An iron ball (coefficient of linear expansion`=1.2xx(10^(-5)//^(@)C`) has a diameter of 6 cm and is 0.010 mm too large to pass through a hole in a brass plate (coefficient of linear expansion`=1.9xx10^-5//^(@)C`) when the ball and the plate are both at a temperature of `30^@C` At what common temperature of the ball and the plate will the ball just pass through the hole in the plate ?A. `23.8^@C`B. `53.8^@C`C. `42.5^@C`D. `63.5^@C`

An iron ball (coefficient of linear expansion`=1.2xx(10^(-5)//^(@)C`)

1.	An iron ball (coefficient of linear expansion`=1.2xx(10^(-5)//^(@)C`) has a diameter of 6 cm and is 0.010 mm too large to pass through a hole in a brass plate (coefficient of linear expansion`=1.9xx10^-5//^(@)C`) when the ball and the plate are both at a temperature of `30^@C` At what common temperature of the ball and the plate will the ball just pass through the hole in the plate ?A. `23.8^@C`B. `53.8^@C`C. `42.5^@C`D. `63.5^@C`
Answer» Correct Answer - B Using the suffixes I and B for the iron ball and the brass plate we have `L_I=6cm`,`L_I-L_B=0.001cm` at `t=30^@C` Heating both the ball and the plate increases the diameters of the ball as well as the hole in the plane, with the hole diameter increasing at a faster rate, since `alpha_Bgtalpha_L`. Now we require that `DeltaL_B-DeltaL_I=0.001cm` at the desired temperature t, with `DeltaL_B=L_Balpha_BDeltat` and `DeltaL_I=L_Ialpha_IDeltat`. Then `DeltaL_B-DeltaL_I=(L_Balpha_B-L_Ialpha_I)DeltaT=0.001cm` `approxL_I(alpha_B-alpha_I)Deltat`, Approximating by putting `LcongL_I` or `(6cm)[1.9xx10^-5-1.2xx10^-5]Deltat=0.001cm` Hence `Deltat=(0.001)/(6[1.9xx10^-5-1.2xx10^-5J]).^@C` `Deltat=23.8^@C` Hence final temperature `=30+23.8^@C=53.8^@C`

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