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QJEEM21Phec4d: JEE Main 2021 Physics - Class 11 JEEM21Phec4d Solution: Two identical metal wires of thermal conductivities K 1 and K 2 …
Two identical metal wires of thermal conductivities K1 and K2 respectively are connected in series. The effective thermal conductivity of the combination is :
Multiple Choice Options
A
2 K 1 K 2 K 1 + K 2 {{2{K_1}{K_2}} \over {{K_1} + {K_2}}} K 1 + K 2 2 K 1 K 2
B
K 1 + K 2 K 1 K 2 {{{K_1} + {K_2}} \over {{K_1}{K_2}}} K 1 K 2 K 1 + K 2
C
K 1 + K 2 2 K 1 K 2 {{{K_1} + {K_2}} \over {2{K_1}{K_2}}} 2 K 1 K 2 K 1 + K 2
D
K 1 K 2 K 1 + K 2 {{{K_1}{K_2}} \over {{K_1} + {K_2}}} K 1 + K 2 K 1 K 2
Step-by-Step Verified Solution
R e q = R 1 + R 2 {R_{eq}} = {R_1} + {R_2} R e q = R 1 + R 2 ⇒ \Rightarrow ⇒ 1 K e q 2 l A = l K 1 A + l K 2 A {1 \over {{K_{eq}}}}{{2l} \over A} = {l \over {{K_1}A}} + {l \over {{K_2}A}} K e q 1 A 2 l = K 1 A l + K 2 A l ⇒ \Rightarrow ⇒ 2 K e q = l K 1 + l K 2 {2 \over {{K_{eq}}}} = {l \over {{K_1}}} + {l \over {{K_2}}} K e q 2 = K 1 l + K 2 l ⇒ \Rightarrow ⇒ 2 K e q = K 1 + K 2 K 1 K 2 {2 \over {{K_{eq}}}} = {{{K_1} + {K_2}} \over {{K_1}{K_2}}} K e q 2 = K 1 K 2 K 1 + K 2 ⇒ \Rightarrow ⇒ K e q = 2 K 1 K 2 K 1 + K 2 {K_{eq}} = {{2{K_1}{K_2}} \over {{K_1} + {K_2}}} K e q = K 1 + K 2 2 K 1 K 2
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