Two identical conducting rods are first connected independently to two vessels, one containing water at `100^@C` and the other containing ice at `0^@C`. In the second case, the rods are joined end to end and connected to the same vessels. Let `q_1 and q_2` gram per second be the rate of melting of ice in the two cases respectively. The ratio `q_1/q_2` is
(a) `1/2` (b)`2/1` (c)`4/1` (d)`1/4`

Explanation

Case I: When the rods are connected end

The rate of flow of heat through a rod of length l, cross sectional area A and temperature difference between its ends `(T_1-T_2)` is
` Q = (KA(T_1 - T_2 ))/(l)`
For the rods connected in series, equivalent conductivity K, is given by
`(l_1 + l_2)/(K_s) = (l_1)/(K_1) + (l_2)/(K_2)`
The rods are identical. `l_1 + l_2 = l` and `K_1 = K_2 = K`
` therefore (2l)/(K_s) = (2l)/(K)`
` therefore K_s = K`
The composite rod is of length 2l and area A.
` therefore ` Rate of heat flow in this case is
` H_1 = (KA(T_1 - T_2))/(2l) ` ....(i)
Case II: When the rods are placed one over another

The equivalent conductivity of the material of the composite rod is
`K_P = K_1 + K_2 = 2K`
The length of the composite rod is l and cross sectional area of the rod is 2A. So, the rate of heat flow in this case is
`H_2 = ((2K)(2A)(T_1 - T_2) )/(l)` ....(ii)
Dividing (i) by (ii), we get
`H_1/H_2 = 1/8`