A steel tyre is 1 m in diameter at `10^@C` . To what temperature it must be raised so that it may just slip on a wheel 1.005 m in diameter ? `alpha` for steel = ` 12 xx 10^(-6)//^@C` .

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 metal sphere 10.01 cm in diameter is placed on a brass ring of internal diameter 10 cm at the same temperature of 12^@C . Find the temperature to which they should be heated together so that the metal sphere just passes through the ring . alpha_"metal" = 12 xx 10^(-6)//^@C and alpha_"brass" = 18 xx 10^(-6)//^@C

A steel rod when measured with a zinc scale (alpha = 26 xx 10^(-6)//^@C) appears to be 1 m long when both rods are at 20^@C . If the zinc scale was graduated at 0^@C , what is the true length of the steel rod at 0^@C ? alpha for steel = 12 xx 10^(-6)//^@C .

A metal scale is graduated at 0^@C . What would be the true length of an object which when measured with this scale at 25^@C , reads 50 cm? alpha for metal = 18 xx 10^(-6)//^@C .

If a cylinder of diameter of 1 cm at 30^@C is fitted into a hole of diameter 0.9997 cm in a steel plate, find the temperature the fitting has to be done ? Co.eff.of linear expansion of steal is 12 xx 10^(-6)//^@ C

A brass scale is correct at 20^@C . The length of a rod measured by it at 35^@C is 60 cm . What is the correct length of rod at that temperature ? ( alpha of brass = 18.9 xx 10^(-6) per^@C) .

What should be the length of steel abd copper rods at 0^@C so that the difference in their lengths is 5.0 cm at all temperatures. Coefficient of linear expansion of copper is 1.7 xx 10^(-5) //^@ C and that of steel is 1.1 xx 10^(-5) //^@ C .

At what temperature will be rate of effusion of N_2 be 1.625 times the rate of effusion of SO_2 at 500^@C :

An iron plate has a circular hole of diameter 10 cm . Find the diameter of the hole when the plate is uniformly heated from 10^@C to 90^@C (alpha = 12 xx 10^(-6)//^@C) .