SOLUTION: A piece of Capillary tubing was Calibrated in the Following Manner. A clean Sample of Mass 3.247 g . A thread of Mercury, drawn into the tube a length of 23.75 mm as observed unde

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Question 1193642: A piece of Capillary tubing was Calibrated in the Following Manner. A clean Sample of Mass 3.247 g . A thread of Mercury, drawn into the tube a length of 23.75 mm as observed under microscope . The mass of Tube and Mercury was 3.489 g. The Density of Mercury is 13.60 g/cm^3 .Assuming that the capillary bore is a uniform cylinder, find the diameter of the bore if the volume of a cylindrical object is V= pie r^2 h
Answer by ikleyn(53937) About Me  (Show Source):
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The mass of the Mercury is the product of the bore volume by the density "d" of Mercury

    3.247 = pi%2Ar%5E2%2Ah%2Ad,

or

    3.247 = 3.14%2Ar%5E2%2A2.375%2A13.6,

where 2.375 is the length of  23.75 mm, converted to centimeters.


From the formula,


    r%5E2 = 3.247%2F%283.14%2A2.375%2A13.6%29 = 0.032015,


     r = sqrt%280.032015%29 = 0.1789 cm = 1.789 mm.    


ANSWER. The diameter of the bore is  2r = 2*1.789 mm = 3.578 mm.

Solved.