In one form of plethysmograph (a device for measuring volume), a rubber capillary tube with an...
In one form of plethysmograph (a device for measuring volume), a rubber capillary tube with an inside diameter of 1.39 mm is filled with mercury at 20
Homework Answers
(a) R= L/A
therefor R=9.4 *10^-7 * 1/ r^2
R= 9.4 *10^-7 / 1.51 * 10^-6
therefor R =0.622 ohm
(b) V = Ai/Li = Af/Lf and Af = Ai(Li/Lf).
Af =1.508 * 10^-6
therefore Fractional change in resistanve = new resistance- initial resistance
new resistance=0.6239 ohm
Fractional change= 0.6239 ohm -0.622 ohm
=0.0019 ohm
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