Two cylindrical rods A and B made of different materials, are joined in a straight line. The ratio of lengths, radii and thermal conductivities of these rods are: fracmathrmL_mathrmAmathrmL_mathrmB = frac12, fracmathrmr_mathrmAmathrmr_mathrmB = 2$\frac{\mathrm{L}_{\mathrm{A}}}{\mathrm{L}_{\mathrm{B}}} = \frac{1}{2}, \frac{\mathrm{r}_{\mathrm{A}}}{\mathrm{r}_{\mathrm{B}}} = 2$ and fracmathrmK_mathrmAmathrmK_mathrmB = frac12$\frac{\mathrm{K}_{\mathrm{A}}}{\mathrm{K}_{\mathrm{B}}} = \frac{1}{2}$ . The free ends of rods A and B are maintained at 400mathrm~K$400\mathrm{~K}$, 200mathrm~K$200\mathrm{~K}$, respectively. The temperature of rods interface is ________ K, when equilibrium is established. [cite: 187, 188, 189, 190, 191, 192, 193]
Numerical Answer Type:
Enter a numerical valueAnswer: 360 to 360+4 marks
Solution & Explanation
### Related Formula
R_textth = fracLKA = fracLK(pi r^2)$R_{\text{th}} = \frac{L}{KA} = \frac{L}{K(\pi r^2)}$ [cite: 817]
fracdQdt = fracDelta TR_textth$\frac{dQ}{dt} = \frac{\Delta T}{R_{\text{th}}}$ [cite: 818]
### Core Logic
At steady state equilibrium, the rate of heat flow through both sections in series must be identical: [cite: 193, 819]
frac400 - TR_1 = fracT - 200R_2 implies frac400 - TT - 200 = fracR_1R_2$\frac{400 - T}{R_1} = \frac{T - 200}{R_2} \implies \frac{400 - T}{T - 200} = \frac{R_1}{R_2}$ [cite: 192, 820]
Let's evaluate the resistance ratio fracR_1R_2$\frac{R_1}{R_2}$ using the dimensional parameters: [cite: 820]
fracR_1R_2 = left(fracL_AL_Bright) cdot left(fracr_Br_Aright)^2 cdot left(fracK_BK_Aright)$\frac{R_1}{R_2} = \left(\frac{L_A}{L_B}\right) \cdot \left(\frac{r_B}{r_A}\right)^2 \cdot \left(\frac{K_B}{K_A}\right)$ [cite: 820]
Substitute the given values: fracL_AL_B = frac12$\frac{L_A}{L_B} = \frac{1}{2}$, fracr_Ar_B = 2 implies fracr_Br_A = frac12$\frac{r_A}{r_B} = 2 \implies \frac{r_B}{r_A} = \frac{1}{2}$, and fracK_AK_B = frac12 implies fracK_BK_A = 2$\frac{K_A}{K_B} = \frac{1}{2} \implies \frac{K_B}{K_A} = 2$ [cite: 189, 190]:
fracR_1R_2 = frac12 times left(frac12right)^2 times 2 = frac14$\frac{R_1}{R_2} = \frac{1}{2} \times \left(\frac{1}{2}\right)^2 \times 2 = \frac{1}{4}$ [cite: 821]
Now link this back into the temperature equation: [cite: 822]
frac400 - TT - 200 = frac14 implies 1600 - 4T = T - 200$\frac{400 - T}{T - 200} = \frac{1}{4} \implies 1600 - 4T = T - 200$ [cite: 822, 823]
5T = 1800 implies T = 360\ textK$5T = 1800 \implies T = 360\ \text{K}$ [cite: 824, 825]
### Pattern Recognition
Thermal conduction processes behave exactly like electric current fields in series connections[cite: 818, 819]. Cross-sectional area scales squarely with radius parameters, which requires extra care during substitution.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Thermal Properties of Matter
Keywords:#steady state interface temperature#JEE Main 2025 Evening Q23#thermal resistance cylinder rods#heat flux continuity series
More Thermal Properties of Matter Previous-Year Questions
Q192025Specific Heat Capacity and Thermal Conductivity
Match List-I with List-II.
List-I:
(A) Heat capacity of body
(B) Specific heat capacity of body
(C) Latent heat
(D) Thermal conductivity
List-II:
(I) mathrmJkg^-1$\mathrm{Jkg^{-1}}$
(II) mathrmJK^-1$\mathrm{JK}^{-1}$
(III) mathrmJkg^-1K^-1$\mathrm{Jkg^{-1}K^{-1}}$
(IV) mathrmJm^-1mathrmK^-1mathrms^-1$\mathrm{Jm}^{-1}\mathrm{K}^{-1}\mathrm{s}^{-1}$
Choose the correct answer from the options given below:
### Related Formula
Governing mathematical equations for thermal properties:
1. Heat capacity: C' = fracDelta QDelta T$C' = \frac{\Delta Q}{\Delta T}$
2. Specific heat capacity: s = fracDelta Qm Delta T$s = \frac{\Delta Q}{m \Delta T}$
3. Latent heat: L = fracDelta Qm$L = \frac{\Delta Q}{m}$
4. Thermal conductivity: K = fracDelta Q cdot LA cdot Delta T cdot t$K = \frac{\Delta Q \cdot L}{A \cdot \Delta T \cdot t}$
### Core Logic
Let's derive the SI units for each property:
- **(A) Heat capacity of body:**
[C'] = fracmathrmJmathrmK = mathrmJK^-1 quad implies text(II)$[C'] = \frac{\mathrm{J}}{\mathrm{K}} = \mathrm{JK^{-1}} \quad \implies \text{(II)}$
- **(B) Specific heat capacity of body:**
[s] = fracmathrmJmathrmkg cdot K = mathrmJkg^-1K^-1 quad implies text(III)$[s] = \frac{\mathrm{J}}{\mathrm{kg \cdot K}} = \mathrm{Jkg^{-1}K^{-1}} \quad \implies \text{(III)}$
- **(C) Latent heat:**
[L] = fracmathrmJmathrmkg = mathrmJkg^-1 quad implies text(I)$[L] = \frac{\mathrm{J}}{\mathrm{kg}} = \mathrm{Jkg^{-1}} \quad \implies \text{(I)}$
- **(D) Thermal conductivity:**
From heat transfer equation fracdQdt = K A fracdTdx$\frac{dQ}{dt} = K A \frac{dT}{dx}$:
K = fracdQ/dtA (dT/dx) implies [K] = fracmathrmJ/smathrmm^2 cdot (mathrmK/m) = mathrmJ cdot m^-1 cdot K^-1 cdot s^-1 quad implies text(IV)$K = \frac{dQ/dt}{A (dT/dx)} \implies [K] = \frac{\mathrm{J/s}}{\mathrm{m}^2 \cdot (\mathrm{K/m})} = \mathrm{J \cdot m^{-1} \cdot K^{-1} \cdot s^{-1}} \quad \implies \text{(IV)}$
### Step 1: Match values
Applying our mappings:
- (A) to$\to$ (II)
- (B) to$\to$ (III)
- (C) to$\to$ (I)
- (D) to$\to$ (IV)
This maps to Option (4).
### Pattern Recognition
Sees: Dimensional unit matching of thermal parameters.
Trap: Confusing Specific Heat Capacity (scaled per mass unit) with Heat Capacity (unscaled entire body parameter).
Shortcut: Specific heat is normalized by mass, meaning its unit must contain kg in the denominator. Heat capacity has no mass constraint, meaning A maps directly to II.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Thermal Properties of Matter
Q242025Radiation
A wire of length 10mathrmcm$10\mathrm{cm}$ and diameter 0.5mathrmmm$0.5\mathrm{mm}$ is used in a bulb. The temperature of the wire is 1727^circmathrmC$1727^{\circ}\mathrm{C}$ and power radiated by the wire is 94.2 W. Its emissivity is fracmathrmx8$\frac{\mathrm{x}}{8}$ where mathrmx =$\mathrm{x} =$
(Given sigma = 6.0 times 10^-8 mathrm~W mathrm~m^-2 mathrm~K^-4$\sigma = 6.0 \times 10^{-8} \mathrm{~W} \mathrm{~m}^{-2} \mathrm{~K}^{-4}$ , pi = 3.14$pi = 3.14$ and assume that the emissivity of wire material is same at all wavelength.)
Numerical Answer.Answer: 5 to 5
Solution
### Related Formula
Stefan-Boltzmann Law for radiated thermal power:
P = epsilon sigma A T^4$P = \epsilon \sigma A T^4$
Surface area A$A$ of a cylindrical wire of diameter d$d$ and length L$L$ is:
A = pi d L$A = \pi d L$
### Core Logic
Convert given parameters to standard SI units:
- L = 10 mathrm~cm = 0.1 mathrm~m$L = 10 \mathrm{~cm} = 0.1 \mathrm{~m}$
- d = 0.5 mathrm~mm = 0.5 times 10^-3 mathrm~m$d = 0.5 \mathrm{~mm} = 0.5 \times 10^{-3} \mathrm{~m}$
- T = 1727 + 273.15 = 2000 mathrm~K$T = 1727 + 273.15 = 2000 \mathrm{~K}$
- P = 94.2 mathrm~W$P = 94.2 \mathrm{~W}$
- sigma = 6.0 times 10^-8 mathrm~W cdot m^-2 cdot K^-4$\sigma = 6.0 \times 10^{-8} \mathrm{~W \cdot m^{-2} \cdot K^{-4}}$
### Step 1: Express Radiated Power and Emissivity
First, calculate the surface area A$A$:
A = 3.14 times (0.5 times 10^-3) times 0.1 = 1.57 times 10^-4 mathrm~m^2$A = 3.14 \times (0.5 \times 10^{-3}) \times 0.1 = 1.57 \times 10^{-4} \mathrm{~m}^2$
Substitute A$A$, sigma$\sigma$, and T$T$ into Stefan's formula:
94.2 = epsilon times (6.0 times 10^-8) times [3.14 times (0.5 times 10^-3) times (10 times 10^-2)] times (2000)^4$94.2 = \epsilon \times (6.0 \times 10^{-8}) \times [3.14 \times (0.5 \times 10^{-3}) \times (10 \times 10^{-2})] \times (2000)^4$
Calculate temperature term:
(2000)^4 = 1.6 times 10^12$(2000)^4 = 1.6 \times 10^{12}$94.2 = epsilon times (6 times 10^-8) times (1.57 times 10^-4) times (16 times 10^12)$94.2 = \epsilon \times (6 \times 10^{-8}) \times (1.57 \times 10^{-4}) \times (16 \times 10^{12})$94.2 = epsilon times 6 times 1.57 times 16 times 1 = epsilon times 150.72$94.2 = \epsilon \times 6 \times 1.57 \times 16 \times 1 = \epsilon \times 150.72$epsilon = frac94.2150.72 = frac58$\epsilon = \frac{94.2}{150.72} = \frac{5}{8}$
Thus, fracx8 = frac58 implies x = 5$\frac{x}{8} = \frac{5}{8} \implies x = 5$.
### Pattern Recognition
Sees: Bulb filament heat radiation equation.
Shortcut: Simplify the multiplication with factors of 10 first. T=2000 mathrm~K$T=2000 \mathrm{~K}$ has four zeros, so T^4$T^4$ adds 10^12$10^{12}$ which cancels the 10^-8$10^{-8}$ and 10^-4$10^{-4}$ from area and constant. The coefficient equation directly yields the ratio epsilon = 5/8$\epsilon = 5/8$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Thermal Properties of Matter
Q192025Newton's Law of Cooling
A cup of coffee cools from 90^circmathrmC$90^{\circ}\mathrm{C}$ to 80^circmathrmC$80^{\circ}\mathrm{C}$ in t$t$ minutes when the room temperature is 20^circmathrmC$20^{\circ}\mathrm{C}$ . The time taken by the similar cup of coffee to cool from 80^circmathrmC$80^{\circ}\mathrm{C}$ to 60^circmathrmC$60^{\circ}\mathrm{C}$ at the same room temperature is :
A.frac135 t$\frac{13}{5} t$
B.frac1013 t$\frac{10}{13} t$
C.frac1310 t$\frac{13}{10} t$
D.frac513 t$\frac{5}{13} t$
Solution
### Related Formula
fracT_i - T_fDelta t = Kleft(fracT_i + T_f2 - T_0right)$\frac{T_i - T_f}{\Delta t} = K\left(\frac{T_i + T_f}{2} - T_0\right)$
### Core Logic
**Case 1** (90^circmathrmC rightarrow 80^circmathrmC$90^{\circ}\mathrm{C} \rightarrow 80^{\circ}\mathrm{C}$ in time t$t$):
frac90 - 80t = Kleft(frac90 + 802 - 20
ight)$\frac{90 - 80}{t} = K\left(\frac{90 + 80}{2} - 20
ight)$frac10t = K(85 - 20) = 65K quad dots (i)$\frac{10}{t} = K(85 - 20) = 65K \quad \dots (i)$
**Case 2** (80^circmathrmC rightarrow 60^circmathrmC$80^{\circ}\mathrm{C} \rightarrow 60^{\circ}\mathrm{C}$ in time t'$t'$):
frac80 - 60t' = Kleft(frac80 + 602 - 20
ight)$\frac{80 - 60}{t'} = K\left(\frac{80 + 60}{2} - 20
ight)$frac20t' = K(70 - 20) = 50K quad dots (ii)$\frac{20}{t'} = K(70 - 20) = 50K \quad \dots (ii)$
Dividing equation (i)$(i)$ by equation (ii)$(ii)$:
frac10/t20/t' = frac65K50K$\frac{10/t}{20/t'} = \frac{65K}{50K}$fract'2t = frac1310 implies t' = frac135t$\frac{t'}{2t} = \frac{13}{10} \implies t' = \frac{13}{5}t$
### Pattern Recognition
Newton's law of cooling in average form uses the arithmetic mean of temperatures to approximate the driving temperature difference over the interval.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Thermal Properties of Matter
Q32025Thermal Expansion
Consider a rectangular sheet of solid material of length ell=9$\ell=9$ cm and width d=4$d=4$ cm. The coefficient of linear expansion is alpha=3.1times10^-5text K^-1$\alpha=3.1\times10^{-5}\text{ K}^{-1}$ at room temperature and one atmospheric pressure. The mass of sheet m=0.1text kg$m=0.1\text{ kg}$ and the specific heat capacity C_v=900text J kg^-1textK^-1$C_{v}=900\text{ J kg}^{-1}\text{K}^{-1}$. If the amount of heat supplied to the material is 8.1times10^2$8.1\times10^{2}$ J then change in area of the rectangular sheet is :-
### Related Formula
Delta Q = m C_v Delta T$\Delta Q = m C_v \Delta T$Delta A = A_0 beta Delta T = A_0 (2alpha) Delta T$\Delta A = A_0 \beta \Delta T = A_0 (2\alpha) \Delta T$
### Core Logic
First, calculate the temperature change using heat supplied:
Delta T = fracDelta Qm C_v$\Delta T = \frac{\Delta Q}{m C_v}$
Substituting the values:
Delta T = frac8.1 times 10^20.1 times 900 = frac81090 = 9text K$\Delta T = \frac{8.1 \times 10^2}{0.1 \times 900} = \frac{810}{90} = 9\text{ K}$
### Step 1: Calculate Change in Area
Initial area A_0 = ell times d = 9text cm times 4text cm = 36text cm^2 = 36 times 10^-4text m^2$A_0 = \ell \times d = 9\text{ cm} \times 4\text{ cm} = 36\text{ cm}^2 = 36 \times 10^{-4}\text{ m}^2$.
Now, substitute into the area expansion formula:
Delta A = 36 times 10^-4 times 2 times (3.1 times 10^-5) times 9$\Delta A = 36 \times 10^{-4} \times 2 \times (3.1 \times 10^{-5}) \times 9$Delta A = 36 times 18 times 3.1 times 10^-9 = 2008.8 times 10^-9text m^2 approx 2.0 times 10^-6text m^2$\Delta A = 36 \times 18 \times 3.1 \times 10^{-9} = 2008.8 \times 10^{-9}\text{ m}^2 \approx 2.0 \times 10^{-6}\text{ m}^2$
### Pattern Recognition
Always remember that areal expansion coefficient beta = 2alpha$\beta = 2\alpha$. Convert geometry dimensions to standard units (1text cm^2 = 10^-4text m^2$1\text{ cm}^2 = 10^{-4}\text{ m}^2$) cleanly before concluding arithmetic.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Thermal Properties of Matter
More Thermal Properties of Matter Questions — jee_main_2025_07_april_evening
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