   Chapter 17, Problem 33P

Chapter
Section
Textbook Problem

In one form of plethysmograph (a device for measuring volume), a rubber capillary tube with an inside diameter of 1.00 mm is filled with mercury at 20°C. The resistance of the mercury is measured with the aid of electrodes sealed into the ends of the tube. If 100.00 cm of the tube is wound in a spiral around a patient’s upper arm, the blood flow during a heart-beat causes the arm to expand, stretching the tube to a length of 100.04 cm. From this observation, and assuming cylindrical symmetry, you can find the change in volume of the arm, which gives an indication of blood flow. (a) Calculate the resistance of the mercury. (b) Calculate the fractional change in resistance during the heartbeat. Take ρHg= 9.4 × 10−7 Ω · m. Hint: Because the cylindrical volume is constant, V = AiLi = AfLf and Af = Ai(Li/Lf).

(a)

To determine

The resistance of mercury.

Explanation

Given Info: The tube with diameter 1.00mm and length 100.00cm is filled with mercury at 20.0°C. The resistivity of mercury is 9.4×107Ωm. Heartbeat causes the stretching of tube into 100.04cm.

Formula to calculate the resistance of mercury is,

R=ρHgLA

• R is the resistance of the mercury,
• ρHg is the resistivity of the mercury,
• A is the area of cross section of tube filled with mercury,
• L is the length of the tube,

Formula to calculate the area of circular cross section of wire is given by

A=πd42

• d is the diameter of the tube filled with mercury

Substitute πd2/4 for A in the R=ρHgL/A to rewrite R.

R=ρHgL(πd24)=4ρHgLπd2

Substitute 9

(b)

To determine

The fractional change in resistance during heartbeat

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