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Thermal Properties of Matter appeared 13 times across 3 years — 1.5% of Physics. This question is from Thermal Conduction in Composite Rods.

Year 2026 2025 2024 Total
Questions 5 7 1 13
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Rods x and y of equal dimensions but of different materials are joined as shown in figure. Temperatures of end points A and F are maintained at 100circC and 40circC respectively. Given the thermal conductivity of rod x is three times of that of rod y, the temperature at junction points B and E are (close to):
Thermal Properties of Matter diagram for Q38 - JEE Main 2026 January Morning
Composite rods x and y joined end-to-end with temperature endpoints.

Solution & Explanation

Related Formula
H = Δ TReq, R = (l)/(KA)
Core Logic

Solution thermal network for Q38 - JEE Main 2026 Morning
Composite rods x and y joined end-to-end with temperature endpoints.

Let thermal resistance of rod y be 3R and rod x be R (since Kₓ = 3Ky).

Using thermal current equations:

H = (100 - 40)/((11R)/(2)) H = (100 - TB)/(R), H = (TE - 40)/(3R)

Solving yields TB = 89circC and TE = 73circC.

Pattern Recognition

Sees: Composite thermal conduction network with series/parallel segments. Shortcut: Calculate equivalent thermal resistances and equate heat currents across junctions. Check: Matches option (1). ✓

Chapter Mix

Class 11 Physics: Thermal Properties of Matter

Solution thermal network for Q38 - JEE Main 2026 Morning
Composite rods x and y joined end-to-end with temperature endpoints.

Reference Study Guides

Thermal Properties Of Matter

More Thermal Properties of Matter Previous-Year Questions — Page 3

Q10 jee_main_2025_24_jan_evening Temperature Scales
Which of the following figure represents the relation between Celsius and Fahrenheit temperatures?
  • A. Graph (1)
  • B. Graph (2)
  • C. Graph (3)
  • D. Graph (4)

Solution

Related Formula
(C)/(5) = (F - 32)/(9)
Core Logic

Rearranging the conversion identity to express C as a function of F:

C = (5)/(9)F - (160)/(9)

This is a straight line equation y = mx + c with:

  • Positive slope m = (5)/(9)
  • Negative y-intercept c = -(160)/(9) (at F=0, C ≈ -17.8°C)
  • Positive x-intercept at C=0, F=32

    • This completely matches the layout shown in Graph (2).

    Correct linear plot for Celsius vs Fahrenheit conversion Q10
    celsius fahrenheit graph, temperature conversion line, linear scaling

Pattern Recognition

0°C = 32°F. Therefore, the line must cross the positive side of the horizontal F axis when C=0.

Chapter Mix

Class 11 Physics: Thermal Properties of Matter

Q16 jee_main_2025_24_jan_evening Newton's Law of Cooling
The temperature of a body in air falls from 40°C to 24°C in 4 minutes. The temperature of the air is 16°C. The temperature of the body in the next 4 minutes will be:
  • A. (14)/(3)°C
  • B. (28)/(3)°C
  • C. (56)/(3)°C
  • D. (42)/(3)°C

Solution

Related Formula
(T₁ - T₂)/(t) = K[(T₁ + T₂)/(2) - Tₛ]
Core Logic

For the first interval (40°C to 24°C in 4 with Tₛ = 16°C):

(40 - 24)/(4) = K[(40 + 24)/(2) - 16] (16)/(4) = K[32 - 16] 4 = 16K K = (1)/(4)

For the next 4 minutes, let the final temperature be T:

(24 - T)/(4) = (1)/(4)[(24 + T)/(2) - 16] 24 - T = (24 + T)/(2) - 16 40 - T = (24 + T)/(2) 80 - 2T = 24 + T 3T = 56 T = (56)/(3)°C
Pattern Recognition

Newton's law of cooling approximation works beautifully when temperature differences are small. Always compute K from the first step and substitute directly into the second.

Chapter Mix

Class 11 Physics: Thermal Properties of Matter

Q jee_main_2024_30_january_evening Heating Curve of Water
A block of ice at -10°C is slowly heated and converted to steam at 100°C. Which of the following curves represent the phenomenon qualitatively:
  • A.
  • B.
  • C.
  • D.

Solution

Core Logic

The heating curve traces temperature vs heat supplied. Stage 1: Ice at -10°C is heated to 0°C (Temperature rises, solid phase). Stage 2: Ice melts at 0°C into water (Temperature remains constant until all ice melts). This is a horizontal plateau. Stage 3: Water is heated from 0°C to 100°C (Temperature rises, liquid phase). Stage 4: Water boils at 100°C into steam (Temperature remains constant). This is a second horizontal plateau.

Option (4) correctly shows this sequential step-like graph.

Pattern Recognition

Heating curves always exhibit horizontal segments during phase changes (latent heat) where temperature remains constant. The slopes of the inclined regions depend on the specific heat capacities of the respective phases.

Chapter Mix

Class 11 Physics: Thermal Properties of Matter

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JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%) | JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%)