Transmission Of Heat Question 338
Question: Two different metal rods of equal lengths & equal cross section area have their ends kept at the same temperatures $ {\theta _{1}}\And {\theta _{2}}. $ If $ K _{1}\And K _{2} $ be the thermal conductivities of rod, $ {\rho _{1}}\And {\rho _{2}} $ are their densities and $ s _{1},s _{2} $ are their specific heats, then the rate of flow of heat in the two rods will be same if
Options:
A) $ \frac{K _{1}}{K _{2}}=\frac{{\rho _{1}}s _{1}}{{\rho _{2}}s _{2}} $
B) $ \frac{K _{1}}{K _{2}}=\frac{{\rho _{1}}s _{2}}{{\rho _{2}}s _{1}} $
C) $ \frac{K _{1}}{K _{2}}=\frac{{\theta _{1}}}{{\theta _{2}}} $
D) $ K _{1}=K _{2} $
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Answer:
Correct Answer: D
Solution:
[d] Rate of heat flow is, $ H=\frac{KA({\theta _{1}}-{\theta _{2}})}{\ell } $ For 1st rod $ H _{1}=\frac{K _{1}A({\theta _{1}}-{\theta _{2}})}{\ell } $ For 2nd rod $ H _{2}=\frac{K _{2}A({\theta _{1}}-{\theta _{2}})}{\ell } $ For $ H _{1}=H _{2},K _{1}=K _{2} $
BETA
