1. Set up the steady-state condition

Let the junction temperature be $T$.

Since the three rods are welded together and the system is in steady state:

• Heat entering the junction through the copper rod
• equals heat leaving the junction through brass and steel rods.

So, $Q_{\text{Cu}} = Q_{\text{Br}} + Q_{\text{St}}$

Using the conduction formula, $Q = \frac{kA\Delta T}{L}$

Given:

• Cross-sectional area of each rod: $A=4\,\text{cm}^2$
• Copper: $k_c=0.92$, $L_c=46\,\text{cm}$, hot end at $100^\circ C$
• Brass: $k_b=0.26$, $L_b=13\,\text{cm}$, cold end at $0^\circ C$
• Steel: $k_s=0.12$, $L_s=12\,\text{cm}$, cold end at $0^\circ C$

2. Write expressions for heat currents

For copper rod: $Q_c=\frac{0.92\cdot 4\cdot (100-T)}{46}$

Since $0.92\cdot 4=3.68,$ $Q_c=\frac{3.68(100-T)}{46}=0.08(100-T)$

For brass rod: $Q_b=\frac{0.26\cdot 4\cdot T}{13}$

Since $0.26\cdot 4=1.04,$ $Q_b=\frac{1.04T}{13}=0.08T$

For steel rod: $Q_s=\frac{0.12\cdot 4\cdot T}{12}$

Since $0.12\cdot 4=0.48,$ $Q_s=\frac{0.48T}{12}=0.04T$

3. Apply heat balance at the junction

$0.08(100-T)=0.08T+0.04T$

$8-0.08T=0.12T$

$8=0.20T$

$T=40^\circ C$

4. Find rate of heat flow through copper rod

$Q_c=0.08(100-40)=0.08\times 60=4.8\,\text{cal/s}$

5. Check options

The correct option is: $\boxed{4.8\,\text{cal/s}}$ which corresponds to Option C.