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

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  .

compass can be used day and night, cloudy or sunny

igneous rocks also form where plates converge. the subducting plate melts as it sinks into the crust of the earth, and the melt rises into the overriding plate forming volcanoes. metamorphic rocks are formed mainly in the lithosphere, wherever there is high pressure and high temperature. -woods