A steel rod is 3-000 cm in diameter at 25-C- A brass ring has an interior diameter of 2-992 cm at 25-C- At what common temperature will the ring just slide on to the rod-steel-12 - 10-6K-1 and -brass-18 - 10-6K-1

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Standard XI

Physics

Thermal Expansion

A steel rod i...

Question

A steel rod is 3.000 cm in diameter at $25^{\circ} C$ . A brass ring has an interior diameter of 2.992 cm at $25^{\circ} C$ . At what common temperature will the ring just slide on to the rod.
$α_{s t e e l} = 12 \times 10^{- 6} K^{- 1}$ and
$α_{b r a s s} = 18 \times 10^{- 6} K^{- 1}$

473^{\circ} C

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373^{\circ} C

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663^{\circ} C

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173^{\circ} C

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Solution

The correct option is A $473^{\circ} C$
Let t be the required common temperature. Then $Δ T = t - 25$ . At the common temperature, both must have the same diameter.
$∴ D = 3.000 (1 + 12 \times 10^{- 6} Δ T) = 2.992 (1 + 18 \times 10^{- 6} Δ T) \Rightarrow Δ T = 448^{\circ} C \Rightarrow t = 448 + 25 = 473^{\circ} C$

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A steel rod is 3.000 cm in diameter at 25∘C. A brass ring has an interior diameter of 2.992 cm at 25∘C. At what common temperature will the ring just slide on to the rod. αsteel=12×10−6K−1 and αbrass=18×10−6K−1

The correct option is A 473∘CLet t be the required common temperature. Then ΔT=t−25. At the common temperature, both must have the same diameter. ∴D=3.000(1+12×10−6ΔT)=2.992(1+18×10−6ΔT)⇒ΔT=448∘C⇒t=448+25=473∘C

A steel rod is 3.000 cm in diameter at $25^{\circ} C$ . A brass ring has an interior diameter of 2.992 cm at $25^{\circ} C$ . At what common temperature will the ring just slide on to the rod.
$α_{s t e e l} = 12 \times 10^{- 6} K^{- 1}$ and
$α_{b r a s s} = 18 \times 10^{- 6} K^{- 1}$