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`