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 100^circtextC$100^{circ}\text{C}$ and 40^circtextC$40^{circ}\text{C}$ 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):

### Related Formula P = fracdmdt S Delta T$$P = \frac{dm}{dt} S \Delta T$$ P_textheater = textRate of combustion times textHeat of combustion$$P_{\text{heater}} = \text{Rate of combustion} \times \text{Heat of combustion}$$ ### Core Logic Water flow rate = 5text l/min = frac560text kg/s = frac112text kg/s$5\text{ l/min} = \frac{5}{60}\text{ kg/s} = \frac{1}{12}\text{ kg/s}$. The power required to heat the water is: P_textheater = left(fracdmdtright)_textwater cdot S cdot Delta T$$P_{\text{heater}} = \left(\frac{dm}{dt}\right)_{\text{water}} \cdot S \cdot \Delta T$$ P_textheater = frac112 times 4200 times (87 - 27) = frac112 times 4200 times 60text W$$P_{\text{heater}} = \frac{1}{12} \times 4200 \times (87 - 27) = \frac{1}{12} \times 4200 \times 60\text{ W}$$ ### Step 1: Solving for Gas Consumption Let the rate of consumption of gas be xtext g/s$x\text{ g/s}$. Heat generated by the gas per second must equal the power required to heat the water: x times 5.0 times 10^4 = frac112 times 4200 times 60$$x \times 5.0 \times 10^{4} = \frac{1}{12} \times 4200 \times 60$$ x times 5.0 times 10^4 = 4200 times 5$$x \times 5.0 \times 10^{4} = 4200 \times 5$$ x = frac4200 times 55 times 10^4 = 4200 times 10^-4$$x = \frac{4200 \times 5}{5 \times 10^{4}} = 4200 \times 10^{-4}$$ x = 0.42text g/s$$x = 0.42\text{ g/s}$$

### Pattern Recognition Equate the thermal power absorbed by fluid (dm/dt cdot c cdot Delta T$dm/dt \cdot c \cdot \Delta T$) with the power output of fuel (dm_textfuel/dt cdot L_textcombustion$dm_{\text{fuel}}/dt \cdot L_{\text{combustion}}$). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Thermal Properties of Matter

### Related Formula Delta l = l_0 alpha Delta T$$\Delta l = l_0 \alpha \Delta T$$ l_textfinal = l_1 + l_2 + Delta l_1 + Delta l_2$$l_{\text{final}} = l_1 + l_2 + \Delta l_1 + \Delta l_2$$ ### Core Logic Let the original lengths be l_0A$l_{0A}$ (aluminium) and l_0S$l_{0S}$ (steel). They have same length, so l_0A = l_0S = 60text cm$l_{0A} = l_{0S} = 60\text{ cm}$. Change in temperature Delta T = 100^circtextC - 30^circtextC = 70^circtextC$\Delta T = 100^{\circ}\text{C} - 30^{\circ}\text{C} = 70^{\circ}\text{C}$. Expansion of Aluminium: Delta l_A = l_0 alpha_A Delta T$\Delta l_A = l_0 \alpha_A \Delta T$ Expansion of Steel: Delta l_S = l_0 alpha_S Delta T$\Delta l_S = l_0 \alpha_S \Delta T$ ### Step 1: Calculate Final Length l_textfinal = l_0(1 + alpha_A Delta T) + l_0(1 + alpha_S Delta T)$$l_{\text{final}} = l_0(1 + \alpha_A \Delta T) + l_0(1 + \alpha_S \Delta T)$$ l_textfinal = l_0 [2 + (alpha_A + alpha_S)Delta T]$$l_{\text{final}} = l_0 [2 + (\alpha_A + \alpha_S)\Delta T]$$ l_textfinal = 60 left[ 2 + (24 times 10^-6 + 12 times 10^-6) times 70 right]$$l_{\text{final}} = 60 \left[ 2 + (24 \times 10^{-6} + 12 \times 10^{-6}) \times 70 \right]$$ l_textfinal = 60 left[ 2 + (36 times 10^-6) times 70 right]$$l_{\text{final}} = 60 \left[ 2 + (36 \times 10^{-6}) \times 70 \right]$$ l_textfinal = 60 [2 + 0.00252] = 120 + 60(0.00252) = 120 + 0.1512 = 120.15text cm$$l_{\text{final}} = 60 [2 + 0.00252] = 120 + 60(0.00252) = 120 + 0.1512 = 120.15\text{ cm}$$

### Pattern Recognition For composite rods connected end-to-end, expansions simply add up. Convert 1.2 times 10^-5$1.2 \times 10^{-5}$ to 12 times 10^-6$12 \times 10^{-6}$ immediately for easy mental addition. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Thermal Properties of Matter

### 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

### 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