The length of sections AB and EF is l 1 in the figure, and their coefficient of thermal expansion is same ( α 1 = 2 × 1 0 − 5 / ∘ C ). The length of section CD is l 2 and its expansion coefficient is α 2 = 1 × 1 0 − 5 / ∘ C . What must be the ratio l 1 / l 2 to ensure that distance AF remains the same at all temperatures? (Points A , B , E and F reside on the same line.)

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Thermal Properties of Matter - Thermal Expansion Question with Solution | TestHub

Thermal Properties of MatterThermal ExpansionThermal StressMedium3 minCT49

Thermal Properties of MatterMediuminteger

The length of sections AB and EF is $l_{1}$ in the figure, and their coefficient of thermal expansion is same ( $α_{1} = 2 \times 1 0^{- 5} /^{\circ} C$ ). The length of section CD is $l_{2}$ and its expansion coefficient is $α_{2} = 1 \times 1 0^{- 5} /^{\circ} C$ . What must be the ratio $l_{1} / l_{2}$ to ensure that distance AF remains the same at all temperatures? (Points $A, B, E$ and F reside on the same line.)

CT49 diagram

Answer:

Solution:

Zero net change in AF requires the two l1 sections to offset section l2: 2α1l1=α2l2.

Thus l1/l2=α2/(2α1)=0.25.

Stream:JEESubject:Thermal Properties of MatterTopic:Thermal ExpansionSubtopic:Thermal Stress

⏱ 3mℹ️ Source: CT49

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