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Exam
NEET
Subject
Physics
Chapter
Thermal Properties of Matter

NEET · Physics · Thermal Properties of Matter

The steady-state temperature ( T ) along a composite rod made of...

The steady-state temperature (TT) along a composite rod made of two materials in series is plotted against the distance (xx) from the hot end. The graph shows a piecewise linear drop such that at x=0x = 0, T=100∘CT = 100^{\circ}\mathrm{C}; at the junction x=Lx = L, T=80∘CT = 80^{\circ}\mathrm{C}; and at the cold end x=3Lx = 3L, T=20∘CT = 20^{\circ}\mathrm{C}. Assuming both materials have the same uniform cross-sectional area, what is the ratio of the thermal conductivity of the first material to that of the second material (K1:K2K_1 : K_2)?

Options

  1. A2:32 : 3
  2. B1:31 : 3
  3. C3:23 : 2
  4. D2:12 : 1

Correct answer

C. 3:23 : 2

Step-by-step solution

In steady state, the rate of heat flow (heat current) is constant throughout the composite rod.

Let H1H_1 and H2H_2 be the heat currents in the first and second materials respectively.

For the first material (from x=0x = 0 to x=Lx = L):
Temperature difference ΔT1=100∘C−80∘C=20∘C\Delta T_1 = 100^{\circ}\mathrm{C} - 80^{\circ}\mathrm{C} = 20^{\circ}\mathrm{C}
Length L1=LL_1 = L
Heat current H1=K1A(20)LH_1 = \dfrac{K_1 A (20)}{L}

For the second material (from x=Lx = L to x=3Lx = 3L):
Temperature difference ΔT2=80∘C−20∘C=60∘C\Delta T_2 = 80^{\circ}\mathrm{C} - 20^{\circ}\mathrm{C} = 60^{\circ}\mathrm{C}
Length L2=3L−L=2LL_2 = 3L - L = 2L
Heat current H2=K2A(60)2L=K2A(30)LH_2 = \dfrac{K_2 A (60)}{2L} = \dfrac{K_2 A (30)}{L}

Since the rods are in series, H1=H2H_1 = H_2:
K1A(20)L=K2A(30)L\dfrac{K_1 A (20)}{L} = \dfrac{K_2 A (30)}{L}

⇒20K1=30K2\Rightarrow 20 K_1 = 30 K_2

⇒K1K2=3020=32\Rightarrow \dfrac{K_1}{K_2} = \dfrac{30}{20} = \dfrac{3}{2}

Thus, the ratio K1:K2K_1 : K_2 is 3:23 : 2.

Answer: 3:23 : 2
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