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3 years ago

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a steel sphere and brass ring have diameter 25cm and 24.9cm at 15°C.If the sphere and the ring are heated together.what is the t

emperature at which the sphere could pass through the ring? alpha steel= 12×10^-6 and alpha brass= 20×10^-6

1 answer:

Oksi-84 [34.3K]3 years ago

6 0

Answer:

Explanation:

Due to heat energy , metal expands . Formula for linear expansion is as follows .

L = l ( 1 + α Δt )

where L is expanded length , l is original length , α is coefficient of linear expansion and Δt is increase in temperature .

To pass the sphere through the ring , the diameter of both ring and sphere should be same after heating . Let after increase of temperature Δt , their diameter becomes same as L . The linear coefficient of brass and steel are

20 x 10⁻⁶ and 12 x 10⁻⁶ respectively .

For steel sphere ,

L = 25 ( 1 + 12 x 10⁻⁶ Δt )

For brass ring

L = 24.9 ( 1 + 20 x 10⁻⁶ Δt )

25 ( 1 + 12 x 10⁻⁶ Δt ) = 24.9 ( 1 + 20 x 10⁻⁶ Δt )

1.004( 1 + 12 x 10⁻⁶ Δt ) = ( 1 + 20 x 10⁻⁶ Δt )

1.004 + 12.0482 x 10⁻⁶ Δt = 1 + 20 x 10⁻⁶ Δt

.004 = 7.9518 x 10⁻⁶ Δt

Δt = 4000 / 7.9518

= 503⁰C.

final temp = 503 + 15 = 518⁰C .

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