Three rods of identical area of cross-section and made from the same metal from the sides of an isosceles triangle. ABC, right angled at B. The points A and B are maintained at temperatures T and `sqrt 2T` RESPECTIVELY. In the steady state the temperature of the point C is `T_(C)`.
Assuming that only heat conduction takes place , `T_(C) / T` is equal to

Options

A. `1 / (sqrt2+1)`
B. `3 / (sqrt2 + 1)`
C. `1 / (2(2sqrt2 - 1)`
D. `1 / (sqrt3(sqrt2 -1)`

Explanation

To solve the problem, we need to analyze the heat conduction through the rods forming the isosceles right triangle ABC, where point A is at temperature T, point B is at temperature √2T, and we need to find the temperature at point C (denoted as TC) in terms of T.

### Step-by-Step Solution:

1. **Identify the Geometry and Temperatures**:
- We have an isosceles right triangle ABC with right angle at B.
- Let the lengths of sides AB and BC be L.
- The length AC (hypotenuse) will be \(L\sqrt{2}\).
- The temperatures are given as:
- \(T_A = T\) (at point A)
- \(T_B = \sqrt{2}T\) (at point B)
- \(T_C = T_C\) (at point C, which we need to find)

2. **Thermal Resistance Calculation**:
- The thermal resistance \(R\) for a rod is given by:
\[ R = \frac{L}{kA} \]
where \(k\) is the thermal conductivity and \(A\) is the cross-sectional area.
- For rod AB (length L), the thermal resistance \(R_{AB} = \frac{L}{kA}\).
- For rod BC (length L), the thermal resistance \(R_{BC} = \frac{L}{kA}\).
- For rod AC (length \(L\sqrt{2}\)), the thermal resistance \(R_{AC} = \frac{L\sqrt{2}}{kA} = \sqrt{2}R\).

3. **Apply the Steady State Condition**:
- In steady state, the heat current through each rod must be equal:
\[ I_{BC} = I_{AC} \]
- The heat current \(I\) can be expressed as:
\[ I = \frac{T_{hot} - T_{cold}}{R} \]
- For rod BC:
\[ I_{BC} = \frac{\sqrt{2}T - T_C}{R} \]
- For rod AC:
\[ I_{AC} = \frac{T_C - T}{\sqrt{2}R} \]

4. **Set the Heat Currents Equal**:
- Equating the two expressions for heat current:
\[ \frac{\sqrt{2}T - T_C}{R} = \frac{T_C - T}{\sqrt{2}R} \]
- Cancel \(R\) from both sides:
\[ \sqrt{2}T - T_C = \frac{T_C - T}{\sqrt{2}} \]

5. **Clear the Fraction**:
- Multiply through by \(\sqrt{2}\) to eliminate the fraction:
\[ 2T - \sqrt{2}T_C = T_C - T \]

6. **Rearrange the Equation**:
- Rearranging gives:
\[ 2T + T = T_C + \sqrt{2}T_C \]
\[ 3T = T_C(1 + \sqrt{2}) \]

7. **Solve for \(T_C\)**:
- Now, solve for \(T_C\):
\[ T_C = \frac{3T}{1 + \sqrt{2}} \]

8. **Find the Ratio \(\frac{T_C}{T}\)**:
- Finally, we find the ratio:
\[ \frac{T_C}{T} = \frac{3}{1 + \sqrt{2}} \]

### Final Answer:
\[ \frac{T_C}{T} = \frac{3}{1 + \sqrt{2}} \]