Two rods of same length and cross section are joined along the length. Thermal conductivities of first and second rod are ${K_1}\,\,{\rm{and}}\,\,{K_2}$. The temperature of the free ends of the first and second rods are maintained at ${\theta _1}\,\,{\rm{and }}{\theta _2}$ respectively. The temperature of the common junction is
$\frac{{{\theta _1} + {\theta _2}}}{2}$
$\frac{{{K_2}{K_2}}}{{{K_1} + {K_2}}}({\theta _1} + {\theta _2})$
$\frac{{{K_1}{\theta _1} + {K_2}{\theta _2}}}{{{K_1} + {K_2}}}$
$\frac{{{K_2}{\theta _1} + {K_1}{\theta _2}}}{{{K_1} + {K_2}}}$
One end of a thermally insulated rod is kept at a temperature $T_1$ and the other at $T_2$. The rod is composed of two sections of lengths $l_1$ and $l_2$ and thermal conductivities $K_1$ and $K_2$ respectively. The temperature at the interface of the two sections is
Three rods made of the same material and having the same cross section have been joined as shown in the figure. Each rod is of the same length. The left and right ends are kept at ${0^o}C$ and ${90^o}C$ respectively. The temperature of the junction of the three rods will be ...... $^oC$
Four rods of silver, copper, brass and wood are of same shape. They are heated together after wrapping a paper on it, the paper will burn first on
Assertion : The equivalent thermal conductivity of two plates of same thickness in contact is less than the smaller value of thermal conductivity.
Reason : For two plates of equal thickness in contact the equivalent thermal conductivity is given by : $\frac{1}{K} = \frac{1}{{{K_1}}} + \frac{1}{{{K_2}}}$
As per the given figure, two plates $A$ and $B$ of thermal conductivity $K$ and $2 K$ are joined together to form a compound plate. The thickness of plates are $4.0 \,cm$ and $2.5 \,cm$ respectively and the area of cross-section is $120 \,cm ^{2}$ for each plate. The equivalent thermal conductivity of the compound plate is $\left(1+\frac{5}{\alpha}\right) K$, then the value of $\alpha$ will be_______