A circular ring (centre O) of radius a, and of uniform cross section is made up of three different metallic rods AB, BC and CA (joined together at the points A, B and C in pairs) of thermal conductivityies `alpha_1`,`alpha_2` and `alpha_3` respectively (see diagram). The junction A, B and C are maintained at the temperatures `100^@C`,`50^@C` and `0^@C`, respectively. All the rods are of equal lengths and cross sections. Under steady state conditions, assume that no heat is lost from the sides of the rods. Let `Q_1`,`Q_2` and `Q_3` be the rates of transmission of heat along the three rods AB, BC and CA. Then

Options

A. (a)`Q_(1)=Q_(2)=Q_(3)` and all are transmitted in the clockwise sense
B. (b)`Q_(1)andQ_(2)` flow in clockwise sense and `Q_(3)` in the anticlockwise sense
C. (c)`Q_(1):Q_(2):Q_(3)::alpha_(1):alpha_(2):2alpha_(3)`
D. (d)`(Q_(1))/(alpha_(1))+(Q_(2))/(alpha_(2))=(Q_(3))/(alpha_(3))`

Answer

BCD

Explanation

Heat flow will always be from higher to lower temperature .
Hence , `Q_(1)` in rod AB, `Q_(2)` in rod BC with both be in clock wise sense while `Q_(3)` in CA will be in anti-clock wise sense .
Now `Q_(1)=(alpha_(1)A(100-50))/(L)=50alpha_(1)(A)/(L),Q_(2)=(alpha_(2)A(50-0))/(L)=50alpha_(2)(A)/(L)`
`Q_(3)=(alpha_(3)A(100-0))/(L)=100alpha_(3)(A)/(L)rArrQ_(1):Q_(2):Q_(3)=alpha_(1):alpha_(2):2alpha_(3)`
Also `(Q_(1))/(alpha_(1))+(Q_(2))/(alpha_(2))=50(A)/(L)+50(A)/(L)=(100A)/(L)=(Q_(3))/(alpha_(3))`