Q.An iron bar ($L_1 = 0.1\ \text{m}$, $A_1 = 0.02\ \text{m}^{2}$, $K_1 = 79\ \text{W m}^{-1}\ \text{K}^{-1}$) and a brass bar ($L_2 = 0.1\ \text{m}$, $A_2 = 0.02\ \text{m}^{2}$, $K_2 = 109\ \text{W m}^{-1}\ \text{K}^{-1}$) are soldered end to end as shown in Fig. 10.16. The free ends of the iron bar and brass bar are maintained at $373\ \text{K}$ and $273\ \text{K}$ respectively. Obtain expressions for and hence compute

Showing the 12 most recent of 42 on this concept.

1. TG EAPCET 2026Set ap-2026-05-04-AN1 markMCQ
   Q.Two rods A and B are made of different materials having thermal conductivities in the ratio 2:3. The lengths of the rods A and B are in the ratio 2:1 and their volumes are in the ratio 1:2. In the steady state, if the temperature differences across the ends of the rods A and B are respectively $60\ ^\circ \mathrm{C}$ and $\Delta \theta$, the ratio of rates of flow of heat through the rods A and B is 1:16, then $\Delta \theta =$ (A) $50\ ^\circ \mathrm{C}$ (B) $120\ ^\circ \mathrm{C}$ (C) $90\ ^\circ \mathrm{C}$ (D) $80\ ^\circ \mathrm{C}$

   ›Reveal solutionSolution

   The key is to use the steady-state heat conduction formula $Q/t = kA(\Delta T)/L$, express the area ratio from the given volume and length ratios, then equate the given heat current ratio to find $\Delta\theta$. The answer is $80\ ^\circ\mathrm{C}$.

   The problem is about steady-state heat conduction through rods. The rate of heat flow (heat current) through a rod in steady state is given by Fourier's law:

   $H = \frac{Q}{t} = \frac{kA(\Delta T)}{L}$

   where $k$ is thermal conductivity, $A$ is cross-sectional area, $L$ is length, and $\Delta T$ is the temperature difference across the ends.

   We are given ratios for $k$, $L$, and volume $V$, and we need to find $\Delta\theta$ such that the ratio of heat currents $H_A : H_B = 1 : 16$.

   Let’s work through it step by step.

   1. Write the given ratios clearly.

      Let $k_A : k_B = 2 : 3$, so $k_A = 2x$, $k_B = 3x$ for some $x$.

      Let $L_A : L_B = 2 : 1$, so $L_A = 2y$, $L_B = y$.

      Let $V_A : V_B = 1 : 2$, so $V_A = V$, $V_B = 2V$.

   2. Find the area ratio from volume and length.

      Volume = area $\times$ length, so $A = V/L$.

      Therefore:

   $A_A = \frac{V_A}{L_A} = \frac{V}{2y}, \quad A_B = \frac{V_B}{L_B} = \frac{2V}{y}$

   The ratio of areas is:

   $\frac{A_A}{A_B} = \frac{V/(2y)}{2V/y} = \frac{1}{4}$

   3. Write the heat current ratio using Fourier’s law.

   $H_A = \frac{k_A A_A (\Delta T_A)}{L_A}, \quad H_B = \frac{k_B A_B (\Delta T_B)}{L_B}$

   Given $\Delta T_A = 60\ ^\circ\mathrm{C}$ and $\Delta T_B = \Delta\theta$, and $H_A : H_B = 1 : 16$, so:

   $\frac{H_A}{H_B} = \frac{1}{16}$

   4. Substitute the ratios.

   $\frac{H_A}{H_B} = \frac{k_A}{k_B} \cdot \frac{A_A}{A_B} \cdot \frac{L_B}{L_A} \cdot \frac{\Delta T_A}{\Delta T_B}$

   Plug in:

   $\frac{1}{16} = \frac{2}{3} \cdot \frac{1}{4} \cdot \frac{1}{2} \cdot \frac{60}{\Delta\theta}$ …

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2. TG EAPCET 2026Set ap-2026-05-05-FN1 markMCQ
   Q.Two rods A and B of equal dimensions having thermal conductivities $K$ and $2K$ respectively are joined in series. Under steady state conditions if the temperature difference between the ends of the rod A is $12\,^\circ\mathrm{C}$, then the temperature difference between the open ends of the rods A and B is (A) $36\,^\circ\mathrm{C}$ (B) $6\,^\circ\mathrm{C}$ (C) $18\,^\circ\mathrm{C}$ (D) $24\,^\circ\mathrm{C}$

   ›Reveal solutionSolution

   In series, the heat current is the same through both rods. Using $Q/t = KA(\Delta T)/L$ for each rod and equating the two currents gives the temperature drop across rod B, and the total temperature difference between the open ends is the sum of the two drops — $18\,^\circ\mathrm{C}$.

   The key idea is that when two rods are joined in series and the system is in steady state, the rate of heat flow (heat current) through each rod must be identical — there is no accumulation of heat anywhere, so what flows into rod A per second must flow out of rod B per second.

   For a rod of length $L$, cross-sectional area $A$, and thermal conductivity $K$, the heat current $H$ is given by Fourier's law:

   $H = \frac{Q}{t} = \frac{K A (\Delta T)}{L}$

   where $\Delta T$ is the temperature difference between its ends.

   Here both rods have equal dimensions — same $L$ and same $A$. Rod A has conductivity $K$, rod B has conductivity $2K$. The temperature difference across rod A alone is given as $\Delta T_A = 12\,^\circ\mathrm{C}$.

   1. Heat current through rod A:

   $H = \frac{K A (12)}{L}$

   2. Heat current through rod B: let $\Delta T_B$ be the temperature difference across rod B:

   $H = \frac{(2K) A (\Delta T_B)}{L}$

   3. Equate the currents (steady state, series connection):

   $\frac{K A (12)}{L} = \frac{2K A (\Delta T_B)}{L}$

   Cancelling $K$, $A$, and $L$ (all non-zero):

   $12 = 2 \Delta T_B \quad\Rightarrow\quad \Delta T_B = 6\,^\circ\mathrm{C}$

   4. Total temperature difference between the open ends: since the rods are in series, the total drop from the free end of A to the free end of B is the sum of the drops across each rod: $\Delta T_{\text{total}} = \Delta T_A + \Delta T_B = 12 + 6 = 18\,^\circ\mathrm{C}$ …

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3. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
   Q.Three rods made of the same material and having the same cross-section have been joined as shown in figure (diagram: three rods meet at a single junction forming a Y-shape; the free end of the single rod on the left is at $0^\circ C$, and the free ends of the other two rods on the right are both at $90^\circ C$). Each rod is of same length. The left and right ends of the arrangement are kept at $0^\circ C$ and $90^\circ C$ respectively. The temperature of the junction of the three rods will be (A) $45^\circ C$ (B) $60^\circ C$ (C) $30^\circ C$ (D) $20^\circ C$

   ›Reveal solutionSolution

   This tests steady-state heat current balance at a junction of identical rods (thermal analogue of Kirchhoff's junction rule). The answer is $60^\circ C$.

   Concept and Intuition

   In steady state, the net heat current flowing into a junction must equal the net heat current flowing out (no heat accumulates at the junction). Since all three rods are identical (same material, length, cross-section), they all have the same thermal resistance $R = \dfrac{L}{kA}$. Two rods at $90^\circ C$ feed heat into the junction; one rod at $0^\circ C$ drains heat away. Setting the total heat current in equal to that out gives the junction temperature.

   Step-by-Step Solution

   1. Let the junction temperature be $T$. Heat current through each of the two 90°C rods (into the junction): $I = \dfrac{90-T}{R}$ each.
   2. Heat current through the single left rod (out of the junction, to the 0°C end): $I' = \dfrac{T-0}{R}$.
   3. Steady-state balance: total current in = current out: $\dfrac{90-T}{R}+\dfrac{90-T}{R} = \dfrac{T}{R}$.
   4. Since $R$ is common to all rods (identical), it cancels: $2(90-T) = T$. …

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4. AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
   Q.Two rods A and B of lengths in ratio of 1 : 2 have thermal conductivities in 1 : 2 ratio and cross sectional areas in 1 : 4 ratio. The temperature difference between the ends of two rods is same. Then the ratio of heat currents is ($H_A : H_B$) (A) 1 : 4 (B) 4 : 1 (C) 2 : 1 (D) 1 : 2

   ›Reveal solutionSolution

   This tests the steady-state conduction formula $H = kA\Delta T/L$ and combining three independent ratios (length, conductivity, area) correctly.

   Concept and Intuition

   In steady-state heat conduction through a rod, the rate of heat flow (heat current) is

   $H = \frac{kA\Delta T}{L},$

   where $k$ is thermal conductivity, $A$ is cross-sectional area, $L$ is length, and $\Delta T$ is the temperature difference across the rod's ends. Since $H$ depends on $k$ and $A$ directly but on $L$ inversely, when comparing two rods you must multiply the conductivity ratio and area ratio, and divide by the length ratio (i.e. multiply by the inverted length ratio).

   Step-by-Step Solution

   1. Given: $L_A:L_B = 1:2$, $k_A:k_B = 1:2$, $A_A:A_B = 1:4$, and $\Delta T$ is the same for both rods.
   2. Write the ratio of heat currents: $\frac{H_A}{H_B} = \frac{k_A A_A \Delta T / L_A}{k_B A_B \Delta T/L_B} = \left(\frac{k_A}{k_B}\right)\left(\frac{A_A}{A_B}\right)\left(\frac{L_B}{L_A}\right).$ …

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5. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
   Q.Two semicircular rods AB and CD each of radius of curvature 14 cm and a straight rod BC of length 22 cm are connected in series. The three rods have equal area of cross-section and the thermal conductivities of the materials of the rods AB, BC and CD are in the ratio 1:2:3. In steady state, if the temperature difference between the ends of the middle rod BC is $30^{0}C$, then the temperature difference between the ends of the rods AB and CD are respectively (A) $120^{0}C, 40^{0}C$ (B) $60^{0}C, 20^{0}C$ (C) $120^{0}C, 60^{0}C$ (D) $60^{0}C, 40^{0}C$

   ›Reveal solutionSolution

   Series conduction means the same heat current flows through all three rods; since $\Delta T \propto L/k$ at fixed $Q$, the longer semicircular rods develop bigger temperature drops than the short straight rod, in proportion to their length-to-conductivity ratio.

   Concept and Intuition

   In a series thermal circuit (just like series resistors carrying the same electric current), the heat current $Q$ through every rod is identical in steady state — heat can't pile up anywhere. Each rod obeys $Q = \dfrac{kA\Delta T}{L}$, i.e. it behaves like a "thermal resistor" $R_{th}=\dfrac{L}{kA}$, and $\Delta T = Q R_{th}$. So knowing $Q$ is common, the temperature drop across each rod is proportional to its thermal resistance $L/k$ (area $A$ is common to all three).

   Step-by-Step Solution

   1. Semicircular rod length: arc length $= \pi r = \dfrac{22}{7}\times 14\text{ cm} = 44$ cm. So $L_{AB}=L_{CD}=44$ cm, $L_{BC}=22$ cm.
   2. Conductivities are in ratio $k_{AB}:k_{BC}:k_{CD}=1:2:3$.
   3. Same $Q$ through all three: $Q = \dfrac{k_{AB}A\Delta T_{AB}}{L_{AB}} = \dfrac{k_{BC}A\Delta T_{BC}}{L_{BC}} = \dfrac{k_{CD}A\Delta T_{CD}}{L_{CD}}$.
   4. Take $k_{AB}=1,\ k_{BC}=2,\ k_{CD}=3$ (units of $k$). From the BC rod, $Q/A = \dfrac{2\times30}{22}=\dfrac{60}{22}$. …

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6. AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
   Q.Two metal rods A and B have lengths in the ratio 1:2, thermal conductivities in the ratio 1:2 and cross sectional areas in the ratio 1:4. The ends of the two rods are maintained between same temperature difference, then the ratio of heat currents $\left(\dfrac{H_A}{H_B}\right)$ is (A) 1 : 4 (B) 4 : 1 (C) 2 : 1 (D) 1 : 2

   ›Reveal solutionSolution

   This tests the heat conduction formula $H=kA\Delta T/L$ applied as a ratio problem. Answer: $H_A:H_B = 1:4$.

   Concept and Intuition

   The rate of heat conduction (heat current) through a rod depends directly on its thermal conductivity and cross-sectional area, and inversely on its length, for a given temperature difference across its ends. Comparing two rods just requires multiplying/dividing the given ratios correctly.

   Step-by-Step Solution

   1. Heat current formula: $H = \dfrac{kA\Delta T}{L}$.
   2. Given ratios: $\dfrac{L_A}{L_B}=\dfrac{1}{2}$, $\dfrac{k_A}{k_B}=\dfrac{1}{2}$, $\dfrac{A_A}{A_B}=\dfrac{1}{4}$, and $\Delta T$ is the same for both.
   3. $\dfrac{H_A}{H_B} = \dfrac{k_A}{k_B}\times\dfrac{A_A}{A_B}\times\dfrac{L_B}{L_A}$. …

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7. AP EAPCET 2026Set ap-2026-05-19-AN1 markMCQ
   Q.Three rods A, B and C of same cross-sectional area having lengths 20 cm, 30 cm and 25 cm respectively are connected in series. If the thermal conductivities of the materials of the rods A, B and C are respectively 400 $\mathrm{Wm^{-1}K^{-1}}$, 600 $\mathrm{Wm^{-1}K^{-1}}$ and 375 $\mathrm{Wm^{-1}K^{-1}}$, then the equivalent thermal conductivity of the combination of the three rods is (A) 450 $\mathrm{Wm^{-1}K^{-1}}$ (B) 500 $\mathrm{Wm^{-1}K^{-1}}$ (C) 550 $\mathrm{Wm^{-1}K^{-1}}$ (D) 475 $\mathrm{Wm^{-1}K^{-1}}$

   ›Reveal solutionSolution

   Treating the three rods as thermal resistances in series and summing $L/K$ for each gives an equivalent conductivity of 450 W/m/K.

   Concept and Intuition

   Just as electrical resistances in series add directly, thermal resistances of rods conducting heat one after another (same cross-sectional area, same heat current through each) also add. Thermal resistance of a rod is $R = \dfrac{L}{KA}$; with a common area $A$, we can drop it from the ratio and work purely with $L/K$ terms.

   Step-by-Step Solution

   1. Individual thermal resistances (dropping the common area $A$): $R_A = L_A/K_A = 20/400 = 0.05$; $R_B = L_B/K_B = 30/600=0.05$; $R_C=L_C/K_C=25/375=\dfrac1{15}\approx0.0667$.
   2. Total resistance: $R_{total} = 0.05+0.05+0.0667 = \dfrac16$ (exactly, since $1/20+1/20+1/15 = 3/60+3/60+4/60=10/60=1/6$).
   3. Total length: $L_{total} = 20+30+25=75\ \text{cm}$. …

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8. AP EAPCET 2026Set ap-2026-05-20-FN1 markMCQ
   Q.A, B and C are the three identical conductors made of different materials. They are kept in contact as shown in the figure. Their thermal conductivities are k, 2k and k/2 respectively. The free end of A is at $100^0C$ and free end of C is at $0^0C$. During the steady state, the temperature of the junction between A and B is nearly [FIGURE] (three conducting rods A, B, C joined end-to-end in a horizontal row inside a rectangular outline; the left end of A is held at 100°C and the right end of C is held at 0°C) (A) $37^0C$ (B) $71^0C$ (C) $29^0C$ (D) $63^0C$

   ›Reveal solutionSolution

   Model the three identical-geometry rods as series thermal resistances $\propto1/k$; solving for the steady-state heat current gives the A–B junction temperature as about $71^\circ$C.

   Concept and Intuition

   For rods in series carrying the same steady heat current, each acts like a thermal resistor $R=\dfrac{L}{kA}$. Since all three rods are identical in length and cross-section, their resistances differ only through $k$: $R\propto1/k$. The same current flows through all three (series), and the temperature drop across each is proportional to its resistance.

   Step-by-Step Solution

   1. Let unit resistance $R_0=\dfrac{L}{kA}$ (rod A's resistance, since A has conductivity $k$).
   2. Rod B (conductivity $2k$): $R_B = \dfrac{L}{2kA}=\dfrac{R_0}{2}$.
   3. Rod C (conductivity $k/2$): $R_C = \dfrac{L}{(k/2)A}=2R_0$.
   4. Total resistance: $R_{tot}=R_0+\dfrac{R_0}{2}+2R_0=3.5R_0$.
   5. Total temperature difference $=100^\circ C$, so heat current $I=\dfrac{100}{3.5R_0}$. …

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9. MHT-CET 2026Set pcm-2026-04-11-M1 markMCQ
   Q.Three identical heat conducting rods are connected in series as shown in the figure. The rods on the side have thermal conductivity 2𝐾 while that in the middle has thermal conductivity 𝐾. The left end of the combination is maintained at temperature 3𝑇 and the right end at 𝑇. The rods are thermally insulated from outside. In steady state, temperature at the left junction is 𝑇1 and that at the right junctions is 𝑇2. The ratio 𝑇1/𝑇2 is (A) 32 (B) 43 (C) 53 (D) 45

   ›Reveal solutionSolution

   Resistances are in ratio 1:2:1, total drop $2T$ splits as $T/2$, $T$, $T/2$, giving $T_1=5T/2$, $T_2=3T/2$ and $T_1/T_2=5/3$ (option C).

   Each rod has the same length $L$ and area $A$. Thermal resistance $R=\frac{L}{kA}$:

   • Outer rods ($2K$): $R_{side}=\dfrac{L}{2KA}\equiv r$
   • Middle rod ($K$): $R_{mid}=\dfrac{L}{KA}=2r$

   Total resistance $=r+2r+r=4r$. The overall temperature difference is $3T-T=2T$, so the steady-state heat current is:

   $Q=\frac{2T}{4r}=\frac{T}{2r}$

   Temperature at the left junction (after the first outer rod): …

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10. MHT-CET 2026Set pcm-2026-04-18-M1 markMCQ
    Q.The ratio of thermal conductivity of two rods of different materials is 4:3. The two rods have same area of cross-section and same thermal resistance. The lengths of rods are in the ratio (A) 5:4 (B) 4:3 (C) 1:5 (D) 7:1

    ›Reveal solutionSolution

    Equal $A$ and equal $R$ force $L \propto k$, so the length ratio equals the conductivity ratio.

    Thermal resistance of a rod is $R = \dfrac{L}{kA}$. Given $R_1=R_2$ and $A_1=A_2$: …

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11. MHT-CET 2026Set pcm-2026-04-20-M1 markMCQ
    Q.Two rods of same length, radius and material transfer a given amount of heat in 't' second when they are joined as shown in fig (1). But when they are joined as shown in fig

    (2) then they will transfer same heat in same condition in time (A) 2𝑡 (B) 3𝑡 (C) 4𝑡 (D) 6𝑡

    ›Reveal solutionSolution

    Series thermal resistance is $4\times$ the parallel resistance for two identical rods; time to conduct the same heat scales with resistance, so $t\to 4t$.

    Let each rod have length $L$, cross-section $A$, thermal conductivity $k$, across the same temperature difference $\Delta T$. Thermal resistance is $R_{th}=\dfrac{\text{length}}{kA}$, and the rate of heat flow is $H=\dfrac{\Delta T}{R_{th}}$; for a fixed amount of heat $Q=H\,t$, the time $t=\dfrac{Q}{H}=\dfrac{Q\,R_{th}}{\Delta T}\propto R_{th}$.

    Parallel arrangement (rods side by side) - fig (1), time $t$: effective length $L$, area $2A$:

    $R_{\parallel}=\frac{L}{k(2A)}=\frac{L}{2kA}.$

    Series arrangement (rods end to end) - fig (2): effective length $2L$, area $A$:

    $R_{s}=\frac{2L}{kA}.$ …

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12. AP EAPCET 2025Set eng-2025-05-27-FN1 markMCQ
    Q.A rectangular slab consists of two cubes of copper and brass of equal sides having thermal conductivities in the ratio 4 : 1. If the free face of brass is at $0\ ^\circ\mathrm{C}$ and that of copper is at $100\ ^\circ\mathrm{C}$, then the temperature of their interface is (A) $80\ ^\circ\mathrm{C}$ (B) $20\ ^\circ\mathrm{C}$ (C) $60\ ^\circ\mathrm{C}$ (D) $40\ ^\circ\mathrm{C}$

    ›Reveal solutionSolution

    Equating steady-state heat currents through the two conductors in series (copper 4× more conductive than brass) gives an interface temperature of 80°C.

    Concept and Intuition

    In steady state, the same rate of heat must flow through both cubes one after another (series conduction) — none is stored at the interface. A better conductor (copper) needs a smaller temperature drop across it to carry the same heat current, so the interface temperature ends up much closer to the copper's own face temperature (100°C) than to brass's (0°C).

    Step-by-Step Solution

    1. Let interface temperature be $\theta$. Both cubes have equal cross-section $A$ and equal length $L$ (same-sized cubes).
    2. Heat current through copper: $H_{Cu} = \dfrac{k_{Cu}A(100-\theta)}{L}$.
    3. Heat current through brass: $H_{brass} = \dfrac{k_{brass}A(\theta-0)}{L}$.
    4. Steady state ⇒ $H_{Cu}=H_{brass}$: $k_{Cu}(100-\theta) = k_{brass}\,\theta$. …

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