} \]

### Step 4: Consider the radiation loss from the sphere
The sphere will lose heat through radiation. The power radiated from the surface of the sphere can be expressed using the Stefan-Boltzmann law:
\[ P_{rad} = \epsilon \cdot \sigma \cdot A_{sphere} \cdot T_S^4 \]
Where:
- \( \epsilon \) is the emissivity of the sphere,
- \( \sigma \) is the Stefan-Boltzmann constant,
- \( A_{sphere} = 4\pi r^2 \) is the surface area of the sphere.

### Step 5: Set the power conducted equal to the power radiated
At steady state, the power conducted to the sphere must equal the power radiated away:
\[ P = \epsilon \cdot \sigma \cdot 4\pi r^2 \cdot T_S^4 \]

### Step 6: Solve for \( T_S \)
Now we have two equations:
1. \( T_S = T - \frac{P \cdot (L + r)}{K \cdot A} \)
2. \( P = \epsilon \cdot \sigma \cdot 4\pi r^2 \cdot T_S^4 \)

Substituting the expression for \( T_S \) from the first equation into the second will allow us to solve for \( T_S \).

### Final Expression
After substituting and rearranging, we can find the final expression for \( T_S \).

### Summary
The temperature of the spherical end of the rod, \( T_S \), can be calculated using the above equations, taking into account both the conduction through the rod and the radiation from the sphere.