Two rods of length d1 and d2 and coefficients of thermal conductivities K1 and K2 are kept touching each other. Both have the same area of cross?section, the equivalent thermal conductivity is

When Two rods are connected in series Q=A (T1−T2)td1K1+d2K1=A (T1−T2)t (d1+d2)/K d 1 + d 2 K = d 1 K 1 + d 2 K 2 K= (d1+d2)d1K1+d2K1

Two rods of length d₁ and d₂ and coefficients of thermal conductivities K₁ and K₂ are kept touching each other. Both have the same area of cross?section, the equivalent thermal conductivity is

Option 1 - <math> <mrow> <msub> <mrow>K</mrow> <mrow> <mn>1</mn> </mrow> </msub> <mo>+</mo> <msub> <mrow>K</mrow> <mrow> <mn>2</mn> </mrow> </msub> </mrow> </math>

Option 2 - <math> <mrow> <msub> <mrow>K</mrow> <mrow> <mn>1</mn> </mrow> </msub> <msub> <mrow>d</mrow> <mrow> <mn>1</mn> </mrow> </msub> <mo>+</mo> <msub> <mrow>K</mrow> <mrow> <mn>2</mn> </mrow> </msub> <msub> <mrow>d</mrow> <mrow> <mn>2</mn> </mrow> </msub> </mrow> </math>

Option 3 - <math> <mrow> <mfrac> <mrow> <msub> <mrow>d</mrow> <mrow> <mn>1</mn> </mrow> </msub> <msub> <mrow>K</mrow> <mrow> <mn>1</mn> </mrow> </msub> <mo>+</mo> <msub> <mrow>d</mrow> <mrow> <mn>2</mn> </mrow> </msub> <msub> <mrow>K</mrow> <mrow> <mn>2</mn> </mrow> </msub> </mrow> <mrow> <msub> <mrow>d</mrow> <mrow> <mn>1</mn> </mrow> </msub> <mo>+</mo> <msub> <mrow>d</mrow> <mrow> <mn>2</mn> </mrow> </msub> </mrow> </mfrac> </mrow> </math>

Option 4 - <math> <mrow> <mfrac> <mrow> <msub> <mrow>d</mrow> <mrow> <mn>1</mn> </mrow> </msub> <mo>+</mo> <msub> <mrow>d</mrow> <mrow> <mn>2</mn> </mrow> </msub> </mrow> <mrow> <mrow> <mo>(</mo> <mrow> <msub> <mrow>d</mrow> <mrow> <mn>1</mn> </mrow> </msub> <mo>/</mo> <msub> <mrow>K</mrow> <mrow> <mn>1</mn> </mrow> </msub> <mo>+</mo> <msub> <mrow>d</mrow> <mrow> <mn>2</mn> </mrow> </msub> <mo>/</mo> <msub> <mrow>K</mrow> <mrow> <mn>2</mn> </mrow> </msub> </mrow> <mo>)</mo> </mrow> </mrow> </mfrac> </mrow> </math>

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Answered by

Raj Pandey | Contributor-Level 9

5 months ago

Correct Option - 4

Detailed Solution:

When Two rods are connected in series

$Q = \frac{A (T_{1} - T_{2}) t}{\frac{d_{1}}{K_{1}} + \frac{d_{2}}{K_{1}}} = \frac{A (T_{1} - T_{2}) t}{(d_{1} + d_{2}) / K}$