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Conduction

Two rods of t...

Question

Two rods of the same length and area of cross section $A_{1}$ and $A_{2}$ have their ends at the same temperature. If $K_{1}$ and $K_{2}$ are the conductivities, $C_{1}$ and $C_{2}$ their specific heats and $ρ_{1}$ , $ρ_{2}$ their densities, then the condition that the rate of flow of heat is same in both the rods is?

\frac{A_{1}}{A_{2}} = \frac{K_{1} ρ_{1}}{K_{2} ρ_{2}}

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\frac{A_{1}}{A_{2}} = \frac{K_{1} C_{1} ρ_{1}}{K_{2} C_{2} ρ_{2}}

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\frac{A_{1}}{A_{2}} = \frac{\frac{K_{1}}{C_{1} ρ_{1}}}{\frac{K_{2}}{C_{2} ρ_{2}}}

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

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Solution

The correct options are
C $\frac{A_{1}}{A_{2}} = \frac{\frac{K_{1}}{C_{1} ρ_{1}}}{\frac{K_{2}}{C_{2} ρ_{2}}}$
D $\frac{A_{1}}{A_{2}} = \frac{K_{2}}{K_{1}}$
For conduction the governing equations are
$Q$ = $K A Δ T$
$Q$ = $M C Δ T$
K=thermal conductivity
A=area of rod
$Δ T$ =change in temperature
now form the given statement we find the following relations by equating the above two equation of conductive heat transfer.
$M C_{1} T = K_{1} A_{1} T$
$ρ_{1} V C_{1} = K_{1} A_{1}$
$A_{1} = \frac{ρ_{1} C_{1} V}{K_{1}}$
SIMILARLY
$A_{2} = \frac{ρ_{2} C_{2} V}{K_{2}}$
SO $\frac{A_{1}}{A_{2}} = \frac{ρ_{1} C_{1} K_{2}}{ρ_{2} C_{2} K_{1}}$
multiply with $K_{1} K_{2}$ on both at there numerator we get
$\frac{A_{1}}{A_{2}} = \frac{\frac{k_{1} k_{2}}{ρ_{1} C_{1} K_{2}}}{\frac{K_{1} K_{2}}{ρ_{2} C_{2} K_{1}}}$
$\frac{A_{1}}{A_{2}} = \frac{\frac{k_{1}}{ρ_{1} C_{1}}}{\frac{K_{2}}{ρ_{2} C_{2}}}$
NOW,from the relation of conduction we know that
$M C T = K_{1} A_{1} T$
$M C T = K_{2} A_{2} T$
by assuming same specific heat and same temperature
$\frac{A_{1}}{A_{2}} = \frac{K_{2}}{K_{1}}$

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Two rods of the same length and area of cross section A1 and A2 have their ends at the same temperature. If K1 and K2 are the conductivities, C1 and C2 their specific heats and ρ1,ρ2 their densities, then the condition that the rate of flow of heat is same in both the rods is?