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Standard X

Physics

Resistivity

Two rods of t...

Question

Two rods of the same length and diameter having thermal conductivities $K_{1} K_{2}$ are joined in parallel. The equivalent thermal conductivity of the combination is:

\frac{K_{1} K_{2}}{K_{1} + K_{2}}

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K_{1} + K_{2}

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\frac{K_{1} + K_{2}}{2}

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\sqrt{K_{1} K_{2}}

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Solution

The correct option is A $\frac{K_{1} K_{2}}{K_{1} + K_{2}}$
$\frac{1}{K_{e q}} = \frac{1}{K_{1}} + \frac{1}{K_{2}}$
$K_{e q} = \frac{K_{1} K_{2}}{K_{1} + K_{2}}$

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Two rods of the same length and diameter having thermal conductivities K1K2 are joined in parallel. The equivalent thermal conductivity of the combination is:

The correct option is A K1K2K1+K21Keq=1K1+1K2Keq=K1K2K1+K2

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