Correct Answer - C

(c) Rate of heat flow is given by,

`Q=(KA( theta_1-theta_2))/l`

Where, K=coefficient of thermal conductivity

l=length of rod and A=Area of cross-section of rod

If the junction temperature is T, then

`Q_(Copper)=Q_(Brass)+Q_(steel)`

`(0.92xx4(100-T))/46=(0.26xx4xx(T-0))/13+`

`(0.12xx4xx(T-0))/12`

`implies 200-2T=2T+T`

`implies T-40^@C`

`:. Q_(Copper)=(0.92xx4xx60)46=4.8cal//s`