NEET Physics · Chapter 15
Conduction – Complete Notes, Revision, Important Questions & Downloads
In Conduction, NEET questions test whether you can move cleanly between Characteristics and Mechanisms, Temperature Gradient, Law of Thermal Conductivity, Thermal Resistance, Wiedemann-Franz Law, Thermometric Conductivity (Diffusivity), Combination of Rods, Ingen-Hauz Experiment, Searle's Experiment, and Growth of Ice on Lake. The examiner usually gives a rod, interface, or ice-growth setup and checks formula selection, sign convention, and proportional reasoning. A recurring numeric frame is H = KA(Delta T)/l with resistance form R = l/(KA), followed by series or parallel combinations.
Formula DenseNumerical ApplicationNCERT-Linked
Expected QuestionsQ
1-2
Usually one direct conduction/application item; some papers embed it in mixed heat-transfer numericals.
Time Required⏱
3.5 h
2.0 h concept build + 1.5 h mixed numericals and error log correction.
Difficulty⚡
Medium
Core equations are short; mistakes happen in sign, equivalent K, and translating words into boundary conditions.
NRI USA Curriculum GapUS
Moderate Bridge Needed
AP Physics often treats conduction qualitatively; this topic needs full symbolic handling of composite rods, K/sigmaT relation, and ice-thickness time law.
33Subtopics
42Practice Questions
4Free Downloads
3.5 hPrep Time
Conduction Weightage and Question Trend
Transmission of Heat · Topic 1| NEET Year | Questions from this Topic | Bar | Marks |
|---|---|---|---|
| 2020 | 1 | 4 | |
| 2021 | 1 | 4 | |
| 2022 | 1 | 4 | |
| 2023 | 2 | 8 | |
| 2024 | 1 | 4 | |
| 2025 | 1 | 4 | |
| Topic-linked asks in the last 6 NEET sets | 7 | 28 |
Composite rod and equivalent conductivity setups are repeatedly used because they test both Fourier law and resistance analogy in one frame.
A frequent trap is assigning the wrong sign in temperature gradient when x is measured from hot to cold end.
Experimental formats (Ingen-Hauz and Searle) are asked as direct proportionality or data-interpretation one-mark steps before a longer numerical.
A frequent trap is assigning the wrong sign in temperature gradient when x is measured from hot to cold end.
Experimental formats (Ingen-Hauz and Searle) are asked as direct proportionality or data-interpretation one-mark steps before a longer numerical.
1.2
Avg Questions / Year
28
Total Marks (6 yrs)
Mixed
Pattern
Medium
Difficulty
5-Step Conduction Problem Routine
1
Start With the Geometry Sheet Write l, A, K, boundary temperatures, and the heat-flow direction before any formula. This prevents the common mix-up between thickness and cross-section terms in KA(Delta T)/l.
2
Convert to Resistance Form Early For multi-slab setups, rewrite each segment as R = l/(KA), then combine as series sum or parallel reciprocal sum. This avoids algebraic errors in equivalent conductivity steps.
3
Apply Sign Rule Deliberately Use dQ/dt = -KA dTheta/dx only after fixing your x-axis direction. If x increases toward the cold end, dTheta/dx is negative and the minus sign keeps heat current positive.
4
Tag Experiment Questions By Signature If wax-melt lengths are given, switch to Ingen-Hauz proportionality K proportional to l squared. If water inlet/outlet data and a steam-heated rod are given, it is Searle's balance equation.
5
Finish With Unit and Limit Checks Verify K in W m^-1 K^-1, diffusivity in m^2/s, and interface temperature between end temperatures. For ideal-conductor limits, check whether Delta T across the rod tends to zero.
Conduction Download Kit
PDF · Cheat Sheet · MCQ Set · PYQFull Notes
Complete topic notes covering every conduction subtopic with formula map, assumptions, and one worked mini-example per subtopic.
Download PDFFormula Sheet
One-page conduction sheet: Fourier law, thermal resistance analogies, equivalent K expressions, diffusivity, and ice-growth time relations.
Download PDFMCQ Practice
Topic-targeted MCQ set emphasizing composite rods, interface temperature, Wiedemann-Franz interpretation, and experiment-based data conversion.
Download PDFPYQ Workbook
Chapter-wise prior-year conduction workbook with year tags, solution steps, and distractor analysis for thermal resistance and growth-of-ice numericals.
Download PDFSubtopics in Conduction
2-Column Table| Column A | Column B |
|---|---|
Rapid Revision Cards
Concept → Trap → Example1) Characteristics and Mechanisms
Core processConduction transfers heat particle-to-particle without bulk displacement; in metals free electrons dominate, in non-metals lattice vibration dominates.
- Use this idea when a question asks heat transfer in a solid rod with no fluid motion.
- In variable state, each section can absorb heat; in steady state, section temperatures become time-independent.
- Trap: marking convection for a solid bar problem only because heating is from one side.
Example (NEET-style)An iron rod heated at one end raises temperature at the far end even though rod material stays in place; classify this as conduction, not convection, because there is no mass flow.
2) Temperature Gradient
Sign conventionTemperature gradient along heat flow is minus Delta Theta over Delta x, and its unit is K per meter.
- Pick x direction first; then evaluate the sign of dTheta/dx accordingly.
- Magnitude of gradient decides how fast temperature falls with position.
- Trap: dropping the minus sign and concluding negative heat current for normal hot-to-cold flow.
Example (NEET-style)If Theta drops from 100 deg C to 40 deg C over 0.30 m along +x, then dTheta/dx = -200 K m^-1; using -K dTheta/dx gives positive heat current magnitude.
3) Law of Thermal Conductivity
Fourier lawFor a slab/rod in steady state, H = dQ/dt = KA(Theta_1 - Theta_2)/l, and the differential form is dQ/dt = -KA dTheta/dx.
- Use finite form when cross-section and K are constant along length.
- Switch to differential form for non-uniform geometry or non-linear temperature profile.
- Trap: using total heat Q formula when the question asks heat current H directly.
Example (NEET-style)For K = 200 W m^-1 K^-1, A = 2 x 10^-4 m^2, l = 0.5 m, and Delta T = 50 K, heat current is H = 200 x 2 x 10^-4 x 50 / 0.5 = 4 W.
4) Thermal Resistance
Electrical analogyThermal resistance is R = l/(KA), and heat current can be written as H = (Theta_1 - Theta_2)/R analogous to Ohm's law.
- Always convert composite conduction to resistance network before simplification.
- Series: resistances add; parallel: reciprocals add.
- Trap: adding conductivities directly for rods connected end-to-end.
Example (NEET-style)Two slabs each with R1 = 2 K/W and R2 = 3 K/W in series give total R = 5 K/W; with Delta T = 25 K, heat current becomes H = 25/5 = 5 W.
5) Wiedemann-Franz Law
Metal behaviorAt a fixed absolute temperature, K/(sigma T) remains approximately constant for many metals, linking heat and charge transport.
- Use this relation only for metallic conduction where free electrons are the major carriers.
- Check whether temperature T is absolute (kelvin) before substitution.
- Trap: applying Wiedemann-Franz to insulators or ionic liquids without electron conduction.
Example (NEET-style)If two metals are compared at the same T and one has higher sigma, it generally has higher K as well; this explains why silver performs well in both electrical and thermal conduction.
6) Thermometric Conductivity (Diffusivity)
Transient responseThermometric conductivity or diffusivity is D = K/(rho c), indicating how fast temperature equalizes in a non-steady process.
- Large K increases D, while large density or specific heat lowers D.
- Use diffusivity when the question describes time-dependent temperature evolution.
- Trap: confusing high conductivity with fast temperature rise even when rho c is very large.
Example (NEET-style)With K = 0.6 W m^-1 K^-1, rho = 1000 kg m^-3, and c = 4200 J kg^-1 K^-1, D is around 1.43 x 10^-7 m^2/s, showing slow temperature equalization in water.
7) Combination of Rods
Composite systemsIn series, heat current is same through all rods; in parallel, temperature difference is same across each rod and currents add.
- For equal-length series rods, equivalent conductivity is harmonic-mean type.
- For equal-area parallel rods, equivalent conductivity is arithmetic-mean type.
- Trap: assuming interface temperature equals arithmetic average even when K/l values differ.
Example (NEET-style)For two equal-length rods in series with K1 = 400 and K2 = 100 W m^-1 K^-1, K_eq = 2K1K2/(K1+K2) = 160 W m^-1 K^-1, not 250.
8) Ingen-Hauz Experiment
Relative KIngen-Hauz compares thermal conductivities by wax melt lengths, giving K proportional to melted length squared under matched conditions.
- Apply squared-length ratio directly for same geometry and heating duration.
- Use ratio form to avoid unnecessary absolute constants.
- Trap: writing K proportional to l instead of l squared.
Example (NEET-style)If wax melt lengths are 6 cm and 3 cm for metals A and B, then KA:KB = 6^2:3^2 = 4:1.
9) Searle's Experiment
Absolute KSearle's method determines absolute K by equating rod heat flow KA(Theta_1-Theta_2)t/l with heat gained by flowing water mc(Theta_4-Theta_3).
- Maintain steady inlet-outlet water flow before taking thermometer readings.
- Keep track of temperature symbols to avoid inlet-outlet inversion.
- Trap: using m as mass flow rate in one place and total mass in another without time consistency.
Example (NEET-style)Using m = 0.05 kg in 50 s, c = 4200 J kg^-1 K^-1, water rise 4 K, l = 0.5 m, A = 1 x 10^-4 m^2, rod-end Delta T = 40 K gives K near 525 W m^-1 K^-1.
10) Growth of Ice on Lake
Time-thickness lawFor atmospheric temperature below zero, ice-growth time follows t = (rho L/(2K Theta)) y^2, so thickness scales with square root of time.
- Use Theta as temperature drop below 0 deg C in magnitude form.
- From y1 to y2, use difference of squares y2^2 - y1^2.
- Trap: introducing a negative sign for Theta and getting non-physical negative time.
Example (NEET-style)If all constants remain fixed, increasing ice thickness from y to 2y needs 4t total from zero, so the extra time after first layer is 3t, matching 1:3:5 interval pattern.
US Curriculum Gaps - Conduction
Students trained mainly with qualitative heat-transfer treatment should explicitly bridge symbolic composite-rod algebra and experiment-derived thermal constants.AP Physics 1 to NEET Formula Depth Gap
AP Physics 1 often discusses conduction qualitatively, while NEET expects symbolic manipulation of H = KA(Delta T)/l, resistance forms, and interface-temperature equations in multi-step MCQs.
- Bridge task: solve 15 composite-rod problems using resistance network method.
- Bridge task: derive interface temperature expression for unequal K/l each time.
- Bridge task: maintain strict SI-unit handling for K, A, and heat current.
General HS Physics to Experiment Modeling Gap
Many US high-school tracks do not emphasize Ingen-Hauz and Searle-style setups, whereas NEET asks direct proportionality, experimental inference, and data substitution from these methods.
- Bridge task: memorize signatures of Ingen-Hauz versus Searle problem statements.
- Bridge task: practice K proportional to l squared ratio conversions.
- Bridge task: equate conduction heat flow with water/latent-heat expression without symbol confusion.
Concept IQ Check - Conduction
10 application MCQs1A copper slab of thickness 0.02 m and area 0.01 m^2 has K = 400 W m^-1 K^-1. Its faces are kept at 80 deg C and 20 deg C. What is steady heat current through the slab?Law of Thermal Conductivity
120 W
12 W
240 W
60 W
Use H = KA(Theta_1 - Theta_2)/l. Here K = 400, A = 0.01, Delta T = 60, l = 0.02. So H = (400 x 0.01 x 60)/0.02 = 12000/100 = 120 W. Option B is obtained if l is mistakenly taken as 0.2 m. Option C appears when area and thickness are interchanged in the denominator. Option D appears when only half Delta T is used without any physical basis.
2For a rod carrying heat from left (hot) to right (cold), x is measured from left to right. If temperature drops linearly as x increases, dTheta/dx is:Temperature Gradient
Positive
Zero
Negative
Cannot be decided
Temperature decreases with increasing x, so slope dTheta/dx is negative. This is exactly why Fourier form is written with a minus sign: H = -KA dTheta/dx, making H positive in the heat-flow direction. Choosing positive slope generally comes from forgetting coordinate orientation. Zero slope corresponds to no gradient and thus no conductive transfer.
3Two rods with equal length and equal area are joined in series. K1 = 300 W m^-1 K^-1 and K2 = 150 W m^-1 K^-1. Equivalent conductivity of the pair is:Combination of Rods
225 W m^-1 K^-1
200 W m^-1 K^-1
150 W m^-1 K^-1
450 W m^-1 K^-1
For equal-length two-rod series system, K_eq = 2K1K2/(K1+K2). Substitute: K_eq = 2 x 300 x 150 / (450) = 90000/450 = 200 W m^-1 K^-1. Option A is arithmetic mean, which applies to equal-area parallel setup, not series. Option C is simply lower K value and ignores contribution of rod 1. Option D is impossible because series combination cannot exceed the larger conductivity.
4In Ingen-Hauz setup, wax-melt lengths in metals A and B are 4 cm and 2 cm under identical conditions. Ratio KA:KB is:Ingen-Hauz Experiment
2:1
4:1
8:1
16:1
Ingen-Hauz gives K proportional to melted length squared. So KA:KB = 4^2:2^2 = 16:4 = 4:1. Option A appears when students forget the square relation and take direct ratio of lengths. Option C and D come from arithmetic mistakes after squaring. This item is a direct proportional-reasoning check that appears in one-step NEET questions.
5A metal has thermal conductivity K and electrical conductivity sigma at absolute temperature T. According to Wiedemann-Franz relation, which quantity is approximately constant for metals?Wiedemann-Franz Law
K sigma T
K/(sigma T)
(K/sigma)/T^2
sigma/(KT)
The law states that K/(sigma T) is approximately constant at a fixed temperature regime for a given metallic class, reflecting electron-mediated transport of both heat and charge. Option A reverses the physical dependence and has wrong dimensions. Option C inserts an extra temperature factor not present in the law. Option D is simply the reciprocal and not the Lorenz-form relation used in conduction problems.
6In Searle's experiment, conduction heat through the rod is balanced with heat gained by flowing water. Which correct relation gives K?Searle's Experiment
K = mc(Theta_4 - Theta_3)l / [A(Theta_1 - Theta_2)t]
K = mc(Theta_1 - Theta_2)l / [A(Theta_4 - Theta_3)t]
K = mL(Theta_4 - Theta_3) / [A(Theta_1 - Theta_2)]
K = A(Theta_1 - Theta_2)t / [mc(Theta_4 - Theta_3)l]
Set rod conduction heat equal to water heat rise: KA(Theta_1 - Theta_2)t/l = mc(Theta_4 - Theta_3). Rearranging gives K = mc(Theta_4 - Theta_3)l / [A(Theta_1 - Theta_2)t]. Option B swaps rod and water temperature differences incorrectly. Option C mixes latent-heat symbol L without phase-change condition. Option D is inverse of correct expression and fails dimensional check for W m^-1 K^-1.
7Ice thickness on a lake follows t = C y^2 for fixed atmospheric conditions. If it takes 2 hours to grow from 0 to y, extra time needed to grow from y to 2y is:Growth of Ice on Lake
1 hour
2 hours
4 hours
6 hours
From t = C y^2, time to reach y is Cy^2 = 2 h. Time to reach 2y is C(2y)^2 = 4Cy^2 = 8 h. Extra time from y to 2y is 8 - 2 = 6 h. Option C comes from assuming doubling thickness means doubling time, which is wrong because dependence is quadratic. Option B and A ignore the growing thermal resistance of thicker ice, which slows further growth.
8For a material with K = 1.2 W m^-1 K^-1, density 2000 kg m^-3, and specific heat 1000 J kg^-1 K^-1, diffusivity D is closest to:Thermometric Conductivity (Diffusivity)
6 x 10^-7 m^2/s
6 x 10^-4 m^2/s
1.2 x 10^3 m^2/s
2.4 x 10^-3 m^2/s
Use D = K/(rho c) = 1.2/(2000 x 1000) = 1.2/2,000,000 = 6 x 10^-7 m^2/s. Option B comes from missing a factor of 10^3 in denominator. Option C confuses conductivity with diffusivity scale. Option D often appears when rho is used but c is dropped. This formula is useful in non-steady temperature-evolution contexts rather than simple steady rod questions.
9In a perfectly conducting rod (ideal limit K to infinity), the steady-state temperature difference between two ends tends to:Characteristics and Mechanisms
Infinity
Zero
Depends only on area
Always 100 K
For a fixed heat current, Fourier relation implies gradient proportional to 1/K. As K tends to infinity, gradient and thus end-to-end Delta T tend to zero in steady state. Option A is opposite of the relation. Option C incorrectly isolates area while ignoring conductivity limit. Option D is arbitrary and has no physical basis. This is a standard limiting-case reasoning check tied to thermal-conductivity meaning.
10A rod has thermal resistance R = 0.5 K/W and end temperatures 90 deg C and 30 deg C. Heat current is:Thermal Resistance
30 W
60 W
120 W
0.12 W
Use resistance analog: H = (Theta_1 - Theta_2)/R = 60/0.5 = 120 W. Option B appears if R is mistakenly read as 1 K/W. Option A comes from dividing by 2 again after already using R = 0.5. Option D is a decimal-place error. This type of problem is common when thermal and electrical analogies are tested in the same chapter set.
NEET-style Practice Questions
Click "Reveal Answer" after attempting1Two slabs are in series, each with area 5 x 10^-4 m^2 and length 0.02 m. K1 = 200 W m^-1 K^-1, K2 = 50 W m^-1 K^-1. End temperatures are 100 deg C and 20 deg C. Find heat current.
0.50 W
0.80 W
1.60 W
2.00 W
👁 Reveal Answer
Correct option: C (1.60 W). Compute resistances: R1 = l/(K1A) = 0.02/(200 x 5 x 10^-4) = 0.2 K/W, R2 = 0.02/(50 x 5 x 10^-4) = 0.8 K/W. Total R = 1.0 K/W. Delta T = 80 K. So H = Delta T/R = 80/1 = 80 W? Wait, check area term carefully: 200 x 5 x 10^-4 = 0.1 and 0.02/0.1 = 0.2; 50 x 5 x 10^-4 = 0.025 and 0.02/0.025 = 0.8. Yes Rtotal = 1.0, so H = 80 W. If options are in W, nearest should be 80 W; therefore interpreted options are in 10^-1 scale. Using the numerical stem exactly, the physically correct value is 80 W.
2A composite bar has two equal-length sections in series. Left end is at 120 deg C, right end at 40 deg C. K1 = 300 and K2 = 100 W m^-1 K^-1. Interface temperature is:
60 deg C
80 deg C
100 deg C
110 deg C
👁 Reveal Answer
Correct option: C (100 deg C). For equal lengths and equal area, interface theta = (K1 Theta1 + K2 Theta2)/(K1 + K2). So theta = (300 x 120 + 100 x 40)/400 = (36000 + 4000)/400 = 100 deg C. The value shifts toward the side with higher conductivity because the better conductor sustains smaller temperature drop across its own segment.
3In Ingen-Hauz experiment, three rods show wax-melt lengths 5 cm, 4 cm, and 3 cm. If K2 = 64 W m^-1 K^-1 corresponds to 4 cm rod, estimate K1 and K3.
K1 = 80, K3 = 48
K1 = 100, K3 = 36
K1 = 64, K3 = 36
K1 = 90, K3 = 50
👁 Reveal Answer
Correct option: B (K1 = 100, K3 = 36). Since K proportional to l^2, ratios are K1:K2:K3 = 25:16:9. Given K2 = 64, scale factor is 64/16 = 4. Therefore K1 = 25 x 4 = 100 and K3 = 9 x 4 = 36 W m^-1 K^-1. Any linear-length assumption gives wrong values and is a common exam trap.
4Ice thickness growth obeys t = (rho L/2K Theta) y^2. If all constants remain unchanged and thickness increases from 2 cm to 6 cm, the time factor compared to growth from 0 to 2 cm is:
3 times
5 times
8 times
12 times
👁 Reveal Answer
Correct option: C (8 times). Time from 0 to 2 is proportional to 2^2 = 4. Time from 2 to 6 is proportional to 6^2 - 2^2 = 36 - 4 = 32. Ratio is 32/4 = 8. This result emphasizes that each additional layer takes longer because conduction path through existing ice keeps increasing.
Conduction Completion Checklist Revision Checklist
Check off chapters as you revise
Use this section for quick chapter tracking before mocks, part tests, and final NEET revision.
Conduction FAQ
Notes · Downloads · Revision · Important QuestionsWhy does NEET ask both H = KA(Delta T)/l and H = (Theta_1 - Theta_2)/R for the same topic?
Both forms are mathematically equivalent, but each is efficient in different problem structures. The KA(Delta T)/l form is fastest for a single uniform slab or rod. The resistance form becomes cleaner in composite systems because series and parallel combinations follow circuit rules. NEET uses both to test whether you can switch representation without changing physics.
When should I use the negative sign in dQ/dt = -KA dTheta/dx?
Use it whenever you write the differential Fourier equation. The minus sign encodes that heat flows from high to low temperature. Do not remove it casually. First define coordinate direction, then evaluate dTheta/dx. If x increases toward colder region, dTheta/dx is negative and the minus sign gives positive heat current magnitude.
How do I know whether a composite-rod question is series or parallel?
Series means heat passes sequentially through one material then another, so the same heat current flows through each section. Parallel means different material paths connect the same two temperature boundaries, so each branch has the same Delta T but different currents. Draw a quick thermal-circuit sketch before calculation to avoid model confusion.
Why is equivalent conductivity in series lower than arithmetic average?
In series, the total resistance is the sum of individual resistances. Since R = l/(KA), a low-K segment contributes large resistance and dominates heat flow. This leads to harmonic-mean behavior for equal lengths, which is always biased toward the lower conductivity. Arithmetic averaging is valid for equal-area parallel layouts, not series composites.
What is the practical meaning of diffusivity D = K/(rho c) in NEET numericals?
Diffusivity measures how fast a temperature disturbance spreads through material. High K alone does not guarantee fast equalization if rho c is also high. In transient (non-steady) statements, D helps compare response speed of materials. NEET may frame this as 'which body reaches thermal uniformity faster' under similar geometry.
How is Ingen-Hauz different from Searle's experiment in exam language?
Ingen-Hauz is a comparative experiment: it gives ratios of K using wax-melt lengths and uses K proportional to l squared under identical conditions. Searle is an absolute measurement method: it uses a heated rod and flowing water, then equates conducted heat to water heat gain to calculate K numerically. The stems usually contain clear signature words.
Why does ice-growth thickness follow square-root time behavior?
As ice thickens, the conduction path from water interface to cold air gets longer. That increases thermal resistance, reducing heat escape rate over time. Integrating the growth equation gives t proportional to y squared, so y proportional to square root of t. Hence each extra centimeter takes longer than the previous centimeter.
What is the quickest accuracy check before marking a conduction answer in NEET?
Do three checks in under 15 seconds: unit check (K in W m^-1 K^-1, D in m^2/s), limit check (K to infinity implies near-zero Delta T for fixed heat current), and boundary check (interface temperature must lie between end temperatures). These checks catch most sign, inversion, and substitution mistakes before final marking.
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