Two identical rods are connected between two containers. One of them is at `100^(@)C` containing water and another is at `0^(@)C` containing ice. If rods are connected in parallel then the rate of melting of ice is `q_(1)g//s` . If they are connected in series then teh rate is `q_(2)g//s` . The ratio `q_(2)//q_(1)` is

Options

A. 2
B. 4
C. `1/2`
D. `1/4`

Answer

D

Explanation

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`