Two rods of equal length and area of cross-section are connected in parallel between temperatures $20^{0} C a n d 80^{0} C$ . If ratio of their thermal conductivities is 3:4,then the ratio of the effective thermal conductivity to that of the first rod thermal conductivity is

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Two rods of length l and 2l with thermal conductivities 2K and K respectively are connected end to end. If cross sectional areas of two rods are equal, then equivalent thermal conductivity of the system is :

The correct option is A 1.2KWe know that for the given condition, equivalent thermal conductivity will be given as: K=K1K2(L1+L2)L1K2+L2K1So putting these value in above formula we find,Keq=6K2(L)5KL=1.2K

Two rods of length $l$ and $2 l$ with thermal conductivities $2 K$ and $K$ respectively are connected end to end. If cross sectional areas of two rods are equal, then equivalent thermal conductivity of the system is :

The correct option is A $1.2 K$
We know that for the given condition, equivalent thermal conductivity will be given as:

$K = \frac{K_{1} K_{2} (L_{1} + L_{2})}{L_{1} K_{2} + L_{2} K_{1}}$
So putting these value in above formula we find,
$K_{e q} = \frac{{6 K}^{2} (L)}{5 K L} = 1.2 K$