Problem 22.67 It?s possible to estimate the percentage of fat n the body by meas

Question

Problem 22.67 It?s possible to estimate the percentage of fat n the body by measuring the resistance of the upper leg rather than the upper arm; the calculation is similar. A person?s leg meas-ures 40 cm between the knee and the hip, with an average leg diameter (ignoring bone and other poorly conducting tissue) of 12 cm . A potential difference of 0.78 V causes a current of 1.6mA. Part A What are the fractions of muscle and fat in the leg? Express your answer using two significant figures separated by a comma.

Explanation / Answer

Given that

Current (I) =1.6mA =01.6*10-3A

Length (L) =40cm =0.4m

The potential diffrerence which causes the current is (V) =0.78V

The average leg diameter is (D) =12cm then radius (r) =6cm =0.06m

Area is given by A =Pir2 =3.14*(0.06)2 =0.0113m2

We have to take the resistivity of muscle and fat in the leg is pm =13ohm-m and pf =25ohm-m

We will now consider the f be the fraction of the fat

The total resistance is Rtotal, can be obtained by resistance of fat and ressitance of the muscle which are in parallel

1/Rtot =I/V =1/Rf+1/Rm

=fA/pfL+(1-f)A/pmL

=A/Lpfpm[(pm-pf)f+pf]

Then fraction (f) =[(I/DeltaV)((L/Apm)-1](pf/pm-pf)

Now f =(0.0016/0.78*0.4/0.0113m2*13-1)(25/13-25)=(-0.9998)(-2.0833)=0.1167=11.676%

Therefore ffat =nearly 12%

fmuslce =88%

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