Two identical rods are connected between two containers. One of them is at 100⁰C. If rods are connected in parallel, then the rate of melting of ice is g₁ gs^-1. If they are connected in series, then the rate is g₂ gs^-1. The ratio g₂/g₁ is

According to the question, When two identical rods are connected in parallel,

So, rate of heat transfer is given by$\frac{dQ}{dt}=\frac{2KA\left(T_1-T_2\right)}{l}$

$=\frac{2KA(100-0)}{l}=\frac{200KA}{l}$

where, net area = 2A
K = thermal conductivity constant,
T₁ = temperature at one end,
T₂ = temperature at other end
and I = length of rod.

As,  $\frac{dQ}{dt}$is rate of heat transfer =  $\frac{mL}{t}$

So,   $\frac{dQ}{dt}=\left(\frac{m}{t}\right)L=\frac{200KA}{l}$

So, for parallel connection,

$\frac{m}{t}=g_1=\frac{200KA}{Ll}$                       …(i)

Similarly, for series connection

So,   $100-T_0=T_0-0\Rightarrow 2T_0=100$

$\Rightarrow T_0=50$

So,  $\frac{dQ}{dt}=\left(\frac{m}{t}\right)L=\frac{KA(50-0)}{l}$

and it can be written as

$g_2=\frac{KA50}{lL}$               …(ii)

On dividing Eq. (i) by Eq. (ii), we get

$\frac{g_1}{g_2}=\frac{200}{50}=4\Rightarrow \frac{g_2}{g_1}=\frac{1}{4}$