Correct Answer - D
According to the question,
When two identical rods are connected in parallel,
So, rate of heat transfer is given by
`(dQ)/(dt) = (2KA(T_(1)-T_(2)))/=(2KA(100-0))/l = (200KA)/l`
where, net area = 2A
K = thermal conductivity constant
`T_(1)` = temperature at one end
`T_(2)`=temperature at other end
l = length of rod.
As, `(dQ)/(dt)` is rate of heat transfer = (mL)/(T)`
So, `g_(1)=(dQ)/(dt) = Q/T rArr (m/T)L = (200KA)/l`
So, for parallel connection,
`(m/T)` = g_(1)/s =(200KA)/(Ll)............(ii)
Similarly, for series connection
So, `(100 - T_(0)=T_(0)-0 rArr 2T_(0)=100`
`rArr T_(0)`=50`
So, `(dQ)/(dt) = (m/l)L=(KA(50-0))/l`
and it can be written as,
`g_(2)(g//s)= (KA50)/(lL)`.............(ii)
Dividing Eqs. (i_) and (ii), we get
`g_(1)/g_(2)=200/50 = 4 rArr g_(2)/g_(1) =1/4`