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Two rods one is semi circular of thermal conductivity K1 and other is straight of thermal conductivity K2 and of same cross-sectional area are joined as shown in figure. The point A and B are maintained at same temperature difference. If rate of flow of heat is same in two rods then \frac{k_1}{k_2} is.

 

 

 

 

Option: 1

\pi : 2


Option: 2

2:\pi


Option: 3

1:2

 


Option: 4

3:2


Answers (1)

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Answer (1)

\left ( \frac{dQ}{dt} \right )_{1}=\left ( \frac{dQ}{dt} \right )_{2}

\frac{K_{1}A\Delta T}{I_{1}}=\frac{K_{2}A\Delta T}{I_{2}}

\frac{K_{1}}{K_{2}}=\frac{I_{1}}{I_{2}}

       =\frac{\pi R}{2R}

       =\pi :2

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Rishi

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