Question

A brass plug is to be placed in a ring made of iron. At 20 ∘C∘C, the diameter of the plug is 8.754 cmcm and that of the inside of the ring is 8.739 cmcm .

1.They must both be brought to what common temperature in order to fit?

Express your answer using two significant figures.

2.

What if the plug were iron and the ring brass?

Express your answer using two significant figures.

Answer 1

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(a)

Just slip over the rod, means

La = Lb after some temperature change

Change in length is given by:

dLa = La*alpha)a*dT

dLb = Lb*alpha)b*dT

La + La*alpha)a*dT = Lb + Lb*alpha)b*dT

dT = Tf - Ti

Tf = Ti + (Lb - La)/[alpha)a*La - alpha)b*Lb]

Iron temperature expansion coefficient = 1.2*10^-5 /C

Brass's temperature expansion coefficient = 1.9*10^-5 /C

Initial temperature = 20 C

diameter of iron = 8.754 cm

diameter of brass = 8.739 cm

Using known values:

Tf = 20 + (8.754 - 8.739)/[1.2*10^-5*8.754 - 1.9*10^-5*8.739]

Tf = -225.9299 C

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(b)

Simply change the value of diameter in above equation:

Tf = Ti + (Lb - La)/[alpha)a*La - alpha)b*Lb]

Using known values:

Tf = 20 + (8.739 - 8.754)/[1.2*10^-5*8.739 - 1.9*10^-5*8.754]

Tf = 264.0691 C

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