Q.These days people use steel utensils with copper bottom. This is supposed to be good for uniform heating of food. Explain this effect using the fact that copper is the better conductor.

Showing the 12 most recent of 13 on this concept.

1. 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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2. 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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3. 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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4. 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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5. 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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6. 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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7. 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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8. AP EAPCET 2024Set eng-2024-05-18-FN1 markMCQ
   Q.The length of a metal bar is 20 cm and the area of cross section is $4 \times 10^{-4}\,m^2$. If one end of the rod is kept in ice at 0°C and the other end is kept in steam at 100°C, the mass of ice melted in one minute is 5 g. The thermal conductivity of the metal in $Wm^{-1}K^{-1}$ is (Latent heat of fusion = 80 cal/gm) (A) 140 (B) 120 (C) 100 (D) 160

   ›Reveal solutionSolution

   Equate the heat conducted through the bar in steady state to the latent heat needed to melt the measured mass of ice, then solve for the thermal conductivity $k$, being careful to convert calories to joules.

   Concept and Intuition

   In steady-state 1-D conduction, the rate of heat flow through a bar is $\dfrac{dQ}{dt}=\dfrac{kA\Delta T}{L}$. Over a time $t$, the total heat conducted is $Q=\dfrac{kA\Delta T\,t}{L}$. This heat, entering at the ice end, goes entirely into melting ice (a phase change, with no temperature rise), so $Q=mL_f$ where $L_f$ is the latent heat of fusion.

   Step-by-Step Solution

   1. Data: $L=20\ \text{cm}=0.2\ \text m$, $A=4\times10^{-4}\ \text{m}^2$, $\Delta T=100-0=100\ ^\circ\text C$, $t=1\ \text{min}=60\ \text s$, $m=5\ \text g$, $L_f=80\ \text{cal/g}$.
   2. Heat needed to melt the ice: $Q=mL_f=5\times80=400\ \text{cal}$.
   3. Convert to joules ($1\ \text{cal}=4.2\ \text J$): $Q=400\times4.2=1680\ \text J$. …

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9. AP EAPCET 2024Set eng-2024-05-20-FN1 markMCQ
   Q.The temperature difference across two cylindrical rods A and B of same material and same mass are 40 °C and 60 °C respectively. In steady state, if the rates of flow of heat through the rods A and B are in the ratio 3:8, the ratio of the lengths of the rods A and B is (A) $1:3$ (B) $5:3$ (C) $4:3$ (D) $2:3$

   ›Reveal solutionSolution

   Combining the heat-conduction formula with the equal-mass (equal-volume, same material) constraint converts the area dependence into a length dependence, giving a length ratio of $4:3$.

   Concept and Intuition

   Heat conduction rate is $H = \dfrac{kA\Delta T}{L}$. Normally $A$ and $L$ are independent, but here the constraint of equal mass and same material (same density) fixes the product $A\times L$ = volume = constant for both rods. This lets us eliminate $A$ in favor of $L$, turning the problem into a pure ratio of lengths.

   Step-by-Step Solution

   1. Equal mass, same material $\Rightarrow$ equal volume: $A_AL_A = A_BL_B = V$ (constant), so $A = V/L$.
   2. Substitute into $H=\dfrac{kA\Delta T}{L}$: $H = \dfrac{k(V/L)\Delta T}{L} = \dfrac{kV\Delta T}{L^2}$, i.e. $H\propto \dfrac{\Delta T}{L^2}$.
   3. Ratio: $\dfrac{H_A}{H_B} = \dfrac{\Delta T_A/L_A^2}{\Delta T_B/L_B^2} = \dfrac{\Delta T_A}{\Delta T_B}\cdot\dfrac{L_B^2}{L_A^2} = \dfrac38$. …

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10. AP EAPCET 2024Set eng-2024-05-21-FN1 markMCQ
    Q.A slab consists of two identical plates of copper and brass. The free face of the brass is at 0 °C and that of copper at 100 °C. If the thermal conductivities of brass and copper are in the ratio 1:4, then the temperature of interface is (A) 20° C (B) 40° C (C) 60° C (D) 80° C

    ›Reveal solutionSolution

    Equal heat current through both identical-thickness plates in series gives interface temperature 80°C — closer to the hotter, better-conducting copper side.

    Concept and Intuition

    In steady-state conduction through plates joined in series, the same heat current $H$ flows through each plate (no heat is stored at the interface). For identical thickness and area, $H\propto k\Delta T$, so the better conductor needs a smaller temperature drop across it to carry the same current — meaning the interface temperature sits closer to the hot end on the good-conductor side.

    Step-by-Step Solution

    1. Let interface temperature be $\theta$. Brass: free face at $0^\circ$C, interface at $\theta$. Copper: interface at $\theta$, free face at $100^\circ$C.
    2. Same heat current through both (steady state, same area $A$, same thickness $L$): $\dfrac{k_{brass}A(\theta-0)}{L} = \dfrac{k_{copper}A(100-\theta)}{L}$.
    3. Cancel $A/L$: $k_{brass}\theta = k_{copper}(100-\theta)$. …

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11. AP EAPCET 2021Set eng-2021-08-19-FN1 markMCQ
    Q.In a steady state, the temperature at the end $A$ and end $B$ of a $20\ cm$ long rod $AB$ are $100^\circ C$ and $0^\circ C$. The temperature of a point, $9\ cm$ from $A$ is _________ (A) $55^\circ C$ (B) $45^\circ C$ (C) $65^\circ C$ (D) $50^\circ C$

    ›Reveal solutionSolution

    In steady state, a uniform rod's temperature gradient is constant, so temperature drops linearly along its length. Answer: $55^\circ\text{C}$.

    Concept and Intuition

    In steady-state heat conduction through a rod with no heat generation and uniform cross-section/material, the heat current $H = \dfrac{kA(T_A-T_B)}{L}$ must be the same through every cross-section (otherwise heat would pile up somewhere, contradicting steady state). This forces the temperature to fall at a constant rate per unit length — i.e., linearly from one end to the other.

    Step-by-Step Solution

    1. Rod length $L = 20\ \text{cm}$, $T_A = 100^\circ\text{C}$ at end A, $T_B = 0^\circ\text{C}$ at end B. …

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12. AP EAPCET 2021Set eng-2021-08-24-FN1 markMCQ
    Q.Two plates of same area are placed in contact. Their thickness as well as their thermal conductivities are in the ratio $2:3$. The outer surface of one plate is maintained at $10^\circ C$ and that of the other at $0^\circ C$. Temperature at the common surface is _______ (A) $0^\circ C$ (B) $25^\circ C$ (C) $5^\circ C$ (D) $6.5^\circ C$

    ›Reveal solutionSolution

    Because thickness and conductivity are in the same ratio (2:3) for the two plates, their thermal resistances are equal, and the common-surface temperature is exactly the midpoint of the two outer temperatures.

    Concept and Intuition

    In steady-state conduction through plates in series, the heat current $\dfrac{kA\Delta T}{d}$ must be the same through each plate. Thermal resistance is $d/(kA)$; when $d$ and $k$ scale together, the resistances of the two plates turn out equal, splitting the temperature drop evenly.

    Step-by-Step Solution

    1. Let $d_1=2x,\,d_2=3x$ (thickness ratio $2:3$) and $k_1=2y,\,k_2=3y$ (conductivity ratio $2:3$).
    2. Steady-state flux continuity: $\dfrac{k_1(10-T)}{d_1}=\dfrac{k_2(T-0)}{d_2}$.
    3. Substitute: $\dfrac{2y(10-T)}{2x}=\dfrac{3y\,T}{3x}\Rightarrow\dfrac{y}{x}(10-T)=\dfrac{y}{x}T$. …

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