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Condider two rods of same length and dif...

Condider two rods of same length and different specific heats `(s_(1), s_(2))`, thermal conductivities `(K_(1), K_(2))` and areas of cross-section `(A_(1), A_(2))` and both having temperatures `(T_(1), T_(2))` at their ends. If their rate of loss of heat due to conduction are equal, then

A

`K_(1)A_(1) = K_(2)A_(2)`

B

`(K_(1)A_(1))/(s_(1)) = (K_(2)A_(2))/(s_(2))`

C

`K_(2)A_(1) = K_(1)A_(2)`

D

`(K_(2)A_(1))/(s_(2)) = (K_(1)A_(2))/(s_(1))`

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To solve the problem, we need to analyze the heat conduction through two rods with given properties. The rods have the same length but different specific heats, thermal conductivities, and areas of cross-section. We are given that the rate of heat loss due to conduction is equal for both rods. ### Step-by-Step Solution: 1. **Understanding the Heat Conduction Formula**: The rate of heat loss (dQ/dt) through conduction can be expressed using Fourier's law of heat conduction: \[ \frac{dQ}{dt} = -K \cdot A \cdot \frac{dT}{dx} \] where: - \( K \) is the thermal conductivity of the material, - \( A \) is the cross-sectional area, - \( \frac{dT}{dx} \) is the temperature gradient. 2. **Setting Up the Equations for Both Rods**: For rod 1 (with thermal conductivity \( K_1 \), area \( A_1 \), and temperature difference \( T_1 - T_2 \)): \[ \frac{dQ}{dt_1} = K_1 \cdot A_1 \cdot \frac{(T_1 - T_2)}{L} \] For rod 2 (with thermal conductivity \( K_2 \), area \( A_2 \), and the same temperature difference \( T_1 - T_2 \)): \[ \frac{dQ}{dt_2} = K_2 \cdot A_2 \cdot \frac{(T_1 - T_2)}{L} \] 3. **Equating the Rates of Heat Loss**: Since the rate of heat loss is equal for both rods, we set the two equations equal to each other: \[ K_1 \cdot A_1 \cdot \frac{(T_1 - T_2)}{L} = K_2 \cdot A_2 \cdot \frac{(T_1 - T_2)}{L} \] 4. **Simplifying the Equation**: Since \( (T_1 - T_2) \) and \( L \) are the same for both rods, they can be canceled out from both sides of the equation: \[ K_1 \cdot A_1 = K_2 \cdot A_2 \] 5. **Final Result**: The final relationship we obtain is: \[ K_1 A_1 = K_2 A_2 \] ### Conclusion: The relationship between the thermal conductivities and areas of cross-section of the two rods is given by \( K_1 A_1 = K_2 A_2 \).
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