EBK PHYSICS FOR SCIENTISTS AND ENGINEER
EBK PHYSICS FOR SCIENTISTS AND ENGINEER
6th Edition
ISBN: 9781319321710
Author: Mosca
Publisher: VST
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Chapter 20, Problem 52P

(a)

To determine

To show that LALB=αAαB .

(a)

Expert Solution
Check Mark

Explanation of Solution

Introduction:

The difference between the two lengths are constant if the length LA and LB is constant. The coefficient of linear expansions are αA and αB .

Write expression for change in length.

  ΔL=LBLA

Write expression for linear expansion.

  ΔL=(LB+αBLBΔT)(LA+αALAΔT)

Solve above expression.

  ΔL=(LBLA)+(αBLBαALA)ΔT

Substitute ΔL for LBLA in above expression and solve.

  ΔL=ΔL+(αBLBαALA)ΔT0=(αBLBαALA)ΔT0=(αBLBαALA)LALB=αBαA

Conclusion:

Thus, the equation is proved.

(b)

To determine

The value of LB .

(b)

Expert Solution
Check Mark

Explanation of Solution

Given:

The material A is brass.

The material B is steel.

The value of LA is 250cm .

Formula used:

Write expression for change in length.

  ΔL=LBLA

Write expression for linear expansion.

  ΔL=(LB+αBLBΔT)(LA+αALAΔT)

Solve above expression.

  ΔL=(LBLA)+(αBLBαALA)ΔT

Substitute ΔL for LBLA in above expression and solve.

  ΔL=ΔL+(αBLBαALA)ΔT0=(αBLBαALA)ΔT0=(αBLBαALA)

Solve above expression.

  LALB=αBαA   ..... (1)

Calculation:

Substitute 250cm for LA , 19×106K-1 for αA and 11×106K-1 for αB in equation (1).

  250cmLB=11× 10 6K -119× 10 6K -1LB=430cm

Conclusion:

Thus, the value of LB is 430cm .

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