Correct Answer - A::C
This problem can be solved like electric current problem.
`R_1`,`R_2`,`R_3`,`R_4`,`R_5`,`R_6` and `R_7` be the rates of heat flow through AE,EB,AC,CD,CE,ED and DB, respectively.
Since `R_1=R_2` `theta_E=50^@C` .(i)
`R_5=R_6` `R_3=R_4+R_5=R_7` .(ii)
`R_4+R_6=R_7`
`(KA(theta_C-50))/(l)=(kA)/(l)(theta_C-50)+(kA)/(l)(theta_C-theta_D)=(kA)/(l)theta_D`
`theta_C+theta_D=100`
`2theta_C-2theta_D=50impliestheta_C=62.5^@C`
`theta_D=37.5^@C`
`theta_Cgttheta_Egttheta_D`