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Standard VII

Physics

Conduction

The two ends ...

Question

The two ends of a rod of length L and a uniform cross-sectional area A are kept at two temperatures $T_{1}$ and $T_{2} (T_{1} > T_{2})$ . The rate of heat transfer, $\frac{d Q}{d t}$ , through the rod in a steady state is given by :

\frac{d Q}{d t} = \frac{k A (T_{1} - T_{2})}{L}

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\frac{d Q}{d t} = \frac{k L (T_{1} - T_{2})}{A}

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\frac{d Q}{d t} = \frac{k (T_{1} - T_{2})}{L A}

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\frac{d Q}{d t} = k L A (T_{1} - T_{2})

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Solution

The correct option is A $\frac{d Q}{d t} = \frac{k A (T_{1} - T_{2})}{L}$
For steady state $\frac{d Q}{d t} = \frac{k A (T_{1} - T_{2})}{L}$

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The two ends of a rod of length L and a uniform cross-sectional area A are kept at two temperatures T1 and T2(T1>T2). The rate of heat transfer, dQdt, through the rod in a steady state is given by :

The correct option is A dQdt=kA(T1−T2)LFor steady state dQdt=kA(T1−T2)L

The two ends of a rod of length L and a uniform cross-sectional area A are kept at two temperatures $T_{1}$ and $T_{2} (T_{1} > T_{2})$ . The rate of heat transfer, $\frac{d Q}{d t}$ , through the rod in a steady state is given by :

The correct option is A $\frac{d Q}{d t} = \frac{k A (T_{1} - T_{2})}{L}$
For steady state $\frac{d Q}{d t} = \frac{k A (T_{1} - T_{2})}{L}$