Thermal Resistance and it's Combination Questions in English

(A) Let $l_1$ be the length of $ADB$ and $l_2$ be the length of $ACB$. Given $\frac{l_2}{l_1} = 3$,so $l_2 = 3l_1$. Let $A$ be the cross-sectional area.
Initially,both parts have conductivity $k$. The thermal resistances are $R_1 = \frac{l_1}{kA}$ and $R_2 = \frac{l_2}{kA} = \frac{3l_1}{kA} = 3R_1$.
The total heat flow $H$ is the sum of heat flows through parallel paths: $H = H_1 + H_2 = \frac{T_1 - T_2}{R_1} + \frac{T_1 - T_2}{R_2} = (T_1 - T_2) \left( \frac{kA}{l_1} + \frac{kA}{3l_1} \right) = (T_1 - T_2) \frac{4kA}{3l_1}$.
When $ADB$ is replaced by a metal of conductivity $k'$,the new heat flow is $H' = 2H = 2 \left( \frac{4kA(T_1 - T_2)}{3l_1} \right) = \frac{8kA(T_1 - T_2)}{3l_1}$.
The new heat flow is $H' = \frac{k'A(T_1 - T_2)}{l_1} + \frac{kA(T_1 - T_2)}{3l_1}$.
Equating the two expressions for $H'$:
$\frac{k'A(T_1 - T_2)}{l_1} + \frac{kA(T_1 - T_2)}{3l_1} = \frac{8kA(T_1 - T_2)}{3l_1}$.
Dividing by $\frac{A(T_1 - T_2)}{l_1}$: $k' + \frac{k}{3} = \frac{8k}{3}$.
$k' = \frac{8k}{3} - \frac{k}{3} = \frac{7k}{3}$.

(B) Let $R$ be the thermal resistance of each rod. The total heat current $H_{total}$ flowing from $B$ to $H$ is given by $H_{total} = \frac{\Delta T}{R_{eq}}$.
For a cube with heat entering at one corner $(B)$ and leaving at the diagonally opposite corner $(H)$,the equivalent thermal resistance is $R_{eq} = \frac{5R}{6}$.
Thus,$H_{total} = \frac{100 - 0}{5R/6} = \frac{600}{5R} = \frac{120}{R}$.
At junction $B$,the heat current splits equally into three identical rods ($BA$,$BC$,$BF$). Therefore,the heat current through rod $BA$ is $H_{BA} = \frac{H_{total}}{3} = \frac{120/R}{3} = \frac{40}{R}$.
Using the heat flow formula for rod $BA$,$H_{BA} = \frac{T_B - T_A}{R}$,where $T_B = 100^{\circ}C$.
$\frac{40}{R} = \frac{100 - T_A}{R} \implies 40 = 100 - T_A \implies T_A = 60^{\circ}C$.

(C) Let the thermal resistance of each rod be $R$. The rods between $B$ and $C$ form two parallel paths,each consisting of two rods in series. The resistance of each path is $R + R = 2R$. The equivalent resistance of these two parallel paths is $R_{eq} = \frac{2R \times 2R}{2R + 2R} = R$.
Now,the system can be simplified as a series combination of three resistances: the rod $AB$ (resistance $R$),the parallel combination $BC$ (resistance $R$),and the rod $CD$ (resistance $R$).
The total resistance between $A$ and $D$ is $R_{total} = R + R + R = 3R$.
The heat current $H$ flowing through the system is $H = \frac{T_A - T_D}{R_{total}} = \frac{200 - 20}{3R} = \frac{180}{3R} = \frac{60}{R}$.
The temperature at junction $B$ can be found using the heat current through rod $AB$:
$H = \frac{T_A - T_B}{R} \implies \frac{60}{R} = \frac{200 - T_B}{R}$.
$60 = 200 - T_B \implies T_B = 200 - 60 = 140^{\circ}C$.

(C) The thermal resistance $R_{th}$ of a spherical shell of inner radius $r_1$,outer radius $r_2$,and thermal conductivity $K$ is given by $R_{th} = \frac{1}{4\pi K} \left( \frac{1}{r_1} - \frac{1}{r_2} \right)$.
For the inner sphere $(R_1 = R, R_2 = 2R)$: $R_{th1} = \frac{1}{4\pi K} \left( \frac{1}{R} - \frac{1}{2R} \right) = \frac{1}{4\pi K} \left( \frac{1}{2R} \right) = \frac{1}{8\pi KR}$.
For the outer sphere $(R_2 = 2R, R_3 = 3R)$: $R_{th2} = \frac{1}{4\pi K} \left( \frac{1}{2R} - \frac{1}{3R} \right) = \frac{1}{4\pi K} \left( \frac{1}{6R} \right) = \frac{1}{24\pi KR}$.
In steady state,the heat current $H$ is the same through both spheres: $H = \frac{T_{int} - 0}{R_{th1}} = \frac{100 - T_{int}}{R_{th2}}$.
Substituting the resistances: $\frac{T_{int}}{1/(8\pi KR)} = \frac{100 - T_{int}}{1/(24\pi KR)}$.
$8\pi KR \cdot T_{int} = 24\pi KR \cdot (100 - T_{int})$.
$T_{int} = 3(100 - T_{int}) \implies T_{int} = 300 - 3T_{int}$.
$4T_{int} = 300 \implies T_{int} = 75^oC$.

(A) In steady state,the rate of heat flow (heat current) through both rods must be the same.
Let $A$ be the cross-sectional area of the rods.
The thermal resistance $R$ of a rod is given by $R = \frac{l}{KA}$.
For rod $A$: $R_1 = \frac{l_1}{K_1 A}$.
For rod $B$: $R_2 = \frac{l_2}{K_2 A}$.
Given $l_2 = 2l_1$ and $K_1 = 2K_2$,we can write $K_2 = \frac{K_1}{2}$.
Substituting these into $R_2$: $R_2 = \frac{2l_1}{(K_1/2)A} = \frac{4l_1}{K_1 A} = 4R_1$.
Since the rods are in series,the heat current $H$ is the same:
$H = \frac{100 - \theta}{R_1} = \frac{\theta - 0}{R_2}$.
Substituting $R_2 = 4R_1$:
$\frac{100 - \theta}{R_1} = \frac{\theta}{4R_1}$.
$4(100 - \theta) = \theta$.
$400 - 4\theta = \theta$.
$5\theta = 400$.
$\theta = 80^{\circ}C$.

(A) The thermal resistance $R$ of a rod is given by the formula $R = \frac{l}{k \cdot A}$,where $l$ is the length,$k$ is the thermal conductivity,and $A$ is the cross-sectional area.
For rod $A$ with thermal conductivity $3K$,length $l$,and area $A$:
$R_A = \frac{l}{(3K) \cdot A} = \frac{l}{3KA}$
For rod $B$ with thermal conductivity $K$,length $l$,and area $A$:
$R_B = \frac{l}{K \cdot A} = \frac{l}{KA}$
Now,calculating the ratio of the thermal resistances:
$\frac{R_A}{R_B} = \frac{\frac{l}{3KA}}{\frac{l}{KA}} = \frac{1}{3}$
Therefore,the ratio $\frac{R_A}{R_B} = \frac{1}{3}$.

(A) For rods connected in series,the equivalent thermal resistance $R_{eq}$ is the sum of individual thermal resistances: $R_{eq} = R_1 + R_2 + R_3$.
Given that all rods have equal length $l$ and cross-sectional area $A$,the thermal resistance of a rod is $R = \frac{l}{KA}$.
Thus,the equivalent thermal resistance is:
$R_{eq} = \frac{l}{(K/2)A} + \frac{l}{(5K)A} + \frac{l}{KA} = \frac{l}{A} \left( \frac{2}{K} + \frac{1}{5K} + \frac{1}{K} \right)$.
Simplifying the expression inside the bracket:
$R_{eq} = \frac{l}{A} \left( \frac{10 + 1 + 5}{5K} \right) = \frac{l}{A} \left( \frac{16}{5K} \right) = \frac{16l}{5KA}$.
For the composite rod of total length $3l$,the equivalent thermal conductivity $K_{eq}$ is given by:
$R_{eq} = \frac{3l}{K_{eq}A} = \frac{16l}{5KA}$.
Solving for $K_{eq}$:
$K_{eq} = \frac{3 \times 5K}{16} = \frac{15K}{16}$.

(A) The thermal circuit can be analyzed by considering the thermal resistances. The rods $AB$ and $BC$ are in series with each other,and this combination is in parallel with rod $AC$.
Thermal resistance of the series branch ($AB$ and $BC$) is $R_1 = 10 + 10 = 20 \text{ units}$.
Thermal resistance of the parallel branch $(AC)$ is $R_2 = 20 \text{ units}$.
The equivalent thermal resistance $R_{eq}$ between points $A$ and $C$ is given by $\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{20} + \frac{1}{20} = \frac{2}{20} = \frac{1}{10}$.
Thus,$R_{eq} = 10 \text{ units}$.
The total rate of heat flow from $A$ to $C$ is $\frac{dH}{dt} = \frac{\Delta T}{R_{eq}} = \frac{100 - 0}{10} = 10 \text{ units}$.
Since the branch $AB+BC$ has a resistance of $20 \text{ units}$ and the branch $AC$ has a resistance of $20 \text{ units}$,the total heat flow splits equally between the two branches.
Therefore,the rate of heat flow through the junction $B$ (which is part of the $AB+BC$ branch) is $\frac{10}{2} = 5 \text{ units}$.

(B) Let $R$ be the thermal resistance of each side of the pentagon.
The path $ABC$ consists of two sides in series,so its resistance is $R_1 = R + R = 2R$.
The path $AEDC$ consists of three sides in series,so its resistance is $R_2 = R + R + R = 3R$.
Let $H$ be the total heat current entering at $A$. The heat current divides into two parallel paths $ABC$ and $AEDC$ such that the temperature drop across both paths is the same $(T_A - T_C)$.
For path $ABC$,the heat current $H_1 = \frac{T_A - T_B}{R} = \frac{T_B - T_C}{R}$. Thus,$T_A - T_C = (T_A - T_B) + (T_B - T_C) = 2(T_B - T_C)$.
For path $AEDC$,the heat current $H_2 = \frac{T_A - T_E}{R} = \frac{T_E - T_D}{R} = \frac{T_D - T_C}{R}$. Thus,$T_A - T_C = (T_A - T_E) + (T_E - T_D) + (T_D - T_C) = 3(T_D - T_C)$.
Equating the temperature differences: $2(T_B - T_C) = 3(T_D - T_C)$.
$2T_B - 2T_C = 3T_D - 3T_C$.
$3T_C - 2T_C = 3T_D - 2T_B$.
$T_C = 3T_D - 2T_B$.

(D) Let the temperature of the common interface be $T$. In a steady state,the rate of heat flow $H$ is the same through both slabs.
The rate of heat flow is given by $H = \frac{KA \Delta T}{l}$.
For the first slab (conductivity $K$,thickness $x$): $H_1 = \frac{KA(T_2 - T)}{x}$.
For the second slab (conductivity $2K$,thickness $4x$): $H_2 = \frac{(2K)A(T - T_1)}{4x} = \frac{KA(T - T_1)}{2x}$.
Since $H_1 = H_2$:
$\frac{KA(T_2 - T)}{x} = \frac{KA(T - T_1)}{2x}$
$T_2 - T = \frac{T - T_1}{2}$
$2T_2 - 2T = T - T_1$
$3T = 2T_2 + T_1 \implies T = \frac{2T_2 + T_1}{3}$.
Substituting $T$ back into the expression for $H_1$:
$H = \frac{KA}{x} \left( T_2 - \frac{2T_2 + T_1}{3} \right)$
$H = \frac{KA}{x} \left( \frac{3T_2 - 2T_2 - T_1}{3} \right)$
$H = \frac{KA}{x} \left( \frac{T_2 - T_1}{3} \right) = \frac{1}{3} \left( \frac{KA(T_2 - T_1)}{x} \right)$.
Comparing this with the given expression $\left( \frac{A(T_2 - T_1)K}{x} \right)f$,we get $f = \frac{1}{3} \approx 0.33$.

(D) Let $K$ be the thermal conductivity,$A$ be the cross-sectional area,and $\ell$ be the length of each rod.
In the first case,the two rods are connected in parallel between the two reservoirs at $T_1 = 100^{\circ}C$ and $T_2 = 0^{\circ}C$.
The rate of heat flow is $\frac{dH}{dt_1} = \frac{KA(T_1-T_2)}{\ell} + \frac{KA(T_1-T_2)}{\ell} = \frac{2KA(T_1-T_2)}{\ell}$.
Since $q_1 \propto \frac{dH}{dt_1}$,we have $q_1 = \frac{2KA(T_1-T_2)}{L_f \ell}$,where $L_f$ is the latent heat of fusion.
In the second case,the rods are connected in series. The equivalent thermal resistance is $R_{eq} = \frac{\ell}{KA} + \frac{\ell}{KA} = \frac{2\ell}{KA}$.
The rate of heat flow is $\frac{dH}{dt_2} = \frac{T_1-T_2}{R_{eq}} = \frac{KA(T_1-T_2)}{2\ell}$.
Thus,$q_2 = \frac{KA(T_1-T_2)}{2 L_f \ell}$.
The ratio $q_2/q_1 = \frac{KA(T_1-T_2) / (2 L_f \ell)}{2KA(T_1-T_2) / (L_f \ell)} = \frac{1/2}{2} = \frac{1}{4}$.

(B) Let the thermal resistance of each rod be $R$.
The total temperature difference between $A$ and $D$ is $\Delta T = 100^\circ C - 0^\circ C = 100^\circ C$.
By symmetry,the points $B$ and $H$ are at the same potential (temperature),and points $C$ and $E$ are at the same potential. Similarly,$G$ and $F$ are at the same potential.
By simplifying the circuit of thermal resistances,we can combine the parallel branches. The equivalent thermal resistance between $A$ and $D$ is calculated as $\frac{7R}{6}$.
However,looking at the simplified circuit path from $A$ to $D$,the total resistance is $\frac{7R}{6}$. The temperature drop across the segment $BC$ can be calculated using the potential divider rule in the equivalent circuit.
The temperature difference across $BC$ is given by $\Delta T_{BC} = \left( \frac{R_{BC}}{R_{total}} \right) \times \Delta T_{AD}$.
Given the equivalent circuit,the resistance between $B$ and $C$ is $\frac{5R}{6}$ and the total resistance is $\frac{7R}{6}$.
Therefore,$\Delta T_{BC} = \left( \frac{5R/6}{7R/6} \right) \times 100^\circ C = \frac{5}{7} \times 100^\circ C = \frac{500}{7} \ ^\circ C$.

(D) Let the thermal resistance of a rod with conductivity $K$ be $R = \frac{l}{Ka}$.
Then,the resistance of a rod with conductivity $2K$ is $R' = \frac{l}{(2K)a} = \frac{R}{2}$.
From the equivalent circuit diagram,the upper branch consists of three parts in series:
$1$. Two rods of $2K$ in parallel: $R_1 = \frac{(R/2)(R/2)}{(R/2) + (R/2)} = \frac{R/4}{1} = \frac{R}{4}$.
$2$. Two rods of $K$ in parallel: $R_2 = \frac{R \cdot R}{R + R} = \frac{R}{2}$.
$3$. Two rods of $2K$ in parallel: $R_3 = \frac{(R/2)(R/2)}{(R/2) + (R/2)} = \frac{R}{4}$.
The total resistance of the upper branch is $R_{upper} = R_1 + R_2 + R_3 = \frac{R}{4} + \frac{R}{2} + \frac{R}{4} = R$.
The lower branch is a single rod of conductivity $K$,so $R_{lower} = R$.
The effective resistance $R_{eq}$ between $A$ and $B$ is the parallel combination of the upper and lower branches:
$R_{eq} = \frac{R_{upper} \cdot R_{lower}}{R_{upper} + R_{lower}} = \frac{R \cdot R}{R + R} = \frac{R}{2}$.
Substituting $R = \frac{l}{Ka}$,we get $R_{eq} = \frac{1}{2} \left( \frac{l}{Ka} \right) = \frac{l}{2Ka}$.

(A) For a composite rod joined in series,the total thermal resistance $R_{eq}$ is the sum of individual thermal resistances.
Thermal resistance $R$ is given by $R = \frac{l}{KA}$,where $l$ is the length,$K$ is the thermal conductivity,and $A$ is the cross-sectional area.
For the two rods in series: $R_{eq} = R_1 + R_2$.
Substituting the expressions: $\frac{l_1 + l_2}{KA} = \frac{l_1}{K_1A} + \frac{l_2}{K_2A}$.
Since the cross-sectional area $A$ is the same for both rods,it cancels out from both sides.
Therefore,$\frac{l_1 + l_2}{K} = \frac{l_1}{K_1} + \frac{l_2}{K_2}$.

(D) Let the square be $ABCD$ with vertices $A, B, C, D$ in order. Let the temperatures at $A$ and $C$ be $T_A = \sqrt{2}\theta$ and $T_C = \theta$. Let the temperatures at $B$ and $D$ be $T_B$ and $T_D$.
Since the rods are identical,the thermal resistance $R$ of each rod is the same. The heat flow is steady.
At node $B$,by Kirchhoff's law for heat current: $\frac{T_A - T_B}{R} + \frac{T_C - T_B}{R} = 0$,which implies $T_B = \frac{T_A + T_C}{2} = \frac{\sqrt{2}\theta + \theta}{2}$.
Similarly,at node $D$: $\frac{T_A - T_D}{R} + \frac{T_C - T_D}{R} = 0$,which implies $T_D = \frac{T_A + T_C}{2} = \frac{\sqrt{2}\theta + \theta}{2}$.
Since $T_B = T_D$,the temperature difference between the other two points $B$ and $D$ is $T_B - T_D = 0$.

(B) Let the temperature at junction $B$ be $T_B$. Since the rods are of the same material and cross-section, their thermal resistance $R$ is proportional to their length $L$ $(R = \frac{L}{kA})$.
Given that there is no heat flow in $AB$, the temperature at $B$ must be equal to the temperature at $A$. Therefore, $T_B = T_A = 20 ^oC$.
Since there is no heat flow in $AB$, all the heat flowing from $D$ to $B$ must flow from $B$ to $C$.
Thus, the rate of heat flow $H_{DB} = H_{BC}$.
Using the formula $H = \frac{\Delta T}{R}$, we have $\frac{T_D - T_B}{R_{BD}} = \frac{T_B - T_C}{R_{BC}}$.
Substituting the values: $\frac{90 - 20}{R_{BD}} = \frac{20 - 0}{R_{BC}}$.
$\frac{70}{R_{BD}} = \frac{20}{R_{BC}} \implies \frac{R_{BD}}{R_{BC}} = \frac{70}{20} = 3.5$.
Since $R \propto L$, the ratio of lengths $\frac{L_{BD}}{L_{BC}} = \frac{R_{BD}}{R_{BC}} = 3.5$.

(B) The thermal resistance $R$ of a layer is given by $R = \frac{L}{KA}$,where $L$ is the thickness,$K$ is the thermal conductivity,and $A$ is the area.
For the first layer (thin layer): $R_1 = \frac{1}{K \cdot A}$.
For the second layer: $R_2 = \frac{4}{3K \cdot A}$.
Since the layers are in series,the heat current $H$ is the same through both layers. The temperature difference $\Delta T$ across a layer is given by $\Delta T = H \cdot R$.
Thus,the ratio of temperature differences is $\frac{\Delta T_1}{\Delta T_2} = \frac{R_1}{R_2} = \frac{1/K}{4/3K} = \frac{3}{4}$.
Given $\Delta T_1 + \Delta T_2 = 50\ ^oC$.
Substituting $\Delta T_2 = \frac{4}{3} \Delta T_1$ into the equation: $\Delta T_1 + \frac{4}{3} \Delta T_1 = 50$.
$\frac{7}{3} \Delta T_1 = 50 \implies \Delta T_1 = \frac{150}{7}\ ^oC$.

(B) For two rods joined in parallel,the equivalent thermal conductivity $K_{eq}$ is given by the formula:
$K_{eq} = \frac{K_1 A_1 + K_2 A_2}{A_1 + A_2}$
Given:
$A_1 = A, A_2 = 2A$
$K_1 = 2K, K_2 = 3K$
Substituting these values into the formula:
$K_{eq} = \frac{(2K)(A) + (3K)(2A)}{A + 2A}$
$K_{eq} = \frac{2KA + 6KA}{3A}$
$K_{eq} = \frac{8KA}{3A}$
$K_{eq} = \frac{8K}{3}$

(C) Let the thermal resistance of each rod be $R$. The system consists of a rod $AB$ of resistance $R$,followed by a parallel combination of two branches between $B$ and $C$,each having a resistance of $2R$. The equivalent resistance of this parallel part is $R_{BC} = \frac{2R \times 2R}{2R + 2R} = R$. Finally,there is a rod $CD$ of resistance $R$.
Thus,the total circuit can be simplified to a series combination of three resistances: $R_{AB} = R$,$R_{BC} = R$,and $R_{CD} = R$. The total resistance is $3R$.
The heat current $H$ flowing through the system is $H = \frac{T_A - T_D}{3R} = \frac{200 - 20}{3R} = \frac{180}{3R} = \frac{60}{R}$.
Now,for junction $B$,the heat current flowing through $AB$ must equal the heat current flowing through the rest of the circuit ($BC$ and $CD$):
$H = \frac{T_A - T_B}{R} = \frac{60}{R}$
$200 - T_B = 60$
$T_B = 200 - 60 = 140^{\circ}C$.

(D) The thermal resistance $R$ of a rod is given by the formula:
$R = \frac{l}{KA}$
where $l$ is the length of the rod,$K$ is the thermal conductivity,and $A$ is the cross-sectional area.
Given:
Length $l = 2 \ m$
Radius $r = 1 \ cm = 10^{-2} \ m$
Thermal conductivity $K = 401 \ W/m-K$
Cross-sectional area $A = \pi r^2 = 3.14 \times (10^{-2})^2 = 3.14 \times 10^{-4} \ m^2$
Substituting the values into the formula:
$R = \frac{2}{401 \times 3.14 \times 10^{-4}}$
$R = \frac{2}{0.125914} \approx 15.88 \ K/W$
Rounding to one decimal place,we get $R \approx 15.9 \ K/W$.

(B) Let $R_A, R_B$ and $R_C$ be the thermal resistances of rods $A, B$ and $C$ respectively. The thermal resistance is given by $R = \frac{l}{KA}$.
$R_A = \frac{2l}{KA}$
$R_B = \frac{l}{(2K)(2A)} = \frac{l}{4KA}$
$R_C = \frac{l}{(4K)(2A)} = \frac{l}{8KA}$
Let $\theta$ be the temperature of the junction. The thermal currents flowing towards the junction must sum to zero (Kirchhoff's junction law for heat flow):
$\frac{100 - \theta}{R_A} + \frac{50 - \theta}{R_B} + \frac{0 - \theta}{R_C} = 0$
Substituting the values of resistances:
$\frac{100 - \theta}{2l/KA} + \frac{50 - \theta}{l/4KA} + \frac{0 - \theta}{l/8KA} = 0$
$\frac{KA}{l} \left[ \frac{100 - \theta}{2} + 4(50 - \theta) - 8\theta \right] = 0$
$50 - 0.5\theta + 200 - 4\theta - 8\theta = 0$
$250 = 12.5\theta$
$\theta = \frac{250}{12.5} = 20^{\circ}C$

(D) Let $R_1$ and $R_2$ be the thermal resistances of the copper and iron wires respectively. Then,
$R_1 = \frac{l}{K_1 A}$ and $R_2 = \frac{l}{K_2 A}$.
From the figure,the arrangement is a Wheatstone bridge network. The arms $AB$ and $BC$ are copper (resistance $R_1$ each),and $AD$ and $DC$ are iron (resistance $R_2$ each). The central arm $BD$ consists of two wires (one iron and one copper) in parallel,but due to the symmetry of the bridge,the potential (temperature) at $B$ and $D$ is the same. Thus,no heat flows through the central arm $BD$,and it can be removed.
The circuit simplifies to two parallel branches: one branch with two copper wires in series $(R_1 + R_1 = 2R_1)$ and one branch with two iron wires in series $(R_2 + R_2 = 2R_2)$.
The equivalent thermal resistance $R_{eq}$ between $A$ and $C$ is given by:
$R_{eq} = \frac{(2R_1)(2R_2)}{2R_1 + 2R_2} = \frac{4R_1 R_2}{2(R_1 + R_2)} = \frac{2R_1 R_2}{R_1 + R_2}$.
Substituting the values of $R_1$ and $R_2$:
$R_{eq} = \frac{2 \left( \frac{l}{K_1 A} \right) \left( \frac{l}{K_2 A} \right)}{\frac{l}{K_1 A} + \frac{l}{K_2 A}} = \frac{2 \frac{l^2}{K_1 K_2 A^2}}{\frac{l(K_1 + K_2)}{K_1 K_2 A}} = \frac{2l}{A(K_1 + K_2)}$.

(A) Let $Q$ be the heat required to melt the ice. The rate of heat flow $P$ is given by $P = \frac{Q}{t}$.
For experiment $I$: $P_1 = \frac{Q}{t_1} = \frac{Q}{20}$.
For experiment $II$: $P_2 = \frac{Q}{t_2} = \frac{Q}{80}$.
In experiment $III$ (series): The equivalent thermal resistance $R_{eq} = R_1 + R_2$. Since $P = \frac{\Delta T}{R}$,we have $\frac{1}{P_{eq}} = \frac{1}{P_1} + \frac{1}{P_2}$.
$\frac{t_{10}}{Q} = \frac{t_1}{Q} + \frac{t_2}{Q} \Rightarrow t_{10} = t_1 + t_2 = 20 + 80 = 100 \text{ minutes}$.
In experiment $IV$ (parallel): The equivalent thermal resistance is $\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}$,which implies $P_{eq} = P_1 + P_2$.
$\frac{Q}{t_{20}} = \frac{Q}{t_1} + \frac{Q}{t_2} \Rightarrow \frac{1}{t_{20}} = \frac{1}{20} + \frac{1}{80} = \frac{4+1}{80} = \frac{5}{80} = \frac{1}{16}$.
Thus,$t_{20} = 16 \text{ minutes}$.
Comparing with the options,the correct statement is that $t_{10} = 100 \text{ minutes}$.

(D) The thermal resistance $R$ of a rod is given by $R = \frac{l}{KA}$.
For rod $A$: $R_A = \frac{2l}{K \cdot A} = \frac{2l}{KA}$.
For rod $B$: $R_B = \frac{l}{(2K) \cdot (2A)} = \frac{l}{4KA}$.
For rod $C$: $R_C = \frac{l}{(4K) \cdot (2A)} = \frac{l}{8KA}$.
Let the junction temperature be $\theta$. According to the principle of conservation of energy (Kirchhoff's junction law for heat flow),the sum of heat currents at the junction must be zero: $\frac{100 - \theta}{R_A} + \frac{50 - \theta}{R_B} + \frac{0 - \theta}{R_C} = 0$.
Substituting the values of resistances:
$\frac{100 - \theta}{2l/KA} + \frac{50 - \theta}{l/4KA} + \frac{0 - \theta}{l/8KA} = 0$.
Multiplying by $\frac{l}{KA}$:
$\frac{100 - \theta}{2} + 4(50 - \theta) + 8(0 - \theta) = 0$.
$50 - 0.5\theta + 200 - 4\theta - 8\theta = 0$.
$250 = 12.5\theta$.
$\theta = \frac{250}{12.5} = 20\,^oC$.
Wait,let's re-evaluate the calculation: $50 - 0.5\theta + 200 - 4\theta - 8\theta = 250 - 12.5\theta = 0 \implies \theta = 20\,^oC$. Checking the options,it seems there might be a typo in the provided options or the setup. Let's re-check the heat flow: $\frac{100-\theta}{R_A} = \frac{\theta-50}{R_B} + \frac{\theta-0}{R_C}$.
$\frac{100-\theta}{2} = 4(\theta-50) + 8(\theta-0)$.
$50 - 0.5\theta = 4\theta - 200 + 8\theta$.
$250 = 12.5\theta \implies \theta = 20\,^oC$. Given the options,if we assume the heat flows from $100$ and $50$ towards the junction and then to $0$,the calculation holds. If the question implies a different configuration,the result might vary. Based on standard interpretation,$\theta = 20\,^oC$ is the result.

(D) The thermal resistance of each rod is given by $R = \frac{L}{KA}$. Since the rods are identical,the thermal resistance $R$ is the same for all three rods.
Using Kirchhoff's law for heat current at the junction,the sum of heat currents entering the junction must be zero:
$i_1 + i_2 + i_3 = 0$
Here,$i_1$ is the heat current from the $30\,^{\circ}C$ end,and $i_2, i_3$ are the heat currents from the two $90\,^{\circ}C$ ends.
$\frac{30 - \theta}{R} + \frac{90 - \theta}{R} + \frac{90 - \theta}{R} = 0$
Multiplying by $R$:
$(30 - \theta) + (90 - \theta) + (90 - \theta) = 0$
$30 + 90 + 90 - 3\theta = 0$
$210 = 3\theta$
$\theta = 70\,^{\circ}C$

(B) In all three cases,the materials are arranged in series between the same two temperatures.
Since the total thermal resistance $R_{eq} = R_1 + R_2 + R_3$ is the same for all three arrangements (as the sum of resistances is independent of the order in series),the total heat current $H = \frac{\Delta T_{total}}{R_{eq}}$ is also the same for all three cases.
For any individual material layer,the heat current is given by $H = \frac{\Delta T_{layer}}{R_{layer}}$.
Since $H$ is constant and the resistance $R_1$ of material $1$ is constant in all three cases,the temperature difference across material $1$ is $\Delta T_1 = H \times R_1$.
Therefore,$\Delta T_a = \Delta T_b = \Delta T_c$.

(B) Let the thermal resistance of each rod be $R$. The circuit can be analyzed as a series-parallel combination of thermal resistances.
$1$. The section between $B$ and $C$ consists of two parallel branches,each containing two rods in series. The resistance of each branch is $R + R = 2R$. The equivalent resistance of the parallel combination between $B$ and $C$ is $R_{BC} = (2R \times 2R) / (2R + 2R) = R$.
$2$. The total thermal resistance of the circuit is $R_{total} = R_{AB} + R_{BC} + R_{CD} = R + R + R = 3R$.
$3$. The total temperature difference is $\Delta T = 200^{\circ}C - 20^{\circ}C = 180^{\circ}C$.
$4$. The heat current $H$ flowing through the circuit is $H = \Delta T / R_{total} = 180 / 3R = 60/R$.
$5$. The temperature drop across $AB$ is $\Delta T_{AB} = H \times R = (60/R) \times R = 60^{\circ}C$. Thus,$T_B = 200^{\circ}C - 60^{\circ}C = 140^{\circ}C$.
$6$. The temperature drop across $BC$ is $\Delta T_{BC} = H \times R_{BC} = (60/R) \times R = 60^{\circ}C$. Thus,$T_C = T_B - 60^{\circ}C = 140^{\circ}C - 60^{\circ}C = 80^{\circ}C$.

(B) The heat flow is given by $\Delta Q = \frac{\Delta T}{R_{eq}} t$,where $R_{eq}$ is the equivalent thermal resistance.
For two identical rods of resistance $R$,when joined in series,the equivalent resistance is $R_{s} = R + R = 2R$.
When joined in parallel,the equivalent resistance is $R_{p} = \frac{R \times R}{R + R} = \frac{R}{2}$.
Given that the heat $\Delta Q$ is the same in both cases and the temperature difference $\Delta T$ is constant:
$\Delta Q = \frac{\Delta T}{2R} \times 12 = \frac{\Delta T}{R/2} \times t$
Simplifying the equation:
$\frac{12}{2R} = \frac{t}{R/2}$
$\frac{6}{R} = \frac{2t}{R}$
$2t = 6$
$t = 3 \, \text{minutes}$.

(B) The thermal resistance of a material is given by $R = \frac{l}{KA}$,where $l$ is the length,$A$ is the area,and $K$ is the thermal conductivity.
$1$. For the first section (left): There are three parallel bricks each of length $l$ and area $A/3$. The resistance of each brick is $R_1 = \frac{l}{K(A/3)} = \frac{3l}{KA}$. Since they are in parallel,the equivalent resistance $R_{eq1}$ is $\frac{1}{R_{eq1}} = \frac{1}{R_1} + \frac{1}{R_1} + \frac{1}{R_1} = \frac{3}{R_1} = \frac{3}{3l/KA} = \frac{KA}{l}$. Thus,$R_{eq1} = \frac{l}{KA}$.
$2$. For the middle section: There is a single brick of length $l$ and area $A$. The resistance is $R_{eq2} = \frac{l}{KA}$.
$3$. For the third section (right): There are two parallel bricks each of length $l$ and area $A/2$. The resistance of each brick is $R_3 = \frac{l}{K(A/2)} = \frac{2l}{KA}$. Since they are in parallel,the equivalent resistance $R_{eq3}$ is $\frac{1}{R_{eq3}} = \frac{1}{R_3} + \frac{1}{R_3} = \frac{2}{R_3} = \frac{2}{2l/KA} = \frac{KA}{l}$. Thus,$R_{eq3} = \frac{l}{KA}$.
$4$. Since these three sections are in series,the total equivalent thermal resistance is $R_{total} = R_{eq1} + R_{eq2} + R_{eq3} = \frac{l}{KA} + \frac{l}{KA} + \frac{l}{KA} = \frac{3l}{KA}$.

(A) Let $R_1$ be the thermal resistance of arc $ADB$ and $R_2$ be the thermal resistance of arc $ACD$. The angle subtended by $ADB$ at the center is $90^\circ$ (or $\pi/2$ radians),and the angle subtended by $ACD$ is $270^\circ$ (or $3\pi/2$ radians).
Since the length of an arc is $L = r\theta$,the length of $ADB$ is $L_1 = r(\pi/2)$ and the length of $ACD$ is $L_2 = r(3\pi/2) = 3L_1$.
Thermal resistance $R = \frac{L}{KA}$. Assuming same cross-sectional area $A$ and same initial conductivity $K$:
$R_1 = \frac{L_1}{KA}$ and $R_2 = \frac{3L_1}{KA} = 3R_1$.
The arcs are in parallel between $A$ and $B$,so the total heat current $H = \frac{\Delta T}{R_{eq}} = \Delta T \left(\frac{1}{R_1} + \frac{1}{R_2}\right) = \Delta T \left(\frac{1}{R_1} + \frac{1}{3R_1}\right) = \frac{4\Delta T}{3R_1}$.
When $ADB$ is replaced with material of conductivity $K'$,the new resistance is $R_1' = \frac{L_1}{K'A} = R_1 \frac{K}{K'}$.
The new heat current is $2H = \Delta T \left(\frac{1}{R_1'} + \frac{1}{R_2}\right) = \Delta T \left(\frac{K'}{R_1 K} + \frac{1}{3R_1}\right)$.
Substituting $H = \frac{4\Delta T}{3R_1}$,we get $2 \left(\frac{4\Delta T}{3R_1}\right) = \Delta T \left(\frac{K'}{R_1 K} + \frac{1}{3R_1}\right)$.
$\frac{8}{3R_1} = \frac{3K' + K}{3KR_1} \implies 8K = 3K' + K \implies 3K' = 7K \implies \frac{K'}{K} = \frac{7}{3}$.

(A) Let $R_0$ be the thermal resistance of a rod of length $L$. The resistance of segment $AP$ (length $L/2$) is $R_0/2$,segment $PQ$ (length $L$) is $R_0$,and segment $QB$ (length $L/2$) is $R_0/2$. The bent rod $PQ$ has a total length of $3L/2$ (two vertical segments of $L/4$ and one horizontal segment of $L$). Its resistance is $R_{bent} = R_0/4 + R_0 + R_0/4 = 1.5R_0$. The segment $PQ$ of the main rod has resistance $R_0$. These two are in parallel,so their equivalent resistance $R_p = \frac{R_0 \times 1.5R_0}{R_0 + 1.5R_0} = \frac{1.5}{2.5}R_0 = 0.6R_0$. The total resistance of the circuit is $R_{eq} = R_{AP} + R_p + R_{QB} = 0.5R_0 + 0.6R_0 + 0.5R_0 = 1.6R_0$. The temperature drop across $PQ$ is $\Delta T_{PQ} = \Delta T_{total} \times \frac{R_p}{R_{eq}} = 120 \times \frac{0.6R_0}{1.6R_0} = 120 \times \frac{6}{16} = 120 \times 0.375 = 45\,^oC$.

(D) When heat flows along the length of the cylinders,the two cylinders act as parallel conductors of heat.
For parallel combination,the equivalent thermal conductance is the sum of individual thermal conductances.
The thermal conductance is given by $C = \frac{KA}{L}$,where $K$ is thermal conductivity,$A$ is the cross-sectional area,and $L$ is the length.
For the inner cylinder,area $A_1 = \pi R^2$.
For the outer cylindrical shell,area $A_2 = \pi (2R)^2 - \pi R^2 = 3\pi R^2$.
The total area of the system is $A = A_1 + A_2 = 4\pi R^2$.
Equating the total conductance to the sum of individual conductances:
$\frac{K_{eff} A}{L} = \frac{K_1 A_1}{L} + \frac{K_2 A_2}{L}$
$K_{eff} (4\pi R^2) = K_1 (\pi R^2) + K_2 (3\pi R^2)$
$4 K_{eff} = K_1 + 3 K_2$
$K_{eff} = \frac{K_1 + 3 K_2}{4}$

(C) For rods connected in series,the equivalent thermal resistance $R_{eq}$ is the sum of individual thermal resistances: $R_{eq} = R_1 + R_2$.
The formula for thermal resistance is $R = \frac{\ell}{KA}$,where $\ell$ is length,$K$ is thermal conductivity,and $A$ is the cross-sectional area.
Substituting this into the series equation:
$\frac{\ell_1 + \ell_2}{K A} = \frac{\ell_1}{K_1 A} + \frac{\ell_2}{K_2 A}$
Since the cross-sectional area $A$ is the same for both rods,we can cancel $A$ from both sides:
$\frac{\ell_1 + \ell_2}{K} = \frac{\ell_1}{K_1} + \frac{\ell_2}{K_2}$

(D) In a steady state,the rate of heat flow through the two sheets connected in series must be the same.
Let $H$ be the rate of heat flow.
$H = \frac{\Delta T_1}{R_1} = \frac{\Delta T_2}{R_2}$
Substituting the temperature values:
$\frac{\theta_1 - \theta}{R_1} = \frac{\theta - \theta_2}{R_2}$
Cross-multiplying to solve for $\theta$:
$(\theta_1 - \theta)R_2 = (\theta - \theta_2)R_1$
$\theta_1 R_2 - \theta R_2 = \theta R_1 - \theta_2 R_1$
$\theta_1 R_2 + \theta_2 R_1 = \theta (R_1 + R_2)$
Therefore,the junction temperature is:
$\theta = \frac{\theta_1 R_2 + \theta_2 R_1}{R_1 + R_2}$

(B) Given that the thermal conductivity $K_A = 2 K_B$.
Since both layers have the same thickness $d$ and cross-sectional area $A$,the thermal resistance $R = \frac{d}{KA}$.
Therefore,$R_A = \frac{d}{K_A A} = \frac{d}{2 K_B A} = \frac{R_B}{2}$.
Let $R_A = R$,then $R_B = 2R$.
In a steady state,the heat current $H$ is the same through both layers connected in series.
The total temperature difference $\Delta T = 36\,^{\circ}C$.
$H = \frac{\Delta T}{R_{total}} = \frac{36}{R_A + R_B} = \frac{36}{R + 2R} = \frac{36}{3R} = \frac{12}{R}$.
The temperature difference across layer $A$ is $\Delta T_A = H \times R_A = \frac{12}{R} \times R = 12\,^{\circ}C$.

(B) Let the length of each rod be $l$,area of cross-section be $A$,and thermal conductivity be $K$. The heat transferred is $Q$ and the temperature difference is $\Delta \theta$.
Case $1$: When joined end to end (series),the total length is $2l$ and the area is $A$. The equivalent thermal resistance is $R_{eq} = R_1 + R_2 = \frac{l}{KA} + \frac{l}{KA} = \frac{2l}{KA}$.
The heat flow is $Q = \frac{\Delta \theta}{R_{eq}} \times t_1 = \frac{\Delta \theta \cdot KA}{2l} \times 12 = \frac{6 KA \Delta \theta}{l} \dots (i)$.
Case $2$: When joined lengthwise (parallel),the length is $l$ and the total area is $2A$. The equivalent thermal resistance is $\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{KA}{l} + \frac{KA}{l} = \frac{2KA}{l}$,so $R_{eq} = \frac{l}{2KA}$.
The heat flow is $Q = \frac{\Delta \theta}{R_{eq}} \times t_2 = \frac{\Delta \theta \cdot 2KA}{l} \times t_2 \dots (ii)$.
Equating $(i)$ and $(ii)$:
$\frac{6 KA \Delta \theta}{l} = \frac{2 KA \Delta \theta}{l} \times t_2$
$6 = 2 t_2 \implies t_2 = 3 \, s$.

(B) The given circuit is a Wheatstone bridge. Since the bridge is balanced (the ratio of resistances in the arms is equal),no heat flows through the central wire.
The circuit simplifies to two parallel branches:
$1$. The upper branch consists of two copper wires in series,each of length $l$. The total resistance is $R_1 = \frac{l}{k_1 A} + \frac{l}{k_1 A} = \frac{2l}{k_1 A}$.
$2$. The lower branch consists of two steel wires in series,each of length $l$. The total resistance is $R_2 = \frac{l}{k_2 A} + \frac{l}{k_2 A} = \frac{2l}{k_2 A}$.
Since these two branches are in parallel,the equivalent thermal resistance $R_{eq}$ is given by:
$\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{k_1 A}{2l} + \frac{k_2 A}{2l} = \frac{(k_1 + k_2)A}{2l}$.
Therefore,$R_{eq} = \frac{2l}{(k_1 + k_2)A}$.

(B) The two plates are connected in parallel,so the thermal resistances $R_1$ and $R_2$ are in parallel.
For parallel combination,the equivalent thermal resistance $R_{eq}$ is given by:
$\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}$
We know that thermal resistance $R = \frac{\ell}{KA}$,where $\ell$ is the thickness.
Substituting the values:
$\frac{K_{eq}(A_1 + A_2)}{\ell} = \frac{K_1 A_1}{\ell} + \frac{K_2 A_2}{\ell}$
Since the thickness $\ell$ is the same for both plates,it cancels out:
$K_{eq}(A_1 + A_2) = K_1 A_1 + K_2 A_2$
$K_{eq} = \frac{K_1 A_1 + K_2 A_2}{A_1 + A_2}$

(C) For rods joined in series,the equivalent thermal resistance $R_{eq}$ is the sum of individual thermal resistances: $R_{eq} = R_1 + R_2$.
The thermal resistance of a rod is given by $R = \frac{\ell}{KA}$,where $\ell$ is the length,$K$ is the thermal conductivity,and $A$ is the cross-sectional area.
Substituting this into the series combination formula:
$\frac{\ell_1 + \ell_2}{KA} = \frac{\ell_1}{K_1 A} + \frac{\ell_2}{K_2 A}$
Since the cross-sectional area $A$ is the same for both rods,we can cancel $A$ from both sides:
$\frac{\ell_1 + \ell_2}{K} = \frac{\ell_1}{K_1} + \frac{\ell_2}{K_2}$

(B) Let $R_1$ and $R_2$ be the thermal resistances of the two rods. The rate of heat flow through the rods in series is given by:
$\frac{100 - 0}{R_1 + R_2} = \frac{100 - 70}{R_1} = \frac{70 - 0}{R_2}$
From this,we get:
$\frac{100}{R_1 + R_2} = \frac{30}{R_1} = \frac{70}{R_2}$
From $\frac{30}{R_1} = \frac{70}{R_2}$,we find $R_2 = \frac{7}{3} R_1$.
Substituting this into $\frac{100}{R_1 + R_2} = \frac{30}{R_1}$:
$\frac{100}{R_1 + \frac{7}{3}R_1} = \frac{30}{R_1} \Rightarrow \frac{100}{\frac{10}{3}R_1} = \frac{30}{R_1} \Rightarrow 30 = 30$ (consistent).
When the rods are interchanged,the new junction temperature $T$ satisfies:
$\frac{100 - T}{R_2} = \frac{T - 0}{R_1}$
$\frac{100 - T}{R_2} = \frac{T}{R_1} \Rightarrow \frac{100 - T}{T} = \frac{R_2}{R_1} = \frac{7}{3}$
$3(100 - T) = 7T \Rightarrow 300 - 3T = 7T \Rightarrow 10T = 300 \Rightarrow T = 30\,^{\circ}C$.

(B) Let the thermal resistance of each rod be $R$.
In figure $(a)$,the rods are in series,so the equivalent thermal resistance is $R_{eq1} = R + R = 2R$.
The rate of heat flow is $\frac{Q_1}{t_1} = \frac{\Delta T}{R_{eq1}} = \frac{100 - 0}{2R} = \frac{50}{R}$.
Given $Q_1 = 10 \, cal$ and $t_1 = 2 \, min$,we have $\frac{10}{2} = \frac{50}{R}$,which gives $R = 10 \, K/cal \cdot min$.
In figure $(b)$,the rods are in parallel,so the equivalent thermal resistance is $R_{eq2} = \frac{R \cdot R}{R + R} = \frac{R}{2}$.
The rate of heat flow is $\frac{Q_2}{t_2} = \frac{\Delta T}{R_{eq2}} = \frac{100 - 0}{R/2} = \frac{200}{R}$.
For the same amount of heat $Q_2 = 10 \, cal$,we have $\frac{10}{t_2} = \frac{200}{R}$.
Substituting $R = 10$,we get $\frac{10}{t_2} = \frac{200}{10} = 20$.
Thus,$t_2 = \frac{10}{20} = 0.5 \, min$.

(D) Let the length of each rod be $l$,area of cross-section be $A$,and thermal conductivity be $k$. The rate of heat flow is given by $H = \frac{kA \Delta T}{l}$.
Case $1$: When joined in parallel,the effective area is $2A$ and effective length is $l$. The rate of heat flow is $H_p = \frac{k(2A) \Delta T}{l} = \frac{2kA \Delta T}{l}$.
Given $H_p = \frac{Q}{12}$,so $\frac{Q}{12} = \frac{2kA \Delta T}{l} \implies Q = \frac{24kA \Delta T}{l}$.
Case $2$: When joined in series,the effective length is $2l$ and effective area is $A$. The rate of heat flow is $H_s = \frac{kA \Delta T}{2l}$.
Given $H_s = \frac{Q}{t}$,so $\frac{Q}{t} = \frac{kA \Delta T}{2l} \implies t = \frac{Q \cdot 2l}{kA \Delta T}$.
Substituting $Q$ from Case $1$ into Case $2$:
$t = \frac{(24kA \Delta T / l) \cdot 2l}{kA \Delta T} = 24 \times 2 = 48 \, s$.

(C) For two plates of equal thickness $d$ and thermal conductivities $K_1$ and $K_2$ placed in series,the total thermal resistance is $R_{eq} = R_1 + R_2$.
Since $R = \frac{d}{KA}$,we have $\frac{2d}{K_{eq}A} = \frac{d}{K_1A} + \frac{d}{K_2A}$.
Simplifying this,we get $\frac{2}{K_{eq}} = \frac{1}{K_1} + \frac{1}{K_2}$,which implies $K_{eq} = \frac{2K_1K_2}{K_1 + K_2}$.
This is the harmonic mean of $K_1$ and $K_2$. The harmonic mean of two numbers is always less than the larger number and greater than the smaller number,but specifically,the equivalent conductivity $K_{eq}$ satisfies $K_{min} < K_{eq} < K_{max}$.
Wait,let's re-evaluate: If $K_1 = 10$ and $K_2 = 2$,then $K_{eq} = \frac{2(10)(2)}{10+2} = \frac{40}{12} \approx 3.33$.
Here $3.33 > 2$ (the smaller value). Thus,the Assertion is incorrect.

(N/A) Given: $L_{1}=L_{2}=L=0.1 \; m$,$A_{1}=A_{2}=A=0.02 \; m^{2}$,$K_{1}=79 \; W m^{-1} K^{-1}$,$K_{2}=109 \; W m^{-1} K^{-1}$,$T_{1}=373 \; K$,$T_{2}=273 \; K$.
Under steady-state conditions,the heat current $(H_{1})$ through the iron bar is equal to the heat current $(H_{2})$ through the brass bar.
$H = H_{1} = H_{2} = \frac{K_{1} A (T_{1}-T_{0})}{L} = \frac{K_{2} A (T_{0}-T_{2})}{L}$
$(i)$ Junction temperature $T_{0}$:
$K_{1}(T_{1}-T_{0}) = K_{2}(T_{0}-T_{2})$
$T_{0} = \frac{K_{1} T_{1} + K_{2} T_{2}}{K_{1} + K_{2}} = \frac{(79 \times 373) + (109 \times 273)}{79 + 109} = \frac{29467 + 29757}{188} = \frac{59224}{188} \approx 315 \; K$.
$(ii)$ Equivalent thermal conductivity $K'$:
For series combination,$\frac{2L}{K'} = \frac{L}{K_{1}} + \frac{L}{K_{2}} \implies \frac{2}{K'} = \frac{1}{K_{1}} + \frac{1}{K_{2}} = \frac{K_{1}+K_{2}}{K_{1} K_{2}}$
$K' = \frac{2 K_{1} K_{2}}{K_{1}+K_{2}} = \frac{2 \times 79 \times 109}{79 + 109} = \frac{17222}{188} \approx 91.6 \; W m^{-1} K^{-1}$.
$(iii)$ Heat current $H$:
$H = \frac{K' A (T_{1}-T_{2})}{2L} = \frac{91.6 \times 0.02 \times (373 - 273)}{2 \times 0.1} = \frac{91.6 \times 0.02 \times 100}{0.2} = \frac{183.2}{0.2} = 916 \; W$.

(N/A) The thermal conductivity of concrete,although much smaller than that of metals,is still not low enough to provide effective thermal insulation. By adding a layer of sand or foam,which are poor conductors of heat (insulators),the overall thermal resistance of the ceiling increases. This significantly reduces the rate of heat transfer from the outside environment into the room,thereby keeping the room cooler during hot weather.

(A) For two wires connected in series,the thermal resistance $R$ is given by $R = \frac{l}{KA}$,where $l$ is the length,$K$ is the thermal conductivity,and $A$ is the cross-sectional area.
Since the wires are identical,they have the same length $l$ and cross-sectional area $A$.
The total thermal resistance $R_{eff}$ of the series combination is the sum of individual resistances:
$R_{eff} = R_{1} + R_{2} = \frac{l}{K_{1}A} + \frac{l}{K_{2}A}$
For the equivalent wire of length $2l$ and thermal conductivity $K_{eq}$,the resistance is:
$R_{eff} = \frac{2l}{K_{eq}A}$
Equating the two expressions for $R_{eff}$:
$\frac{2l}{K_{eq}A} = \frac{l}{K_{1}A} + \frac{l}{K_{2}A}$
$\frac{2}{K_{eq}} = \frac{1}{K_{1}} + \frac{1}{K_{2}} = \frac{K_{1} + K_{2}}{K_{1}K_{2}}$
$K_{eq} = \frac{2K_{1}K_{2}}{K_{1} + K_{2}}$

(C) Since the two insulating sheets are in series,the rate of heat flow $(H)$ through both sheets must be the same in the steady state.
The rate of heat flow is given by $H = \frac{\Delta \theta}{R}$.
For the first sheet (bottom),the heat flow is $H = \frac{\theta - \theta_{1}}{R_{1}}$.
For the second sheet (top),the heat flow is $H = \frac{\theta_{2} - \theta}{R_{2}}$.
Equating the two rates:
$\frac{\theta - \theta_{1}}{R_{1}} = \frac{\theta_{2} - \theta}{R_{2}}$
$R_{2}(\theta - \theta_{1}) = R_{1}(\theta_{2} - \theta)$
$R_{2}\theta - R_{2}\theta_{1} = R_{1}\theta_{2} - R_{1}\theta$
$R_{2}\theta + R_{1}\theta = R_{1}\theta_{2} + R_{2}\theta_{1}$
$\theta(R_{1} + R_{2}) = R_{1}\theta_{2} + R_{2}\theta_{1}$
$\theta = \frac{R_{1}\theta_{2} + R_{2}\theta_{1}}{R_{1} + R_{2}}$
Thus,the correct option is $C$.

(A) Since the rods are identical,the thermal resistance of each rod is $R_{AB} = R_{CD} = 10 \; K/W$.
$C$ is the midpoint of $AB$,so the thermal resistance of each half is $R_{AC} = R_{CB} = 5 \; K/W$.
Let the temperature at the junction $C$ be $T$.
Applying the junction rule for heat current (sum of heat currents entering = sum of heat currents leaving):
$\frac{200 - T}{5} = \frac{T - 125}{10} + \frac{T - 100}{5}$
Multiplying the entire equation by $10$:
$2(200 - T) = (T - 125) + 2(T - 100)$
$400 - 2T = T - 125 + 2T - 200$
$400 - 2T = 3T - 325$
$5T = 725$
$T = 145^{\circ}C$
The heat current $P$ in rod $CD$ is given by:
$P = \frac{T - 125}{R_{CD}} = \frac{145 - 125}{10} = \frac{20}{10} = 2 \; W$.
Thus,the value of $P$ is $2$.

(B) The rate of heat flow is given by $\frac{\Delta Q}{\Delta t} = \left(\frac{1}{R}\right) \Delta T$,where $R$ is the thermal resistance.
For a plate of thickness $L$,thermal conductivity $K$,and area $A$,the thermal resistance is $R = \frac{L}{KA}$.
For plate $A$: $R_{1} = \frac{L_{1}}{K_{1}A} = \frac{4.0}{K(120)}$.
For plate $B$: $R_{2} = \frac{L_{2}}{K_{2}A} = \frac{2.5}{(2K)(120)}$.
Since the plates are in series,the equivalent thermal resistance is $R_{\text{eq}} = R_{1} + R_{2}$.
The equivalent thermal conductivity $K_{\text{eq}}$ for a compound plate of total thickness $L_{\text{eq}} = L_{1} + L_{2}$ and area $A$ is given by $\frac{L_{\text{eq}}}{K_{\text{eq}}A} = \frac{L_{1}}{K_{1}A} + \frac{L_{2}}{K_{2}A}$.
Substituting the values: $\frac{4.0 + 2.5}{K_{\text{eq}}(120)} = \frac{4.0}{K(120)} + \frac{2.5}{2K(120)}$.
$\frac{6.5}{K_{\text{eq}}} = \frac{4}{K} + \frac{1.25}{K} = \frac{5.25}{K} = \frac{21/4}{K} = \frac{21}{4K}$.
Therefore,$K_{\text{eq}} = \frac{6.5 \times 4K}{21} = \frac{26}{21}K = \left(1 + \frac{5}{21}\right)K$.
Comparing this with $\left(1 + \frac{5}{\alpha}\right)K$,we get $\alpha = 21$.

(D) Let the junction temperature be $T^{\circ} C$.
Since the rods are made of the same material,have the same cross-sectional area $A$,and the same length $l$,their thermal resistance $R = \frac{l}{kA}$ is the same for all rods.
According to the principle of steady-state heat flow,the total heat inflow through the three forked rods must equal the heat outflow through the fourth rod (the handle).
Heat inflow from three rods = $3 \times \frac{kA}{l}(100 - T)$
Heat outflow through the fourth rod = $\frac{kA}{l}(T - 0)$
Equating the two:
$3 \frac{kA}{l}(100 - T) = \frac{kA}{l}(T - 0)$
$3(100 - T) = T$
$300 - 3T = T$
$4T = 300$
$T = 75^{\circ} C$