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NEET
Exam
NEET
Subject
Physics
Chapter
Thermal Properties of Matter

NEET · Physics · Thermal Properties of Matter

Two rods A and B of equal length L and identical cross-sectional...

Two rods A and B of equal length LL and identical cross-sectional area are joined in series. The thermal conductivities of rods A and B are 3K3K and 2K2K respectively, and their coefficients of linear expansion are 2α2\alpha and 3α3\alpha respectively. Initially, the entire composite rod is at a uniform temperature of 0∘C0^\circ\text{C}. The free end of rod A is then maintained at 100∘C100^\circ\text{C} and the free end of rod B is maintained at 0∘C0^\circ\text{C}. Assuming no heat loss from the sides, the total increase in length of the composite rod in the steady state is

Options

  1. A200αL200\alpha L
  2. B300αL300\alpha L
  3. C260αL260\alpha L
  4. D250αL250\alpha L

Correct answer

D. 250αL250\alpha L

Step-by-step solution

First, find the junction temperature TJT_J in the steady state by equating the rate of heat flow.

3K⋅A⋅(100−TJ)L=2K⋅A⋅(TJ−0)L\dfrac{3K \cdot A \cdot (100 - T_J)}{L} = \dfrac{2K \cdot A \cdot (T_J - 0)}{L}

300−3TJ=2TJ300 - 3T_J = 2T_J

5TJ=300⇒TJ=60∘C5T_J = 300 \Rightarrow T_J = 60^\circ\text{C}

In steady state, the temperature gradient is linear. The average temperature of each rod determines its thermal expansion.

Average temperature of rod A, TA=100+602=80∘CT_A = \dfrac{100 + 60}{2} = 80^\circ\text{C}

Average temperature of rod B, TB=60+02=30∘CT_B = \dfrac{60 + 0}{2} = 30^\circ\text{C}

Change in length of rod A:
ΔLA=L⋅αA⋅ΔTA=L⋅(2α)⋅(80−0)=160αL\Delta L_A = L \cdot \alpha_A \cdot \Delta T_A = L \cdot (2\alpha) \cdot (80 - 0) = 160\alpha L

Change in length of rod B:
ΔLB=L⋅αB⋅ΔTB=L⋅(3α)⋅(30−0)=90αL\Delta L_B = L \cdot \alpha_B \cdot \Delta T_B = L \cdot (3\alpha) \cdot (30 - 0) = 90\alpha L

Total increase in length = ΔLA+ΔLB=160αL+90αL=250αL\Delta L_A + \Delta L_B = 160\alpha L + 90\alpha L = 250\alpha L

Using the end temperatures instead of average temperatures gives 200αL200\alpha L, and using the junction temperature for the entire length gives 300αL300\alpha L.

Answer: 250αL250\alpha L
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