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A steel rod is 3.000 cm 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? |
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Answer» `alpha_("steel")=12xx10^-6K^-1` and `alpha_("brass")=18xx10^-6K^-1` Let t be the required common temperature Then `:. DeltaT=t-25`. At the common temperature, both must have the same diameter. `D=3.000(1+12xx10^-6DeltaT)` `=2.992(1+18xx10^-6DeltaT)` `DeltaT=448^@Cimpliest=448+25` `=473^@C` |
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