Two cylindrical rods A and B are joined in series. Rod A has length L and radius r , while rod B has length 2L and radius 2r . The free end of rod A is maintained at 100^ and the free end of rod B is maintained at 0^ .

Question

Quantrex Academy · Accepted Answer

Correct answer: C. 8

In steady state, the rate of heat flow through both rods is equal.

H_A = H_B

\dfrac{K_A A_A \Delta T_A}{L_A} = \dfrac{K_B A_B \Delta T_B}{L_B}

Substituting the given values:

A_A = \pi r^2

,

A_B = \pi (2r)^2 = 4\pi r^2

\Delta T_A = 100^\circ\text{C} - 80^\circ\text{C} = 20^\circ\text{C}

\Delta T_B = 80^\circ\text{C} - 0^\circ\text{C} = 80^\circ\text{C}

\dfrac{K_A (\pi r^2) (20)}{L} = \dfrac{K_B (4\pi r^2) (80)}{2L}

20 K_A = 160 K_B

\dfrac{K_A}{K_B} = \dfrac{160}{20} = 8

If the radius is not squared, the ratio obtained is

4

. If the lengths are swapped, the ratio obtained is

32

.

Answer:

8

Two cylindrical rods A and B are joined in series. Rod A has length...

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