Three rods of Copper, Brass and Steel are welded together to form a Y-shaped structure. Area of cross section of each rod = 4cm². End of copper rod is maintained at 100°C whereas ends of brass and steel are kept at 0°C. Lengths of the copper, brass and steel rods are 46, 13 and 12 cms respectively. The rods are thermally insulated from surroundings except at ends. Thermal conductivities of copper, brass and steel are 0.92, 0.26 and 0.12 CGS units respectively. Rate of heat flow through copper rod is:

Let dQ1/dt be the rate of heat flow through the copper rod, dQ2/dt be the rate of heat flow through the brass rod, and dQ3/dt be the rate of heat flow through the steel rod.

According to the principle of thermal conductivity, the rate of heat flow is given by:
dQ/dt = KA(ΔT/L)
where:
K = thermal conductivity
A = area of cross-section
ΔT = temperature difference
L = length of the rod

For the copper rod:
dQ1/dt = K1A(100 - T)/L1
For the brass rod:
dQ2/dt = K2A(T - 0)/L2
For the steel rod:
dQ3/dt = K3A(T - 0)/L3

Since the rods are welded together, the rate of heat flow through the copper rod is equal to the sum of the rates of heat flow through the brass and steel rods:
dQ1/dt = dQ2/dt + dQ3/dt

Substituting the expressions for dQ1/dt, dQ2/dt, and dQ3/dt, we get:
K1A(100 - T)/L1 = K2A(T)/L2 + K3A(T)/L3

Given values:
K1 = 0.92
K2 = 0.26
K3 = 0.12
A = 4 cm²
L1 = 46 cm
L2 = 13 cm
L3 = 12 cm

Substituting these values into the equation, we get:
0.92(4)(100 - T)/46 = 0.26(4)(T)/13 + 0.12(4)(T)/12
0.08(100 - T) = 0.08T + 0.04T
8 - 0.08T = 0.12T
8 = 0.2T
T = 40°C

Now we can find the rate of heat flow through the copper rod:
dQ1/dt = K1A(100 - T)/L1 = 0.92 * 4 * (100 - 40) / 46 = 0.92 * 4 * 60 / 46 = 4.8 cal/s

Therefore, the rate of heat flow through the copper rod is 4.8 cal/s.