Electrical Analogy For Thermal Conduction - Practice Questions & MCQ

The temperature $\theta$ at the junction of two insulting sheets, having thermal resistances $R_1$ and $R_2$ as well as top and bottom temperatures $\theta_1$ and $\theta_2$ (as shown in figure) is given by :

Two metallic blocks $\mathrm{M}_1$ and $\mathrm{M}_2$ of same area of cross-section are connected to each other(as shown in figure).If the thermal conductivity of $\mathrm{M}_2$ is K then the thermal conductivity of $\mathrm{M}_1$ will be : [Assume steady state heat conduction]

As per the given figure, two plates $A$ and $B$ of thermal conductivity $
\mathrm{K} \text { and } 2 \mathrm{~K}
$ are joined together to form a compound plate. The thickness of the plates is $
4.0 \mathrm{~cm} \text { and } 2.5 \mathrm{~cm}
$ respectively and the area of the cross-section is $120 \mathrm{~cm}^2$  for each plate. The equivalent thermal conductivity of the compound plate is $\left(1+\frac{5}{\alpha}\right) \mathrm{K}$, then the value of $\alpha$ will be_____________.

If $\mathrm{K}_1$ and $\mathrm{K}_2$ are the thermal conductivities, $\mathrm{L}_1$ and $\mathrm{L}_2$ are the lengths and $\mathrm{A}_1$ and $\mathrm{A}_2$ are the cross sectional areas of steel and copper rods respectively such that $\frac{\mathrm{K}_2}{\mathrm{~K}_1}=9, \frac{\mathrm{~A}_1}{\mathrm{~A}_2}=2, \frac{\mathrm{~L}_1}{\mathrm{~L}_2}=2$. Then, for the arrangement as shown in the figure, the value of temperature T of the steel - copper junction in the steady state will be :

A rod CD of thermal resistance $10.0 \mathrm{KW}^{-1}$ is joined at the middle of an identical rod AB as shown in the figure. The ends $A, B$ and $D$ are maintained at $200^{\circ} \mathrm{C}, 100^{\circ} \mathrm{C}$ and $125^{\circ} \mathrm{C}$ respectively. The heat current in $C D$ is $P$ watt. The value of $P$ is $\qquad$

Heat is transmitted through a glass window of area 1$m^2$ and thickness 0.2cm. If the temperature difference between the two sides is $40^{\circ} \mathrm{C}$ then the calculation of heat current through the glass is [K for glass 0.2cal/m $^{\circ} \mathrm{C}$ s]

Two rod of same length and different material are arrange as shown . At steady state the temperature of middle point is - K₁ =300 unit, K₂ =200 unit

Two rod of same material and cross- section are arranged as shown in figure . The temperature at mid point B at steady state is 

Two materials having coefficients of thermal conductivity ' $3 K^{\prime}$ and ' $K^{\prime}$ and thickness ' $d$ ' and ' $3 d^{\prime}$, respectively are joined to form a slab as shown in the figure. The temperatures of the outer surfaces are ' $\theta_2^{\prime}$ and ' $\theta_1^{\prime}$ respectively , $\left.\theta_2>\theta_1\right)$.The temperature at the interface is :