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?

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 onto the rod ? (Coefficient of linear expansion of steel= 12 xx 10^(-6) per k and coefficient of linear expansion of brass= 18 xx 10^(-6) per k) .

A steel rod is 3.000cm in diameter at 25^(@)C . A brass ring has an interior diameter of 2.9912cm at 25^(@)C . At what common temperature will the ring just slide onto the rod? alpha_("brass")=18xx10^(-6)K^(-1) and alpha_("steel")=12xx10^(-6)K^(-1)

A steel rod is 3 cm in diameter at -10.00^@C . A brass ring has an interior diameter of 2.992 cm at -10.00^@C . At what common temperature will the ring just slide onto the rod ?

A steel rod is 4000cm in diameter at 30^(@)C A brass ring has an interior diameter of 3.992cm at 30^(@) in order that the ring just slides onto the steel rod the common temperature of the two should be nearly (alpha_(steel)=11xx10^(-6)/(^(@)C) and alpha_(brass)=19xx10^(-6)/(^(@)C)

A steel rod is 4.00cm in diameter at 30^(@)C A brass ring has an interior diameter of 3.992cm at 30^(@) in order that the ring just slides onto the steel rod the common temperature of the two should be nearly (alpha_(steel) = 11xx10^(-6)(^(@)C) and alpha_(brass)=19xx10^(-6)(^(@)C)

A cylinder of diameter exactly 1 cm at 30^@C is to be slided into a hole in a steel plate. The hole has a diameter of 0.99970 cm at 30^@C . To what temperature must the plate be heated? For steel alpha = 1.1 xx 10^-5 ^@C^-1