Two cylindrical rods of the same substance have diameters $d_{1}$ and $d_{2}$ . The amounts of heat conducted by these two rods, for same temperature difference between two ends will be equal if their lengths are related by

\frac{l_{1}}{l_{2}} = \frac{d_{1}}{d_{2}}

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\frac{l_{1}}{l_{2}} = {(\frac{d_{1}}{d_{2}})}^{2}

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\frac{l_{1}}{l_{2}} = \frac{d_{2}}{d_{1}}

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\frac{l_{1}}{l_{2}} = {(\frac{d_{2}}{d_{1}})}^{2}

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Solution

The correct option is D $\frac{l_{1}}{l_{2}} = {(\frac{d_{1}}{d_{2}})}^{2}$
From the fourier's law we know that heat transfer through a plane wall is given as:
$\Rightarrow Q = \frac{K A Δ T}{l}$
$\Rightarrow Q \propto \frac{d^{2}}{l}$
$\Rightarrow \frac{l_{1}}{l_{2}} = (\frac{d_{1}}{d_{2}})^{2}$

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Two cylindrical rods of the same substance have diameters d1 and d2 . The amounts of heat conducted by these two rods, for same temperature difference between two ends will be equal if their lengths are related by

The correct option is D l1l2=(d1d2)2From the fourier's law we know that heat transfer through a plane wall is given as:⇒Q=KAΔTl⇒Q ∝d2l⇒l1l2=(d1d2)2

Two cylindrical rods of the same substance have diameters $d_{1}$ and $d_{2}$ . The amounts of heat conducted by these two rods, for same temperature difference between two ends will be equal if their lengths are related by

The correct option is D $\frac{l_{1}}{l_{2}} = {(\frac{d_{1}}{d_{2}})}^{2}$
From the fourier's law we know that heat transfer through a plane wall is given as:
$\Rightarrow Q = \frac{K A Δ T}{l}$
$\Rightarrow Q \propto \frac{d^{2}}{l}$
$\Rightarrow \frac{l_{1}}{l_{2}} = (\frac{d_{1}}{d_{2}})^{2}$