Between two similar thermometers, one is filled with mercury and another with alcohol of the same volume at 0^(@)C. The gap for each degree in the mercury thermometer is l and that in the alcohol thermometer is l^('). Show that l/l^(')=(gamma-3alpha)/(gamma_(1)-3alpha) "where " gamma= coefficient of real expansion of mercury,gamma_(1)= coefficient of real expansion of alcohol and alpha= coefficient of linear expansion of glass.

Between two similar thermometers, one is filled with mercury and anoth

1.	Between two similar thermometers, one is filled with mercury and another with alcohol of the same volume at 0^(@)C. The gap for each degree in the mercury thermometer is l and that in the alcohol thermometer is l^('). Show that l/l^(')=(gamma-3alpha)/(gamma_(1)-3alpha) "where " gamma= coefficient of real expansion of mercury,gamma_(1)= coefficient of real expansion of alcohol and alpha= coefficient of linear expansion of glass.
Answer» Solution :In both thermometers, let the area of cross-section of the TUBE be A and the volume of the BULB be V. For `1^(@)` rise in temperature in MERCURY thermometer, apparent expansion of mercury = volume of length l of the tube i.e. `""V=(gamma-3alpha) times 1=l times A " " [ therefore gamma^(.)=gamma-gamma_(g)] "...(1)"` Similarly, for `1^(@)` rise in temperature in alcohol thermometer, `""V(gamma_(1)-3alpha) times 1=l^(.) times A"...(2)"` Dividing (1) by (2) we GET, `(l times A)/(l^(.) times A)=(V(gamma-3alpha))/(V(gamma_(1)-3alpha)) " or, "l/l^(.)=(gamma-3alpha)/(gamma_(1)-3alpha).`