Eleven identical rods are arranged as shown in Fig. Each rod has length `l`, cross sectional area A and thermal conductivity of material k. Ends A and F are maintained at temperatures `T_1` and `T_2(ltT_1)`, respectively. If lateral surface of each rod is thermally insulated, the rate of heat transfer `((dQ)/(dt))` in each rod is

Options

A. `((dQ)/(dt))_(AB)=((dQ)/(dt))_(CD)`
B. `((dQ)/(dt))_(BE)=(2)/(7)((T_1-T_2)KA)/(l)`
C. `((dQ)/(dt))_(CH)ne((dQ)/(dt))_(DG)`
D. `((dQ)/(dt))_(BC)=((dQ)/(dt))_(DC)`

Answer

B, C, D

Explanation

Resistance of each rod `R=(l)/(kA)`
In steady state `T_B=T_D`
`T_E=T_G`
Thermal current `((dQ)/(dt))=i`
`i_1R+i_3R+i_1R=(T_1-T_2)`
`2i_1+i_3=((T_1-T_2))/(R )` ..(i)
`2i_1+i_3=(T_1-T_2)/( R)` .(ii)
`i_1=i_2`
For the path ABCHGF
`i_1R+(i_1-i_3)R+2(i_1+i_3)R+(i_1-i_3)R+i_1R=(T_1-T_2)`
`6i_i-4i_3=((T_1-T_2))/(R )=2i_i+i_3` .(iv)
`4i_i=5i_3impliesi_3=(4)/(5)i_i` .(v)
From eqs (ii) and (v)
`2i_i+(4)/(5)i_1=((T_1-T_2))/(R )`
`(14i_i)/(5)=((T_1-T_2))/(R )`
Equivalent thermal resistance `((T_1-T_2))/(2i_i)=(7)/(5)R`
`((dQ)/(dt))_(AB)=i_1=(5(T_1-T_2)KA)/(14l)`
`((dQ)/(dt))_(BE)=(2)/(7)((T_1-T_2)KA)/(l)`
`((dQ)/(dt))_(BC)=((T_1-T_2)KA)/(14l)`
`((dQ)/(dt))_(CH)=((T_1-T_2)KA)/(7l)`