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Question

Physics Question on thermal properties of matter

Two rods AA and BB of different materials are welded together as shown in figure. Their thermal conductivities are K1K_1 and K2K_2. The thermal conductivity of the composite rod will be :

A

3(K1+K2)2\frac{3(K_1 + K_2)}{2}

B

K1+K2K_1 + K_2

C

2(K1+K2)2( K_1 + K_2)

D

K1+K22\frac{K_1 + K_2}{2}

Answer

K1+K22\frac{K_1 + K_2}{2}

Explanation

Solution

R1=dk1AR _{1}=\frac{ d }{ k _{1} A }
R2=dK2AR _{2}= d K _{2} A
Req=R1R2R1+R2\operatorname{Req}=\frac{ R _{1} R _{2}}{ R _{1}+ R _{2}}
dkeq(2A)=(dk1A)(dk2A)A(1k1+1k2)\frac{ d }{ keq (2 A )}=\frac{\left(\frac{ d }{ k _{1} A }\right)\left(\frac{ d }{ k _{2} A }\right)}{ A }\left(\frac{1}{ k _{1}}+\frac{1}{ k _{2}}\right)
12keq=1k1k2k1+k2k1k2=1k1+k2\frac{1}{2 keq }=\frac{\frac{1}{ k _{1} k _{2}}}{\frac{ k _{1}+ k _{2}}{ k _{1} k _{2}}}=\frac{1}{ k _{1}+ k _{2}}
keq=k1+k22keq =\frac{ k _{1}+ k _{2}}{2}